Answers
Answer:
The fundamental counting principle states that if there are p ways to do one thing, and q ways to do another thing, then there are p×q ways to do both things. possible outcomes of the experiment. The counting principle can be extended to situations where you have more than 2 choices.
Step-by-step explanation:
Related Questions
PLEASE HELP!!!!!!!!!!!
Answers
Answer:
The answer is B. [tex]\frac{2x+1}{x^{2}-7} ,x\neq \sqrt{7}[/tex]
Step-by-step explanation:
[tex]f(x)=2x+1\\g(x)=x^{2} -7[/tex]
Focusing on the denominator makes it equal to = 0
[tex]x^{2} -7=0\\x^{2} =7\\x=+\sqrt{7},-\sqrt{7}[/tex]
If the denominator equals zero then the equation ends up undefined. A number over zero, zero can not go divided into a number.
Hello Calculus!
Find the value
[tex]\\ \rm\Rrightarrow {\displaystyle{\int\limits_3^5}}(e^{3x}+7cosx-3tan^3x)dx[/tex]
Note:-
Answer with proper explanation required and all steps to be mentioned .
Answers
Answer:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \frac{e^{15} - e^9}{3} + 7 \bigg( \sin 5 - \sin 3 \bigg) - 3 \bigg( \frac{\sec^2 5 - \sec^2 3}{2} - \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \bigg)[/tex]
General Formulas and Concepts:
Calculus
Differentiation
DerivativesDerivative NotationDerivative Property [Multiplied Constant]:
[tex]\displaystyle (cu)' = cu'[/tex]
Derivative Property [Addition/Subtraction]:
[tex]\displaystyle (u + v)' = u' + v'[/tex]
Derivative Rule [Basic Power Rule]:
f(x) = cxⁿf’(x) = c·nxⁿ⁻¹Integration
IntegralsIntegration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]
Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]
Integration Method: U-Substitution + U-Solve
Step-by-step explanation:
Step 1: Define
Identify given.
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx[/tex]
Step 2: Integrate Pt. 1
[Integral] Rewrite [Integration Rule - Addition/Subtraction]:[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + \int\limits^5_3 {7 \cos x} \, dx - \int\limits^5_3 {3 \tan^3 x} \, dx[/tex][Integrals] Rewrite [Integration Property - Multiplied Constant]:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + 7 \int\limits^5_3 {\cos x} \, dx - 3 \int\limits^5_3 {\tan^3 x} \, dx[/tex][3rd Integral] Rewrite:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + 7 \int\limits^5_3 {\cos x} \, dx - 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx[/tex]
Step 3: Integrate Pt. 2
Identify variables for u-substitution and u-solve.
1st Integral
Set u:[tex]\displaystyle u = 3x[/tex][u] Differentiate [Derivative Properties and Rules]:
[tex]\displaystyle du = 3 \, dx[/tex][Bounds] Swap:
[tex]\displaystyle \left \{ {{x = 5 \rightarrow u = 3(5) = 15} \atop {x = 3 \rightarrow u = 3(3) = 9}} \right.[/tex]
3rd Integral
Set v:[tex]\displaystyle v = \sec x[/tex][v] Differentiate [Trigonometric Differentiation]:
[tex]\displaystyle dv = \sec x \tan x \, dx[/tex][dv] Rewrite:
[tex]\displaystyle dx = \frac{1}{\sec x \tan x} \, dv[/tex]
Step 4: Integrate Pt. 3
Let's focus on the 3rd integral first.
Apply Integration Method [U-Solve]:[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \int\limits^{x = 5}_{x = 3} {\frac{v^2 - 1}{v}} \, dv[/tex][Integral] Rewrite [Integration Property - Addition/Subtraction]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \int\limits^{x = 5}_{x = 3} {v} \, dv - \int\limits^{x = 5}_{x = 3} {\frac{1}{v}} \, dv \Bigg)[/tex][Integrals] Apply Integration Rules [Reverse Power Rule and Logarithmic Integration]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{v^2}{2} \bigg| \limits^{x = 5}_{x = 3} - \ln | v | \bigg| \limits^{x = 5}_{x = 3} \Bigg)[/tex][v] Back-Substitute:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{\sec^2 x}{2} \bigg| \limits^{5}_{3} - \ln | \sec x | \bigg| \limits^{5}_{3} \Bigg)[/tex]Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{\sec^2 5 - \sec^2 3}{2}- \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \Bigg)[/tex]
Step 5: Integrate Pt. 4
Focus on the other 2 integrals and solve using integration techniques listed above.
1st Integral:
[tex]\displaystyle\begin{aligned}\int\limits^5_3 {e^{3x}} \, dx & = \frac{1}{3} \int\limits^5_3 {3e^{3x}} \, dx \\& = \frac{1}{3} \int\limits^{15}_9 {e^{u}} \, du \\& = \frac{1}{3} e^u \bigg| \limits^{15}_9 \\& = \frac{1}{3} \bigg( e^{15} - e^9 \bigg)\end{aligned}[/tex]
2nd Integral:
[tex]\displaystyle\begin{aligned}7 \int\limits^5_3 {\cos x} \, dx & = 7 \sin x \bigg| \limits^5_3 \\& = 7 \bigg( \sin 5 - \sin 3 \bigg)\end{aligned}[/tex]
Step 6: Integrate Pt. 5
[Integrals] Substitute in integrals:[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \frac{e^{15} - e^9}{3} + 7 \bigg( \sin 5 - \sin 3 \bigg) - 3 \bigg( \frac{\sec^2 5 - \sec^2 3}{2} - \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \bigg)[/tex]
∴ we have evaluated the integral.
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Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Answer:
1086950.36760
Formula's used:
[tex]\rightarrow \sf \int sin(ax+b)=-\dfrac{1}{a} cos(ax+b)+c[/tex]
[tex]\rightarrow \sf \int cos(ax+b)=\dfrac{1}{a} sin(ax+b)+c[/tex]
[tex]\rightarrow \sf \int \dfrac{1}{ax+b} =\dfrac{1}{a} ln|ax+b|+c[/tex]
[tex]\rightarrow \sf \int e^{ax+b}=\dfrac{1}{a} e^{ax+b} + c[/tex]
[tex]\rightarrow \bold{ ln|a| - ln|b| = ln|\frac{a}{b} | }[/tex]
Explanation:
[tex]\dashrightarrow \sf \int \left(e^{3x}+7cos\left(x\right)-3tan^3\left(x\right)\right)[/tex]
apply sum rule: [tex]\bold{\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx}[/tex]
[tex]\dashrightarrow \sf \int \:e^{3x}dx+\int \:7\cos \left(x\right)dx-\int \:3\tan ^3\left(x\right)dx[/tex]
Integrate simple followings first, using formula's given above
[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-\int 3tan^3x[/tex]
Breakdown the component
[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\int tan^2x(tanx)[/tex]
[ tan²x = sec²x - 1 ][tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\int (sec^2x-1)(tanx)[/tex]
===========================================================
for integration of [tex]\bold{\int (sec^2x-1)(tanx)}[/tex]
apply substitution ... u[tex]\dashrightarrow \int \dfrac{-1+u^2}{u}[/tex]
[tex]\dashrightarrow \sf \int \:-\dfrac{1}{u}+udu[/tex]
[tex]\dashrightarrow \sf - \int \dfrac{1}{u}du+\int \:udu[/tex]
[tex]\dashrightarrow -\ln \left|u\right|+\dfrac{u^2}{2}[/tex]
substitute back u = sec(x)[tex]\dashrightarrow \sf-\ln \left|\sec \left(x\right)\right|+\dfrac{\sec ^2\left(x\right)}{2}[/tex]
================================================= insert back
[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\left(-\ln \left|\sec \left(x\right)\right|+\dfrac{\sec ^2\left(x\right)}{2}\right)[/tex] outcome after integrating
Now apply the given limits
[tex]\sf \hookrightarrow \sf \dfrac{1}{3}e^{3(5)}+7\sin \left(5\right)-3\left(-\ln \left|\sec \left(5\right)\right|+\dfrac{\sec ^2\left(5\right)}{2}\right) - (\sf \dfrac{1}{3}e^{3(3)}+7\sin \left(3\right)-3\left(-\ln \left|\sec \left(3\right)\right|+\dfrac{\sec ^2\left(3\right)}{2}\right))[/tex]
simplify
[tex]\sf \hookrightarrow \sf \dfrac{1}{3}e^{15}+7\sin \left(5\right)-3\left(-\ln \left|\sec \left(5\right)\right|+\dfrac{\sec ^2\left(5\right)}{2}\right) - (\sf \dfrac{1}{3}e^{9}+7\sin \left(3\right)-3\left(-\ln \left|\sec \left(3\right)\right|+\dfrac{\sec ^2\left(3\right)}{2}\right))[/tex]
and group the variables
[tex]\sf \hookrightarrow \dfrac{e^{15}-e^9}{3}-\dfrac{3}{2\cos ^2\left(5\right)}+\dfrac{3}{2\cos ^2\left(3\right)}+7\sin \left(5\right)-7\sin \left(3\right)+3\ln \left(\dfrac{1}{\cos \left(5\right)}\right)-3\ln \left(-\dfrac{1}{\cos \left(3\right)}\right)[/tex]
value:
[tex]\sf \hookrightarrow 1086950.36760[/tex]
NEED HELP ASAP!!!! Will give brainiest
Answers
Which value is not a solution of the inequality x-4 symbol 2
Answers
The inequality x -4 > 2 uses a greater than symbol
All numbers lesser or equal to 6 are not a solution of the inequality x -4 > 2
How to determine the value not in the solution?The inequality is given as:
x -4 > 2
Add 4 to both sides of the inequality
x - 4 + 4 > 2 + 4
Evaluate the sum
x > 6
The above means that only numbers greater than 6 are in the solution of the inequality.
Since the options are not given, I will give a general solution that all numbers lesser or equal to 6 are not a solution of the inequality x -4 > 2
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8. Daisies and tulips are planted in a
garden. There are 11 fewer tulips
planted than daisies.
a. Write an expression that represents
the number of tulips in terms of the
number of daisies. Define any
variables used.
b. If 18 daisies are planted, how many
tulips are planted?
Answers
Answer:
a) [tex]d-11 = t[/tex]
b) 7
Step-by-step explanation:
Let [tex]d[/tex] = daisies
Let [tex]t[/tex] = tulips
a) 11 fewer tulips than daisies: [tex]d-11 = t[/tex]
b) Substitute 18 into [tex]d[/tex] and solve for [tex]t[/tex].
[tex]18-11= t[/tex]
[tex]7 = t[/tex]
The circumference of a circle is 3π cm. What is the area of the circle?
Question 3 options:
9πcm2
1.5πcm2
6πcm2
2.25πcm2
Answers
circumference of circle: 2πr
2πr = 3πr = 1.5 cmarea of circle: πr^2
πr^2π(1.5)^22.25πWhat are range, index of qualitative variation (IQV), interquartile range (IQR), standard deviation, and variance
Answers
Answer:
To find the interquartile range (IQR), first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.
Step-by-step explanation:
Wind is
• air moving from areas of high pressure to areas of low pressure.
• air moving from areas of low pressure to areas of high pressure.
air moving from areas of high temperature to areas of low temperature.
Answers
Wind is caused by differences in the atmospheric pressure. When a difference in atmospheric pressure exists, air moves from the higher to the lower pressure area, resulting in winds of various speeds. On a rotating planet, air will also be deflected by the Coriolis effect, except exactly on the equator.
Hope this helped!
Samir recorded the grade-level and instrument of everyone in the middle school School of Rock below. Seventh Grade Students Instrument # of Students Guitar 6 Bass 4 Drums 6 Keyboard 7 Eighth Grade Students Instrument # of Students Guitar 9 Bass 9 Drums 9 Keyboard 10 Based on these results, express the probability that a seventh grader chosen at random will play an instrument other than drums as a fraction in simplest form.
Answers
Using it's concept, it is found that there is a [tex]\frac{17}{23}[/tex] probability that a seventh grader chosen at random will play an instrument other than drums.
What is a probability?A probability is given by the number of desired outcomes divided by the number of total outcomes.
In this problem:
There is a total of 6 + 4 + 6 + 7 = 23 seventh graders.Of those, 23 - 6 = 17 play instruments that are not the drum.Hence:
[tex]p = \frac{17}{23}[/tex]
There is a [tex]\frac{17}{23}[/tex] probability that a seventh grader chosen at random will play an instrument other than drums.
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What are the solutions of the equation (x + 2)2 + 12(x + 2) – 14 = 0? Use u substitution and the quadratic formula to
solve.
-8+5√2
O x=-6252
O x=-4+5√2
x=-2 +5√2
Answers
Answer:
-8+5√2
Step-by-step explanation:
(x+2)^2+12(x+2)–14=0
(x+2)^2=(x+2)(x+2)=x^2+4+4x
12(x+2)=12x+24
x^2+4+4x+12x+24-14=0
x^2+4x+12x+4+24-14=0
x^2+16x+14=0
quadratic formula
x = {-b +- square root of (b^2 – 4ac)} ÷ {2a}
a= 1
b = 16
c = 14
x = {-16 +- square root of (16^2 – 4*1*14)} ÷ {2*1}
x = {-16 +- square root of (256 – 56)} ÷ {2*1}
x = ((-16 +- square root of (200)) ÷ (2)
x = ((-16 +- 10√2)) ÷ (2)
x= -8+-5√2
7. What value of c will make x2 – 20x + c
a perfect square trinomial?
Answers
∠A and \angle B∠B are vertical angles. If m\angle A=(7x-6)^{\circ}∠A=(7x−6)
∘
and m\angle B=(8x-27)^{\circ}∠B=(8x−27)
∘
, then find the measure of \angle B∠B
Answers
keeping in mind that vertical angles are always congruent.
[tex]\stackrel{\measuredangle A}{7x-6}~~ = ~~\stackrel{\measuredangle B}{8x-27}\implies -6=x-27\implies 21=x~\hfill \underset{\measuredangle B}{\stackrel{8(21)~~ - ~~27}{141}}[/tex]
help me please need help
Answers
Answer:
Step-by-step explanation:
1. x -> opposite side of 48°
o → hypotenuse
b → adjacent side of 48°
[tex]\sf Sin \ 48^\circ = \dfrac{opposite \ side }{hypotenuse}\\\\\\0.7431 = \dfrac{15}{o}\\\\\\0.74 * o = 15\\\\\\ o = \dfrac{15}{0.74}\\\\\\[/tex]
o = 20.27
[tex]\sf cos \ 48^\circ = \dfrac{adjacent \ side }{hypotenuse}\\\\\\0.67 =\dfrac{b}{o}\\\\\\0.67=\dfrac{b}{20.27}[/tex]
b = 0.67*20.27
b = 13.58
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
2) i → opposite side of 25°
n → adjacent side of 25°
[tex]\sf Sin \ 25 =\dfrac{i}{t}\\\\\\0.42=\dfrac{i}{30}\\\\\\0.42*30=i[/tex]
i = 12.6
[tex]\sf Cos \ 30^\circ =\dfrac{n}{t}\\\\0.91=\dfrac{n}{30}\\\\\\0.91*30 = n[/tex]
n = 27.3
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
3) a → opposite side of 70°
e → adjacent side of 70°
[tex]Sin \ 70^\circ =\dfrac{a}{l}\\\\0.94 =\dfrac{a}{25}\\\\0.94*25=a[/tex]
a = 23.5
[tex]\sf Cos \ 70^\circ =\dfrac{e}{l}\\\\0.34=\dfrac{e}{25}\\\\0.34*25=e[/tex]
e = 8.5
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
4)
[tex]\sf Sin \ 52^\circ = \dfrac{x}{75}\\\\0.79*75=x\\[/tex]
x = 59.25
[tex]\sf Cos \ 52^\circ = \dfrac{z}{75}\\\\0.62*75 =z[/tex]
z = 46.5
A kitchen can be broken into 2 rectangles. One rectangle has a base of 7 feet and height of 5 feet. The second rectangle has a base of 2 feet and height of 2 feet. One package of tile will cover 3 square feet. How many packages of tile will she need? 8 13 15 39
Answers
Answer:
its 13 or B
Step-by-step explanation:
Find the circumference of the circle of 13 inches use 3.14 for pi and round to the nearest whole number
Answers
Answer:
ohh just use this formula
this is for Area- A= π r^2
this is for Circumference- C= 2 π r
Step-by-step explanation:
The circumference of the circle of the radius of 13 inches is, 41 inches
What is circumference ?Circumference is the distance around the perimeter of a circular object. It is defined as the length of the circle that is found by multiplying the diameter of the circle by π (pi), which is approximately equal to 3.14.
The formula for the circumference of a circle is given by: C = 2πr,
Given that,
The diameter of the circle is 13 inches,
The radius of the circle can be calculated as follows:
r = d/2
= 13 inches / 2
= 6.5 inches
Using the formula for the circumference, we can calculate the circumference as follows:
C = 2πr
= 2 × 3.14 × 6.5 inches
= 40.76 inches
Rounding to the nearest whole number, the circumference of the circle is 41 inches.
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The length of a rectangle is 7 cm less than four times its width. The area of the rectangle is 36 square cm
Answers
Answer:
W = 4 cm and L = 9 cm
Step-by-step explanation:
I don't see a question, but will assume the problem wants the length(L) and width(W) of the described rectangle.
Let L and W stand for Length and Width.
Area of a rectangle is given by L*W
We are told that L*W = 36 cm^2
We are also told that L = 4W-7 ["length of a rectangle is 7 cm less than four times its width"]
Substituting the second into the first equation:
L*W = 36 cm^2
(4W-7)*W = 36 cm^2 [L = 4W-7]
4W^2-7W - 36 cm^2 = 0
(W-4)(4W+9) = 0
The roots are: 4 and -(9/4)
We'll use the positive value: W = 4
Since L = 4W-7:
L = 4(4)-7
L = 16-7
L = 9 cm
If -3=4x+7 what is x?
Answers
4x=-3-7
4x=-10
X=-10/4
X=-2.5
Answer: x is -2.5 or [tex]-2\frac{1}{2}[/tex]
Step-by-step explanation:
We need to solve for x
-3=4x+7
Step 1) Subtract 7 from both sides
-3-7=4x+7-7
-10=4x
Step 2) Divide both sides by 4 to isolate x
-10=4x
[tex]\frac{-10}{4} =\frac{4x}{4} \\-2.5=x[/tex]
Nada drives her car at an average speed of 60 miles per hour. She is planning a drive between 150 and 180 miles. Write and solve an inequality to model how many hours Nada will be driving
Answers
The inequality model that can be used to solve the number of hours it would take Nada to drive is 2.5 ≤ x ≤3.
What is the number of hours it would take Nada to drive?
Average speed is the total distance travelled per time.
Average speed = total distance / total time
Time = total distance / average speed.
Time if she drives 150 miles = 150 / 60 = 2.5 hours
Time if she drives 180 miles = 180 / 60 = hours
2.5 ≤ x ≤3.
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Help Me write the inequality that represents the number line! Sorry the picture is bad quality!
Answers
Answer:
x ≤ -6
Step-by-step explanation:
The line is moving towards the left, which means that the numbers are getting smaller. X is going to be less than something.
The dot on -6 is colored in. This means that -6 is included in our inequality. [X could equal -6]
So, we could write the equation x ≤ -6
What’s the answer for this ?
Answers
Step-by-step explanation:
1)-1, 0, 1... do like this
Which pair of expressions has equivalent values?
1^13 and 1^15
6^1and 9^1
7^8and 8^7
9- and 4^3
Answers
Answer:
1^13 and 1^15
Step-by-step explanation:
1 raised to anything is still just 1
so, 1^13 = 1 and 1^15 =1
A large multiplex movie house has many theaters. The largest theater has 41 rows. There are 19 seats in the first row. Each row has two seats more than the previous row. How many total seats are there in this theater?
Answers
Answer:
101
Step-by-step explanation:
I time 41x2 than add 19
A right triangle includes one algae that measures 14º. what is the measure of the third angle
A 14º C 90º
B 76º D 104º
Answers
Answer: B. 76
Step-by-step explanation: A right triangle is 180 degrees. It has an angle of 90 since it is a right triangle. 90 + 14 = 104. 180 - 104 = 76
Answer:
B. 76 degrees
Step-by-step explanation:
EVERY triangle's angles add up to 180 degrees. We already know that since it's a right triangle, one of the angles equals 90 degrees (that's a right angle) and they give us the second angle measurement, 14 degrees. If we add those two angle measures together and subtract them from 180, we should get the measure of the third angle as our answer.
14 + 90 = 104
180 - 104 = 76
Therefore, the third and final angle in this right triangle equals 76, so your answer is B. I hope this helps! Have a lovely day!! :)
Identify the surface area of the cylinder to the nearest tenth. Use 3.14 for π.
Answers
Answer:
967.6
Step-by-step explanation:
967.6
967.12 in
Step-by-step explanation:
the formula for the area (surface area) of a cylinder is: A=2πrh+2πr2
to solve we need to determine the values
R= radius = half the diameter = 14/2 =7
D= diameter =14inches
H= height= 15 inches
plug in
A=2πrh+2πr2 = A=2π([tex]\frac{d}{2}[/tex])h+2π([tex]\frac{d}{2}[/tex])^2
= 2[tex]\pi[/tex]([tex]\frac{14}{2}[/tex])(15) + 2[tex]\pi[/tex]([tex]\frac{14}{2}[/tex])^2
= 2[tex]\pi[/tex](7)(15) + 2[tex]\pi[/tex](7)^2
=[tex]\pi[/tex]((2x7x15)+(2x7^2))
=[tex]\pi[/tex](210+98)
=[tex]308\pi[/tex]
=967.12 in
I need this for school, please help!!
Answers
A plane leaves an airport traveling at 400 mph in the direction n 45° e. a wind is blowing at 40 mph in the direction n 45° w. what is the ground speed of the plane?
Answers
The ground speed of the plane is 402 mph
We have given,
A plane leaves an airport traveling at 400 mph in the direction n 45° e. a wind is blowing at 40 mph in the direction n 45° w.
Now, Draw a right triangle with 400 and 40 as the two legs.
This is a right angle because 45 degrees NE is perpendicular to 45 degrees NW.
So, the two side lengths are 400 and 40
Using the Pythagorean theorem,
What is the Pythagorean theorem?(a^2 + b^2 = c^2)
[tex]400^2+40^2=c^2\\\\c^2=160000+1600\\c^2=161600\\c=401.99[/tex]
Therefore the answer is 401.995, or 402mph.
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Which expression is equal to 0.75×0.09
Answers
Arrange the expressions below in order from least to greatest. place the least at the top and greatest at the bottom. ( 72 ÷ 8 ) − 2 × 3 1 72 ÷ ( 8 − 2 ) × 3 1 72 ÷ ( 8 − 2 ) × ( 3 1 ) 72 ÷ 8 − 2 × ( 3 1 )
Answers
The expressions in order from least to greatest 72 / 8 - 2 x (3 + 1), (72 / 8) - 2 x 3 + 1, 72 / (8 - 2) x 3 + 1 and 72 / (8 - 2) x (3 + 1).
What is BODMAS?BODMAS stands for B - Bracket, O - order of Power, D - Division, M - Multiplication, A - Addition, and S - Subtraction.
To Arrange the expressions below in order from least to greatest. place the least at the top and the greatest at the bottom
72 / 8 - 2 x (3 + 1) equals 1
(72 / 8) - 2 x 3 + 1 equals 4
72 / (8 - 2) x 3 + 1 equals 37
72 / (8 - 2) x (3 + 1) equals 48
Thus, The expressions in order from least to greatest 72 / 8 - 2 x (3 + 1), (72 / 8) - 2 x 3 + 1, 72 / (8 - 2) x 3 + 1 and 72 / (8 - 2) x (3 + 1).
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The temperature is dropping at a rate of five degrees per hour.
Let d represent the number of degrees the temperature drops.
Let t represent the number of hours that pass.
Which is the dependent variable?
Answers
Answer:
The number of degrees the temperature drops°
Step-by-step explanation:
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pls mark brainliest
15 × [-5] +15 × [-3] = solution
Answers
[tex]Heyo![/tex]
SaddySokka is here to help!!
Let's do this step-by-step explanation!
[tex](15)(-5)+(15)(-3)[/tex]
[tex]=-75+(15)(-3)[/tex]
[tex]=-75+-45[/tex]
[tex]=-120[/tex]
Answer:
[tex]-120[/tex]
Hopefully, this helps you!!
Have a great day!!
SaddySokka~
Answer:
-120
Step-by-step explanation:
15×[-5]+15×[-3]
Use BODMAS
-75+-45
-120