30 POINTS
How many cubic feet of water canthi in-ground swimming pool design below hold?
A) 520 ft3
B) 528 ft3
D) 548 ft3
C) 550 ft3

The equation that models the path of the football released from the given position is h(t) = -10 + 25.37t - 16.1t².

The initial velocity of the ball is determined by applying third kinematic equation as shown below;

Vf² = V₀² - 2gh

0 = V₀² - 2gh

V₀² = 2gh

V₀ = √(2gh)

V₀ = √(2 x 32.17 x 10)

V₀ = 25.37 ft/s

The equation of the ball's motion can be modelled as follows;

h = V₀t - ¹/₂gt²

10 = 25.37t - ¹/₂(32.17)t²

10 = 25.37t - 16.1t²

h(t) = -10 + 25.37t - 16.1t²

Thus, the equation that models the path of the football at any given position is h(t) = -10 + 25.37t - 16.1t².

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Answer:

Multiply the moduli and add the arguments.Step-by-step explanation:First write out the two terms being multiplied: z1z2= [r1(cos(theta 1) + isin(theta 1))][r2(cos(theta 2)+ isin(theta 2))] z1z2= r1r2 (cos(theta 1) + isin(theta 1))(cos(theta 2)+ isin(theta 2)) z1z2= r1r2(cos(theta 1)cos(theta 2))+(icos(theta 1)sin(theta 2))+(icos(theta 2)sin(theta 1))+(i^2 sin(theta 1)sin(theta 2)) Next group and rewrite the like term with the sums and differences. Your answer is the multiplication of the moduli and the addition of the arguments: z1z2= r1r2[cos(theta 1 + theta 2) + i sin (theta 1 and theta 2)].

The product complex number is [tex]z_1z_2[/tex]= [tex]r_1r_2[/tex][cos (θ1+θ2) + i sin (θ1+θ2)].

A real number and an imaginary number are effectively combined to create a complex number. The complex number is written as a+ib, where a and ib are real and imaginary numbers, respectively.

We have,

z1=r1 (cos θ1 +i sin θ1)

z2=r2 (cos θ2 +i sin θ2)

where r-radius/radii

and, m-modulus/moduli

Now, the product is

[tex]z_1z_2[/tex] = r1 (cos θ1 +i sin θ1) x r2 (cos θ2 +i sin θ2)

[tex]z_1z_2[/tex] = r1 r2(cos θ1 . cos θ2 +i cos θ1 sin θ2 + i sin θ1 cos θ2 + i² sin θ1 sin θ2)

[tex]z_1z_2[/tex] = r1 r2(cos θ1 . cos θ2 +i cos θ1 sin θ2 + i sin θ1 cos θ2 - sin θ1 sin θ2)

We know,

cos (a + b)= cos a cos b - sin a sin b

sin (a+b) = sin a cos b + cos a sin b

Thus, [tex]z_1z_2[/tex] = [tex]r_1r_2[/tex][cos (θ1+θ2) + i sin (θ1+θ2)].

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Answer: -10 or D

Step-by-step explanation: simplify radical then combine

Answer:

  (a)  <-3, 14>

Step-by-step explanation:

Addition and subtraction of vectors and matrices is done on a term-by-term basis.

__

The difference is ...

  r - s

  = <4, 9> -<7, -5>

  = <4 -7, 9 -(-5)>

  = <-3, 14> . . . . . . . matches the first choice

Answer:

Step-by-step explanation:

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.Solution:Probability of an event, P(E) = Number of occurrences where the event takes place / Total number of occurrencesThe probability of 2 heads coming up when three coins are tossed simultaneously is given by the ratio of the number of times a particular outcome occurs to the total number of tosses.Total number of tosses = 200Number of 2 heads outcomes = 72Probability of 2 heads outcomes = Number of 2 heads outcomes / Total number of tosses= 72/200= 9/25

Answer:

First box, second from the left,

Second box, first on the left.

Third box, third from the left.

Fourth box, Last on the right.

Step-by-step explanation:

Welcome.

Answer:

4x + 15 = 6x - 3

Step-by-step explanation:

Four times an unknown number is 4x, add 15 like it says, then it says three less so it would be - 3 and then it says than six times the [unknown] number, which remains being x

hope this helps!

Answer:

1 hour and 51 minutes

Step-by-step explanation:

Find the difference between the time.

Answer: 256,1

Step-by-step explanation:

[tex]\sqrt{x} - \sqrt{y} =15\\\sqrt{y} = 49-3\sqrt{x} \\\\\sqrt{x} -(49-3\sqrt{x} ) =15 \\so, x=256\\substitute \{x}\ :\\\\\sqrt{y} = 49-3\sqrt{256} =y=1[/tex]

so, 256=x

and, y=1

Hope this Helped!

Answer:

1, 4 or -1, -4

Step-by-step explanation:

Let the two numbers be x and such that x > y.

According to the first condition:

[tex]x^2 -y^2= 15[/tex]

[tex]\implies x^2 =y^2+15[/tex]......(1)

According to the second condition:

[tex]3x^2 +y^2= 49[/tex]

[tex]\implies 3(y^2+15) +y^2= 49[/tex]

(From equation 1)

[tex]\implies 3y^2+45 +y^2= 49[/tex]

[tex]\implies 4y^2 =49-45[/tex]

[tex]\implies 4y^2 =4[/tex]

[tex]\implies y^2 =\frac{4}{4}[/tex]

[tex]\implies y^2 =1[/tex]

[tex]\implies y =\pm 1[/tex]

When y = 1

[tex]\implies x^2 =(1)^2+15=1+15=16[/tex]

[tex]\implies x =\pm 4[/tex]

When y = -1

[tex]\implies x^2 =(-1)^2+15=1+15=16[/tex]

[tex]\implies x =\pm 4[/tex]

Thus, the required numbers are either 1, 4 or -1, -4

Barbara's net loss on the sale of the stocks she purchased last year is 1245.63.

A net loss is earned when the cost price of purchasing the stock is greater than the selling price.

Net loss = cost price - selling price

Cost price = 8500 + (8500 x 1.25%) = 8606.25

Selling price = 7324 + (7324 x 0.5%) = 7360.62

Net loss = 8606.25 - 7360.62 = 1245.63

Here is the complete question: barbara purchased stock last year for 8500 and paid a 1.25% broker fee. she sold it for 7324 and had to pay a .5% broker fee. Compute her net loss.

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[tex]\angle F+ \angle E = 180^{\circ}\\\\ \implies \angle F + 164^{\circ} = 180^\circ\\\\\implies \angle F = 180^\circ - 164^\circ\\\\\implies \angle F= 16^\circ[/tex]

Answer:

How to solve your problem 42+36+814r2+36r+81

Use the sum-product pattern 42+36+814r2+36r+8142+18+18+81

Answer:

because a= 1,1 then 1+1 = 2

Answer:

angle 2 and angle 4

Answer:

Factoring out Greatest Common Factor (GCF)

[tex]6xy+12x^2y^2-4x^3y^3[/tex]

Factor out greatest common term [tex]2xy[/tex]:

[tex]\implies 2xy(3+6xy-2x^2y^2)[/tex]

Factoring by Grouping

[tex]20x^2+11x-3[/tex]

[tex]\implies a=20, b=11, c=-3[/tex]

[tex]\implies ac=20 \cdot -3=-60[/tex]

Find 2 two numbers that multiply to ac (-60) and sum to b (11)

Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

Therefore, the two numbers are 15 and -4

Rewrite b as the sum of these 2 numbers:

[tex]\implies 20x^2+15x-4x-3[/tex]

Factorize the first two terms and the last two terms separately:

[tex]\implies 5x(4x+3)-1(4x+3)[/tex]

Factor out the common term (4x + 3):

[tex]\implies (5x-1)(4x+3)[/tex]

Factoring by Grouping

[tex]3x^2+3xa-2x-2a[/tex]

Factorize the first two terms and the last two terms separately:

[tex]\implies 3x(x+a)-2(x+a)[/tex]

Factor out the common term (x + a):

[tex]\implies (3x-2)(x+a)[/tex]

Answer:

1/81

Step-by-step explanation:

First lets simplfy the equation :

(3^2)^-2

(remember that you can mutiply the exponent outside the paranthesis by the exponent inside so -2*2=-4)

3^-4

Now lets evaluate this further!

3^-4

(negative exponets mean that you have to raise the number but as the denominator of a fraction)

1/3^4

Now we can solve it fully

1/3^4

1/81

Hey there!

7t - 3 = -59

ADD 3 to BOTH SIDES

7t - 3 + 3 = -59 + 3

CANCEL out: -3 + 3 because it give you 0

KEEP: -59 + 3 because it help solve for the t-value

NEW EQUATION: 7t = -59 + 3

SIMPLIFY IT!

7t = -56

DIVIDE 7 to BOTH SIDES

7t/7 = -56/7

CANCEL out: 7/7 because it give you 1

KEEP: -56/7 because it help solve for the t-value

NEW EQUATION: t = -56/7

SIMPLIFY IT!
t = -8

Therefore, your answer is: t = -8

Good luck on your assignment & enjoy your day!

~Amphitrite1040:)

i)

Factor using completing square method

[tex]\dashrightarrow \sf y = -x^2+16x-64[/tex]

[tex]\dashrightarrow \sf y = -(x^2-16x+64)[/tex]

[tex]\dashrightarrow \sf y = -(x^2-8x-8x+64)[/tex]

[tex]\dashrightarrow \sf y = -(x(x-8)-8(x-8))[/tex]

[tex]\dashrightarrow \sf y = -((x-8)(x-8))[/tex]

ii)

Find zeros of a function, f(x) = 0

[tex]\dashrightarrow \sf -((x-8)(x-8))=0[/tex]

[tex]\dashrightarrow \sf (x-8)=0, \ (x-8)=0[/tex]

[tex]\dashrightarrow \sf x = 8[/tex]

iii)

In order to find vertex use the formulae : x =  -b/2a

[tex]\dashrightarrow \sf x = \dfrac{-(16)}{2(-1)}[/tex]

[tex]\dashrightarrow \sf x = 8[/tex]

Then find y:

[tex]\dashrightarrow \sf y = -(8)^2+16(8)-64[/tex]

[tex]\dashrightarrow \sf y = 0[/tex]

coordinates: (8, 0)

iv) Sketched Below:

#1

#2

Zeros are the x intercepts

Here they are

#3

Vertex form of parabola:-y=a(x-h)²+k

#4

Attached

The slope of the line is -2.5 and the y-intercept is 2. This shows that the Quantity B is greater that A

The intercept is the point where the line intersects

Determine the slope of the graph using the coordinate points (0, 2) and (0.8, 0)
Slope = -2/0.8
Slope = -2.5

The slope of the line is -2.5 and the y-intercept is 2

This shows that the Quantity B is greater than A

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Answer:

The distance betweem the two points is about 14

Step-by-step explanation:

The distance formula is [tex]d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/tex]

According to the problem: [tex]x_{2}=8[/tex], [tex]x_{1} =1[/tex], [tex]y_{2} =-5[/tex], and [tex]y_{1} =7[/tex]

Substitute these numbers into the problem to get [tex]d=\sqrt{(8-1)^{2}+(-5-7)^{2}}[/tex]

[tex](8-1)^2=(7)^2=49[/tex]

[tex](-5-7)^2=(-12)^2=144[/tex]

[tex]\sqrt{49+144}=\sqrt{193} =13.8923349894[/tex]

You can round this number to approximately 14.

6x + 3 = 2

its the answer

easy peasy

Answer:

3 + 6(x) =  2

Step-by-step explanation:

Given statement:

3 more than the product of a number and six is equal to 2​

Converting the statement into an equation:

{3} {more} than {the product of a number and six} is {equal to} {2​}

↓      ↓                                          ↓                                      ↓           ↓

3      +                                        6 × x                                   =           2

⇒ Equation formed: 3 + (6 × x) =  2

⇒ Equation formed: 3 + 6(x) =  2

Answer:

9.22

I really hope it helps

let me know am i right to be shure

Answer:

[tex]9.22[/tex]

Step-by-step explanation:

You would need to use the distance formula to find the approximate distance between points A and B.

[tex]distance=\sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1})^2}[/tex]

[tex]\sqrt{((-2)-4)^{2} +((-4)-3)^2}[/tex]

[tex]\sqrt{(-6)^{2} +(-7)^2}\\[/tex]

[tex]\sqrt{36 +49}[/tex]

[tex]\sqrt{85}[/tex]

[tex]\sqrt{85}=9.219544457=9.22[/tex]

Therefore, the approximate distance between points A and B is 9.22.

Step-by-step explanation:

7x² + 9x² - x²

when you wrote the expression you used the "multiplication ×" and not "x".

anyway, when you have terms of the same variable of the same exponent, then the combination variable and exponent become an item. and the same items can simply be added in an expression.

this is similar to

7 apples + 9 apples - 1 apple

what would you do there ?

see ?

and the same we are doing now for the given expression : we are adding the terms up :

7x² + 9x² - x² = (7 + 9 - 1)x² = 15x²

and in the middle I gave you even the reason why we can do that.

please, let me know, if this is still unclear.

Answer:

6!/3!=6x5x4x3x2/3x2=6x5x4=120

Step-by-step explanation:

There are L, O, O, P, H, O, L, E; 8 letters in LOOPHOLE. However, three letters repeat: OOO. That means we cannot count them as different and use ! for them.

L,P,H,L,E,O are the letters not double counted. There are SIX letters, so 6!, however, you still have to divide since there are the Os you have to consider.

You have to divide by 3! as they repeat 3 times.

Thus, 6!/3!

"There is a subset of permutations that takes into account that there are double objects or repetitions in a permutation problem. In general, repetitions are taken care of by dividing the permutation by the factorial of the number of objects that are identical" -www. ck12. org > probability > lesson > permutation

The answer would be ⁻[tex]\frac{19}{x-5\\}[/tex]

Answer:

[tex]\underline{\boxed{\sf \cfrac{-19}{x-5}}}[/tex]

Step-by-step explanation:

[tex]\sf \cfrac{x^2+10+25}{x+5}-\cfrac{x^2-6}{x-5}[/tex]

First, Let's factor the expressions that are not already factored in x^2+10+25/x+5

[tex]\sf \cfrac{(x+5)^2}{x+5}-\cfrac{x^2-6}{x-5}[/tex]

Now, cancel out x+5 in both numerator and denominator:-

[tex]\sf x+5-\cfrac{x^2-6}{x-5}[/tex]

Expand equations to make their denominators the same before adding or subtracting them. Multiply x+5 * x-5/x-5:-

[tex]\sf \cfrac{(x+5)(x-5)}{x-5}-\cfrac{x^2-6}{x-5}[/tex]

Now, Since [tex]\sf \frac{(x+5)(x-5)}{x-5}[/tex] and [tex]\sf \frac{x^2-6}{x-5}[/tex] and similar denominator, we'll subtrac their numerators.

[tex]\sf \cfrac{(x+5)(x-5)-(x^2-6)}{x-5}[/tex]

[tex]\sf \cfrac{x^2-5x+5x-25-x^2+6}{x-5}[/tex]

Combine like tems in [tex]\sf x^2-5x+5x-25-x^2+6[/tex] :-

[tex]\sf \cfrac{-19}{x-5}[/tex]

Therefore, your answer is D.

Answer:

Hi mate

The right angled triangle is the best name for it

Because it has an right angle or 90 degree angke

Step-by-step explanation:

Answer:

[tex] \longrightarrow \sf \purple{ \boxed{ \bold{ \: 12x² + 12x – 18}}}[/tex]

Step-by-step explanation:

Simplify

10x² + 3x – 6 + 2x² + 9x – 12

Steps

1Ox² + 3x- 6 - 2x² + 9x - 12

Step 1:- Group like terms:

= 10x² + 2x²+ 3x + 9x - 6 - 12

Step 2:- Subtract the numbers: – 6–12 = -18

= 10x² + 2x² + 3x + 9x – 18

Step 3:- Add similar elements: 10x² + 2x² – 12x²

= 12x² + 3x + 9x – 18

Step 4:- Add similar elements: 3x + 9x – 12x

= 12x² + 12x – 18

Answer:

choice f

Step-by-step explanation:

take ∆ abc

ab^2 = ac^2 + ab^2

400 = 225 + ab^2

ab =√ 175

∆aDB

bd = 20-x

ad^2 = 175 -(20-x)^2

∆acd

225 = x^2 + 175 - 400+ 40x - x^2

225 = -225 + 40x

450 = 40x

x = 45/4