A bouncy ball is dropped. The function h(n)=300(3/4)^n−1 models the maximum height of the ball, in centimeters, after its nth bounce.

Select EACH true statement about the scenario.

[] The function represents a geometric sequence.
[] The first term in the sequence is 300.
[] The height of the ball after the first bounce is 300 cm.
[] The ball is dropped from a height of 300 cm.

*Multiple Choice*

Answers

Answer 1

The true statements about the scenario are;

The function represents a geometric sequence. The first term in the sequence is 300.

The standard form of a geometric sequence is expressed as:

[tex]Tn = ar^{n-1}[/tex]

a is the first term

r is the common ratio

n is the number of terms

Given the expression [tex]h(n)=300(\frac{3}{4} )^{n-1}[/tex]

Compare and contrast

a = 300

This shows that the first term of the sequence is 300.

Also, a function represents a geometric sequence

Hence the true statements about the scenario are;

The function represents a geometric sequence. The first term in the sequence is 300.

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Write 3/6/7 in an improper fraction in the lowest term

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27/7

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Mr. Evans prepared 5 rows for 6 tomato plants in each
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1) [tex]y=\frac{3}{2}x+8[/tex]

2)[tex]y=\frac{13}{11}x +\frac{12}{11}[/tex]

3)[tex]y=\frac{9}{7}x+1[/tex]

4)[tex]y=\frac{1}{3}x-2[/tex]

5)[tex]y=-\frac{6}{5}x-3[/tex]

6)[tex]y=4x-1[/tex]

7)[tex]y=\frac{11}{4}x-8[/tex]

8)[tex]y=\frac{11}{8}x+6[/tex]

Answer:

[tex]\sf 1) 3x-2y=-16[/tex]

subtract both sides by 3x:

[tex]\sf -2y=-16-3x[/tex]

[tex]\sf \cfrac{-2y}{-2} =\cfrac{3x-16}{-2}[/tex]

[tex]\boxed {\sf y=\frac{3}{2}x +8}[/tex]

________________

[tex]\sf 2) 13x-11y=-12[/tex]

Subtract 13x from both sides:

[tex]\sf 13x+11y-13x=-12-13x[/tex]

[tex]\sf \cfrac{-11y}{-11} =\cfrac{-12-13x}{-11-11}[/tex]

[tex]\boxed {\sf y=\cfrac{13}{11} \:x+\cfrac{12}{11} }[/tex]

___________________

[tex]\sf 3) 9x-7y=-7[/tex]

Subtract 9x from both sides:

[tex]\sf \cfrac{-7y}{-7} =\cfrac{9x-7}{-7}[/tex]

[tex]\boxed {\sf y=\cfrac{9}{7} \:x+1}[/tex]

___________________

[tex]\sf 4)x-3y=6[/tex]

Add 3y from both sides:

[tex]\sf x-3y+3y=6+3y[/tex]

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[tex]\sf \cfrac{x}{3}-\cfrac{6}{3} =\cfrac{3y}{3}[/tex]

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[tex]\boxed {\sf y=\cfrac{1}{3} \:x-2}[/tex]

____________________

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Subtract 6x from both sides:

[tex]\sf \cfrac{5y}{5}=\cfrac{-6x-15}{5}[/tex]

[tex]\boxed {\sf y=-\cfrac{6}{5} \:x-3 }[/tex]

_____________________

[tex]\sf 6) 4x-y=1[/tex]

Subtract 4x from both sides:

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[tex]\boxed {\sf y=4x-1}[/tex]

_______________________

[tex]\sf 7)11x-4y=32[/tex]

Subtract 11x from both sides:

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[tex]\boxed {\sf y=\frac{11}{4} \:x-8}[/tex]

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[tex]\sf 8) 11x-8y=-48[/tex]

Subtract 11x from both sides:

[tex]\sf \cfrac{-8y}{-8} =\cfrac{11-48}{-8}[/tex]

[tex]\boxed {\sf y=\frac{11}{8} \:x+6}[/tex]

________________________________

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Answer:

(-7, -7)

Step-by-step explanation:

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Alyssa makes $14 baby sitting on Monday, $15.50 baby sitting on Tuesday, spends $24.35 at the
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Answer:

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Last year, Deandre had $10,000 to invest. He invested some of it in an account that paid 7% simple interest per year, and he invested the rest in an account that paid 10% simple interest per year. After one year, he received a total of $790 in interest. How much did he invest in each account?
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He had $10,000 in the first account & $900 in the second account.

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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Answers

Answer:

answer for 10th question is 121

Step-by-step explanation:

2(a+b)+a+b/2+(a+b)^2-4

=2(10)+10/2+(10)^2-4

=20+5+100-4

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Answer:

10.  121      11.  Hamburger = $1.43

Step-by-step explanation:

2(a + b) + (a + b)/2 + (a + b)² - 4

2(10) + (10)/2 + (10)² - 4

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25 + 100 - 4

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      2 H + 3D = $5.26

H + D = $2.23        Solve for D

D = $2.23 - H        

2H + 3D = $5.26     Now substitute the value of H into the second equation.

2H + 3($ 2.23 - H) = $5.26

2H + $6.69 - 3H = $5.26

 -H + $6.69 = $5.26

 -H = - $1.43

  H = $1.43

 

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Answers

Answer:

y ≤ 20-8x

Step-by-step explanation:

Darren needs to spend no more than 15 hours on cardio exercise (y). To figure out how much time he has we need to do 20-8x. Any leftover hours are the hours left for him to spend doing his exercises. Since y has to be less than or equal to 20-8x, the system of linear inequalities should be:

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Answer:

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Step-by-step explanation:

If cosec ∅ = √2, find the value of 1+sin²∅+cos²∅/1+sec²∅-tan²∅.​

Answers

Given: [tex]cosec[/tex] [tex]A[/tex]  [tex]=\sqrt{2}[/tex]

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[tex]sin[/tex] [tex]A[/tex] [tex]=sin[/tex] [tex]45^{0}[/tex] = [tex]\frac{1}\sqrt{2}[/tex]

[tex]cot[/tex] [tex]A[/tex] [tex]=cot[/tex] [tex]45^{0}[/tex] [tex]=1,[/tex]

[tex]tan[/tex] [tex]A[/tex] [tex]=[/tex] [tex]tan[/tex] [tex]45^{0}[/tex] [tex]=1[/tex]

[tex]cos[/tex] [tex]A[/tex] [tex]=[/tex] [tex]cot[/tex] [tex]45^{0}[/tex] [tex]=\frac{1}{\sqrt{2} }[/tex]

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[tex]\large\underline{\sf{Solution-}}[/tex]

METHOD 1: Using Trigonometric values.

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[tex]\sf\cosec\theta=\sqrt2[/tex]

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We need to find:

[tex]\sf\dfrac{1+\sin^2\theta+\cos^2\theta}{1+\sec^2\theta-\tan^2\theta}[/tex]

So,

[tex]\sf\longmapsto\dfrac{1+\sin^2(45^{\circ})+\cos^2(45^{\circ})}{1+\sec^2(45^{\circ})-\tan^2(45^{\circ})}[/tex]

[tex]\sf\longmapsto\dfrac{1+\left(\frac{1}{\sqrt2}\right)^2+\left(\frac{1}{\sqrt2}\right)^2}{1+(\sqrt2)^2-(1)^2}[/tex]

[tex]\sf\longmapsto\dfrac{1+\frac{1}{2}+\frac{1}{2}}{1+2-1}[/tex]

[tex]\sf\longmapsto\dfrac{1+1}{2}[/tex]

[tex]\sf\longmapsto\dfrac{2}{2}[/tex]

[tex]\sf\longmapsto\bf1[/tex]

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We need to find:

[tex]\implies\dfrac{1+\sin^2\theta+\cos^2\theta}{1+\sec^2\theta-\tan^2\theta}[/tex]

But, we know that,

[tex]\sf\red⇛ \sin^2\phi+\cos^2\phi=1[/tex]

And,

[tex]\sf\red⇛ \sec^2\phi-\tan^2\phi=1[/tex]

So,

[tex]\sf\longmapsto\dfrac{1+(\sin^2\theta+\cos^2\theta)}{1+(\sec^2\theta-\tan^2\theta)}[/tex]

[tex]\sf\longmapsto\dfrac{1+1}{1+1}[/tex]

[tex]\sf\longmapsto\dfrac{2}{2}[/tex]

[tex]\sf\longmapsto\bf1[/tex]

Therefore, by both methods the answer is 1.

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What is 300251 take away 21596 what round it to the highest place value then subtract​

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we need to subtract

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