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1 year ago

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6.Jay jumped 16 feet on his first long jump attempt and 22 feet on his second attempt. By what percent did his second jump incre

ase? Fill in your answer on the grid provided. Be sure to use the correct place value?

1 answer:

MrMuchimi1 year ago

8 0

His second jump increased by 37.5%.

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Pls help i need this done as quick as possible

Anvisha [2.4K]

Answer:

Infinite Solutions.

Step-by-step explanation:

$m+7=3+m+4$

Subtract m from both sides:

$m+7-m=3+m+4-m$

$7=3+4$

$7=7$

Both sides are equal.

This means that the value of m does not affect the end result of 7 = 7, which is true. Thus, there are infinite solutions as any value of m will still make the equation true.

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Step-by-step explanation:

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2 years ago

ASHLEY IS GOING ON A TRIP TO LONDON. SHE SAVED $100.00 IN SPENDING MONEY. WHEN SHE ARRIVES IN ENGLAND, SHE GOES TO A BANK TO CHA

Ilya [14]

Answer:

65.93

Step-by-step explanation:

Okay, first lets convert the money.

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69.93 - 4 = 65.93

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Hope this helps! Brainliest would be appreciated. :)

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2 years ago

What would be equation would be perpendicular to p(1,3), y = 2x - 1?

Nutka1998 [239]

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2 years ago

Find the taylor polynomial t3(x) for the function f centered at the number

inysia [295]

Answer:

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

Step-by-step explanation:

We are given that

$f(x)=7tan^{-1}(x)$

a=1

$T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}$

Substitute n=3 and a=1

$t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}$

$f(x)=7tan^{-1}(x)$

$f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}$

Where $tan^{-1}(1)=\frac{\pi}{4}$

$f'(x)=\frac{7}{1+x^2}$

Using the formula

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$

$f'(1)=\frac{7}{2}$

$f''(x)=\frac{-14x}{(1+x^2)^2}$

$f''(1)=-\frac{7}{2}$

$f''(x)=-14x(x^2+1)^{-2}$

$f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})$

By using the formula

$(uv)'=u'v+v'u$

$f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}$

$f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}$

$f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}$

Substitute the values

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3$

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

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2 years ago