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3 years ago

14

A weight is attached to the end of a spring. Its height after t seconds is given by the equation h(t)=5−2sin((2π(t+1))/7)

1 answer:

taurus [48]3 years ago

3 0

Answer:

실례합니다 ? 이 질문입니까?

Step-by-step explanation:

스프링 끝에 추가 부착되어 있습니다. t초 후의 높이는 h(t)=5−2sin((2π(t+1))/7) 방정식으로 주어집니다. 무게가 처음 최대 높이에 도달하는 시점은 언제입니까? 정확한 답을 주십시오. 오른쪽 ?

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Answer:

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3 years ago

3/2 times 2/3 in the simplest form

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3/2 x 2/3 = 1

Step-by-step explanation:

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2 years ago

A person standing at the edge of a cliff throws a rock upward from a height of 180ft above ground level with an initial velocity

taurus [48]

Answer: 8.73 ft/s

Step-by-step explanation:

We have the following equation that models the motion of the rock:

$h(t)=-16t^{2}+64t+180$

Where $h(t)$ is the height of the rock at a time $t$ .

Now, if $h(t)=48 ft$ we can find $t$ and then the velocity of the rock at that height:

$48=-16t^{2}+64t+180$

Rearranging the equation:

$-16t^{2}+64t+132=0$

Multiplying the equation by -1:

$16t^{2}-64t-132=0$

Solving with the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ , where $a=16$ , $b=-64$ , $c=-132$ .

$t=\frac{-(-64)\pm\sqrt{(-64)^{2}-4(16)(-132)}}{2(16)}$

Choosing the positive result of the equation:

$t=5.5 s$

Since velocity $V$ is defined as the traveled distance $h(t)$ in a given time $t$ , we have:

$V=\frac{h(t)}{t}$

$V=\frac{48 ft}{5.5 s}$

$V=8.727 ft/s \approx 8.73 ft/s$ This is the velocity of the rock at the height of 48 feet

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3 years ago

Which expressions are equivalent to 2^−4·2^2?

Ede4ka [16]

Answer: the answer is 6)

Step-by-step explanation:

Because $2x^{-4}$ . $2x^{2}$ = $2x^{-4+2}$ = $2x^{-2}$

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2 years ago

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leonid [27]

Answer:

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Step-by-step explanation:

44/4 = 22

circumfrence of half circle is 3.14r = 22

22/3.14 is about 7

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2 years ago