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andrew-mc [135]

3 years ago

10

Find the average speed. 700 feet in 1 minute 10 seconds

2 answers:

allsm [11]3 years ago

6 0

116.666666667 in ten seconds

tresset_1 [31]3 years ago

4 0

I believe that the answer is 20.45

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An object dropped from rest from the top of a tall building on Planet X falls a distance feet in the first t seconds. Find the a

ch4aika [34]

Complete Question:

An object dropped from rest from the top of a tall building on Planet X falls a distance d(t)=13t^2 feet in the first t seconds. Find the average rate of change of distance with respect to time as t changes from t1=2 to t2=10. This rate is known as the average velocity, or speed.

The average velocity as t changes from 2 to 10 seconds is _______ feet/sec?

Answer:

$v = 156$

Step-by-step explanation:

Given

$d(t)=13t^2$

Time Interval: (a,b)

$a = 2$

$b = 10$

Required

Determine the average rate of change (v)

This is calculated as thus:

$v = \frac{d(b) - d(a)}{b - a}$

Substitute values for b and a

$v = \frac{d(10) - d(2)}{10 - 2}$

$v = \frac{d(10) - d(2)}{8}$

Calculate d(10): Substitute 10 for t

$d(t)=13t^2$

$d(10) = 13 * 10^2$

$d(10) = 13 * 100$

$d(10) = 1300$

Calculate d(2): Substitute 2 for t

$d(t)=13t^2$

$d(2) = 13 * 2^2$

$d(2) = 13 * 4$

$d(2) = 52$

The equation $v = \frac{d(10) - d(2)}{8}$ becomes

$v = \frac{1300 - 52}{8}$

$v = \frac{1248}{8}$

$v = 156$

Hence, the average rate of change is 156ft/s

7 0

2 years ago

A positive real number is 2 less than another. When 4 times the larger is added to the square of the smaller, the result is 49.

Kobotan [32]

Answer:

The numbers are

$-2+3\sqrt{5}$ and $3\sqrt{5}$

Step-by-step explanation:

Let

x -----> the smaller positive real number

y -----> the larger positive real number

we know that

A positive real number is 2 less than another

so

$x=y-2$

$y=x+2$ ----> equation A

When 4 times the larger is added to the square of the smaller, the result is 49

so

$4y+x^2=49$ ----> equation B

substitute equation A in equation B

$4(x+2)+x^2=49$

solve for x

$x^2+4x-41=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^2+4x-41=0$

so

$a=1\\b=4\\c=-41$

substitute in the formula

$x=\frac{-4\pm\sqrt{4^{2}-4(1)(-41)}} {2(1)}$

$x=\frac{-4\pm\sqrt{180}} {2}$

$x=\frac{-4\pm6\sqrt{5}} {2}$

$x=-2\pm3\sqrt{5}$

so

The positive real number is

$x=-2+3\sqrt{5}$

Find the value of y

$y=x+2$

$y=-2+3\sqrt{5}+2$

$y=3\sqrt{5}$

6 0

3 years ago

Can y = sin(t2) be a solution on an interval containing t = 0 of an equation y + p(t) y + q(t) y = 0 with continuous coefficient

JulsSmile [24]

Answer:

Step-by-step explanation:

y = sin(t^2)

y' = 2tcos(t^2)

y'' = 2cos(t^2) - 4t^2sin(t^2)

so the equation become

2cos(t^2) - 4t^2sin(t^2) + p(t)(2tcos(t^2)) + q(t)sin(t^2) = 0

when t=0, above eqution is 2. That is, there does not exist the solution. so y can not be a solution on I containing t=0.

7 0

2 years ago

Please help me in this question.I need it right now!!!

zheka24 [161]

Answer:

X=102 y=288 =390

Step-by-step explanation:

X=50+32=82=180-82=102

Y= 50+32=82=360-82=288

Total answer is =390

6 0

2 years ago

the two triangles below are similar use what you know about similar triangles to find the missing side. show your work

lara31 [8.8K]

If both triangles are similar the ratio of sides will be same

$\\ \sf\longmapsto \frac{x}{24} = \frac{6}{12} \\ \\ \sf\longmapsto 12x = 6 \times 24 \\ \\ \sf\longmapsto 12x = 144 \\ \\ \sf\longmapsto x = \frac{144}{12} \\ \\ \sf\longmapsto x = 12$

7 0

2 years ago