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2 years ago

9

28 first graders and 12 other students attended a school assembly what percentage of the students at the assembly were first gra

ders

2 answers:

Hitman42 [59]2 years ago

8 0

Answer:

70%

Step-by-step explanation:

28+12= 40

28/40= 0.7

0.7x100= 70

Harlamova29_29 [7]2 years ago

3 0

Answer:

70%

Step-by-step explanation:

Find the total students who attended a school assembly:

28 + 12 = 40

100% = 40 students

1% = 40 ÷ 100 = 0.40

28 ÷ 0.40 = 70

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The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)/g(

Rus_ich [418]

U(x) = f(x).(gx)
v(x) = f(x) / g(x)

Use chain rule to find u(x) and v(x).

u '(x) = f '(x) g(x) + f(x) g'(x)

v ' (x) = [f '(x) g(x) - f(x) g(x)] / [g(x)]^2

The functions given are piecewise.

You need to use the pieces that include the point x = 1.

You can calculate f '(x) and g '(x) at x =1, as the slopes of the lines that define each function.

And the slopes can be calculated graphycally as run / rise of each graph, around the given point.

f '(x) = slope of f (x); at x = 1, f '(1) = run / rise = 1/1 = 1

g '(x) = slope of g(x); at x = 1, g '(1) = run / rise = 1.5/ 1 = 1.5

You also need f (1) = 1 and g(1) = 2

Then:

u '(1) = f '(1) g(1) + f(1) g'(1) = 1*2 + 1*1.5 = 2 + 1.5 = 3.5

v ' (x) = [f '(1) g(1) - f(1) g(1)] / [g(1)]^2 = [1*2 - 1*1.5] / (2)^2 = [2-1.5]/4 =

= 0.5/4 = 0.125

Answers:
u '(1) = 3.5
v '(1) = 0.125

3 0

3 years ago

A ball is thrown into the air and its position is given by h ( t ) = − 6.3 t 2 + 53 t + 24 where h is the height of the ball in

nordsb [41]

Answer:

$h'(t) = -12.6 t +53$

Now we can set up the derivate equal to 0 and we have:

$-12.6 t +53 = 0$

And solving for t we got:

$t = \frac{53}{12.6}= 4.206$

For the second derivate respect the time we got:

$h''(t) = -12.6$

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

$h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft$

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

Step-by-step explanation:

For this case we have the following function:

$h(t) = -6.3t^2 +53 t+24$

In order to maximize this function we need to take the first derivate respect the time and we have:

$h'(t) = -12.6 t +53$

Now we can set up the derivate equal to 0 and we have:

$-12.6 t +53 = 0$

And solving for t we got:

$t = \frac{53}{12.6}= 4.206$

For the second derivate respect the time we got:

$h''(t) = -12.6$

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

$h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft$

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

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3 years ago

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djverab [1.8K]

We are given the area of the region under the curve of the function f(x) = 5x + 7 with an interval [1, b] which is 88 square units where b > 1

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5/2 x^2 + 7x (limits: 1, b)

substitute the limits:

5/2 (1^2) + 7 (1) - 5/2 b^2 + 7b = 0

solve for b

Then after solving for b, this would be your interval input with 1: [1, b].<span />

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2 years ago

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Answer:

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