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3 years ago

10

PLEASE HELP 3/4M -12 but m = 32

1 answer:

slava [35]3 years ago

4 0

Answer:

Replace the variable m with 32

in the expression.

3/4⋅(32)−12 Simplify each term.

24−12 Subtract 12 from 24.

12

Step-by-step explanation:

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-16t squared + 16t + 480 t=time after jumping in seconds h=height in feet How many seconds did did it take her to hit the water

Agata [3.3K]

That is height
h=16t^2+16t+480
when h=0 is when it hits the water

0=16t^2+16t+480
undistribbute 16 for ease
0=16(t^2+t+30)
16*0=0 so assume
t^2+t+30=0
factor
(t+6)(t-5)=0
set to zero

t+6=0
minus 6
t=-6
false, since you cannot have negative time

t-5=0
add 5
t=5

at t=5 she hits the water

5 seconds

4 0

3 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

Imena's math test scores are 83, 93, 78, 89, and 83. She has one more math test this semester. What score most she get on the te

marshall27 [118]

87 is the median because 83 and 87's average is 85.

8 0

3 years ago

Read 2 more answers

There are 42 sixth-grade students and 60 fifth grade students. The students will be put in to the largest possible equal groups

scoundrel [369]

Answer:51 Groups

Step-by-step explanation:

60 students + 42 students = 102

102 divided by 2 = 51 Total groups.

5 0

3 years ago

The distance on the number line from -10 to -2 equals 5x – 12. What is the value of x?

Troyanec [42]

It would be: - 2 - (-10) = 5x - 12

= -2 + 10 = 5x - 12

= 8 + 12 = 5x

= x = 20 / 5

= x = 4

4 0

3 years ago