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3 years ago

Please help!! I'm super confused

Mathematics

1 answer:

Katyanochek1 [597]3 years ago

8 0

Answer:

The tom brady question is

2.5

Step-by-step explanation:

40/16 = 2.5

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Jeff bought 10 cookies and ate 4 of them. He ate

zmey [24]

he ate 40 percent of the cookies

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3 years ago

Read 2 more answers

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed

Olegator [25]

Answer:

The length of shadow decrease at 0.48 m/s.

Step-by-step explanation:

The distance of man from building = x

Thus initially, x = 12

It is given that dx/dt= -1.3 m/s

Let the shadow height = Y

Now solve for the $dy/dt$ when x=4

from the diagram (triangle) it can be seen similar triangles with similar base/height ratios.

$\frac{2}{12-x} = \frac{y}{12} \\Y = \frac{24}{12 - x} \\ \frac{dy}{dt} = \frac{24}{(12-x)^{2}} \times \frac{dx}{dt} \\at x=4, dy/dt \ becomes: \\\frac{24}{(12-4)^2} \times -1.3 \\= -0.48 m/s$

8 0

3 years ago

Direct variation or not a direct variation<br><br> y=5/x

MrRissso [65]

Answer:

not direct variation

Step-by-step explanation:

The equation for x and y in direct variation is

y = kx ← k is the constant of variation

y = $\frac{5}{x}$ is not in this form, thus not direct variation.

The equation for x and y varying inversely is

y = $\frac{k}{x}$ ← k is the constant of variation

y = $\frac{5}{x}$ is in this form and represents inverse variation

7 0

3 years ago

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

7 0

3 years ago

In the following. Diagram line is parallel to line be find the measure of angle X

Mrac [35]

Isjwkddjkwhdjxjwkhsidhwjhaosjshwuwjw

7 0

3 years ago