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Leto [7]
3 years ago
15

If y varies directly as x , and y = 6 when x = 15, find y when x = 25 .k = ______y = ______

Mathematics
2 answers:
viva [34]3 years ago
6 0
At x = 25, y equals to 10.4.
dsp733 years ago
6 0

3

y varies inversely as x

y = K / x , where K is any constant

y = 15 when x = 5

15 ( 5 ) = K = 75

y = 75 / x

When x = 25 ,  

y = 75 / 25

y = 3

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Natasha2012 [34]

Answer:

4fg^{2}

Step-by-step explanation:

1) Remove parentheses.

fg^{2} *4

2) Regroup terms.

4fg^{2}

5 0
2 years ago
What is the area of this composite plane figure??
Ghella [55]
A = 16 \times 6 + \sqrt{3 + 6} = 96 + 9 = \boxed{105}
5 0
3 years ago
Verify that -tan2x = (2tanx) / (sec^2x - 2)
sashaice [31]

Answer:

Below.

Step-by-step explanation:

- tan 2x =   -2tanx / (1 - tan^2x)

Using the identity  tan^2x = sec^2x - 1 and substituting for tan^2x:

- tan 2x = -2 tanx / (1 - (sec^2x - 1))

=   2 tanx / ( - 1(sec^2x + 2))

= 2 tan x / (sec^2 x - 2)

8 0
3 years ago
Y=25x+35
kow [346]

Answer:

The equation is in the form y=mx+b, where m=slope, b=y-intercept.

The y-intercept is 35: so the line goes through the y-axis at (0,35).

The slope is 25: so we need to count out the slope as rise over run, or 25 up and 1 to the right, starting from the y-intercept.

8 0
3 years ago
Read 2 more answers
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
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