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3 years ago

Which situation is best model by the expression 1/2÷5

Mathematics

1 answer:

ahrayia [7]3 years ago

3 0

I can give 1/2 of a cookie to all 5 kids so how many cookies do i need to get?

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F(x) = 3x + 2 replace f(x) with y

kirill [66]

Answer:

-1 x - 2

f (x) = -----------

Step-by-step explanation:

Then we have y = 3x + 2.

Next, interchange x and y, obtaining x = 3y + 2

Next, solve this result for y: 3y = x - 2; then divide both sides by 3:

y = (x-2)/3

Last, replace "y" with the inverse function label:

-1 x - 2

f (x) = -----------

5 0

3 years ago

suppose a population of 40 crickets doubles in size every month. How many crickets will there be after 2 years?

blondinia [14]

Answer:1920

Step-by-step explanation:

You can simplify it by multiplying the 24 by 2 to get 48 then get 48 time 40 which is 1920

7 0

3 years ago

-18-3 what's the answer

Llana [10]

The answer is -21. Rule is that you follow the sign of the larger and add the opposite

4 0

4 years ago

Read 2 more answers

Find 4 consecutive odd numbers such that twice the sum of the least 2 numbers is 40 less than three times the sum of the greates

ANTONII [103]

4 consecutive odd numbers....x, x + 2, x + 4, x + 6

2(x + x + 2) = 3(x + 4 + x + 6) - 40
2(2x + 2) = 3(2x + 10) - 40
4x + 4 = 6x + 30 - 40
4x + 4 = 6x - 10
4x - 6x = -10 - 4
-2x = - 14
x = -14/-2
x = 7

x + 2 = 7 + 2 = 9
x + 4 = 7 + 4 = 11
x + 6 = 7 + 6 = 13

ur numbers are : 7,9,11,13 <==

4 0

4 years ago

If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ? sin Θ = square root 2 ov

densk [106]

Answer:

$\huge\boxed{\sin\theta=-\dfrac{\sqrt2}{2};\ \tan\theta=-1}$

Step-by-step explanation:

We have:

$\\cos\theta=\dfrac{\sqrt2}{2},\ \dfrac{3\pi}{2}$

For sine use:

$\sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x$

Substitute:

$\sin^2\theta=1-\left(\dfrac{\sqrt2}{2}\right)^2\\\\\sin^2\theta=1-\dfrac{(\sqrt2)^2}{2^2}\\\\\sin^2\theta=1-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4}{4}-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4-2}{4}\\\\\sin^2\theta=\dfrac{2}{4}\to\sin\theta=\pm\sqrt{\dfrac{2}{4}}\\\\\sin\theta=\pm\dfrac{\sqrt2}{\sqrt4}\\\\\sin\theta=\pm\dfrac{\sqrt2}{2}$

θ in IV quadrant, therefore sine is negative.

$\sin\theta=-\dfrac{\sqrt2}{2}$

For tangent use:

$\tan x=\dfrac{\sin x}{\cos x}$

Substitute:

$\tan\theta=\dfrac{-\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=-\dfrac{\sqrt2}{2}\cdot\dfrac{2}{\sqrt2}=-1$

8 0

3 years ago