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3 years ago

Please help!! I need it

Mathematics

1 answer:

anygoal [31]3 years ago

3 0

Answer:

4th

Step-by-step explanation:

make me as branalist

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Find the midpoint of the coordinates (-2,1) and (4,-4)

kow [346]

(-2,1)
(4,-4)

$\\ \rm\longmapsto P(x,y)$

$\\ \rm\longmapsto \left(\dfrac{-2+4}{2},\dfrac{1-4}{2}\right)$

$\\ \rm\longmapsto \left(\dfrac{2}{2},\dfrac{-3}{2}\right)$

$\\ \rm\longmapsto \left(1,\dfrac{-3}{2}\right)$

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3 years ago

Find the value of x.

Pepsi [2]

I would the answer to x is 12

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3 years ago

Type the correct answer in each box. Functions h and K are inverse functions, and both are defined for all real numbers Using th

kompoz [17]

Answer:

(h o k) (3) = 3

(k o h) (-4b) = -4b

Step-by-step explanation:

An inverse function is the opposite of a function. An easy way to find inverse functions is to treat the evaluator like another variable, then solve for the input variable in terms of the evaluator. One property of inverse functions is that when one finds the composition of inverse functions, the result is the input value, no matter the order in which one uses the functions in the combination. This is because all terms in a function and their inverse cancel each other and the result is the input. Thus, when one multiplies two functions that are inverse of each other, no matter the input, the output will always be the input value.

This holds true in this case, it is given that (h) and (k) are inverses. While one is not given the actual function, one knows that the composition of the functions (h) and (k) will result in the input variable. Therefore, even though different numbers are being evaluated in the composition, the output will always be the input.

7 0

3 years ago

Need help simplifying this

diamong [38]

The simplified answer is $\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$ .

<u>Step-by-step explanation:</u>

$$\frac{3 y+2 x}{z+2 x}-\frac{2 y-3 x}{3 x+y}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}$

To add or subtract denominators of the fraction must be same.

If it is not the same, we must take LCM of the denominators. and so we can add the fractions.

To make the denominator same multiply the 1st term $(\frac{3x+y}{3x+y})$ and 2nd term by $(\frac{z+2x}{z+2x})$

= $\frac{(3 y+2 x)(3 x+y)}{(z+2 x)(3 x+y)}-\frac{(2 y-3 x)(z+2 x)}{(3 x+y)(z+2 x)}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}$

LCM of the denominators is 6x²+ 3xz + 2xy +yz.

Multiply the factors in the numerator.

= $\frac{\left(6 x^{2}+3 y^{2}+11 x y\right)}{(z+2 x)(3 x+y)}-\frac{\left(2 y z+4 x y-3 x z-6 x^{2}\right)}{(3 x+y)(z+2 x)}-\frac{2 z y+6 x z}{6 x^{2}+3 x z+2 x y+y z}$