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3 years ago

Please help me!!!!!!!!!

Mathematics

1 answer:

bagirrra123 [75]3 years ago

5 0

F(-2)+g(5) = 3(-2) -1 + (-5) + 6
f(-2)+g(5) = -6 -1 - 5 + 6
f(-2)+g(5) = -6

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5(x-y)-5(y+x)<br> pls help with this

anastassius [24]

Answer:

-10y

Step-by-step explanation:

=5x-5y-5y-5x

=5x-5x-5y-5y

=-10y

6 0

3 years ago

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Im literally sitting here crying my eyes out because i have so much homework due tomorrow someone please just help me with this

skelet666 [1.2K]

Answer:

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3 years ago

What is f–1(8)??<br><br> Please select the best answer

notka56 [123]

Answer:

$f^{-1}(8)=\frac{3}{2}$

Step-by-step explanation:

Given f(x)= $2x+5$

We have to find $f^{-1}(8)$

In order to find $f^{-1}(x)$ we have to make x as the subject of the formula

Let us assume f(x)=y

⇒ $2x+5=y$

Subtracting both sides by 5

$2x+5-5=y-5$

$2x=y-5$

Dividing both sides by 2

$\frac{2x}{2}=\frac{y-5}{2}$

⇒ $x=\frac{y-5}{2}$

Now substituting y with x

we have $f^{-1}(x)=\frac{x-5}{2}$

Now $f^{-1}(8)=\frac{8-5}{2}$

= $\frac{3}{2}$

So option (ii) is correct

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3 years ago

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Are you supposed to put a period at the end of a verbal expression in algebra?

ivann1987 [24]

Nope :) hope this helps!

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3 years ago

Select two polynomials that have a sum of 16n^3+13n^2+7n-10 A. 8n^3 - 5n +11n^2 - 5 B. 7n^2 + 8n - n^3 + 2 C. 15n^2 - 10n + 3n^3

andreyandreev [35.5K]

Answer:

Step-by-step explanation:

We have the polynomial

$16n^3+13n^2+7n-10$ ( 1 )

To solve this problem we have to take into account that only we can sum term with the same order in the variable. We have the polynomials

$8n_{3}-5n+11n^{2}-5\\7n^2 + 8n - n^3 + 2\\15n^2 - 10n + 3n^3 - 4\\2n^2 + 12n - 5 + 8n^3$

we can note (by summing term by term) that only the sum of the first and the fourth equation correspond to the given polynomial ( 1 ) of the problem. If we organize these polynomials (that is, write the equation down in a form where higher order appears first ) we have

$8n^3 +11n^2 - 5n - 5\\ 8n^3+2n^2 + 12n - 5\\$

and if we sum we obtain

$16n^3+13n^2+7n-10$

that is what we was looking for

I hope this is useful for you

regards

7 0

3 years ago