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2 years ago

HELP PLZ THIS IS DUE TODAY

Mathematics

1 answer:

STatiana [176]2 years ago

3 0

Answer:

A.−3x^4+2x^2−4x

B.4p^7−10p^5+2p^4

Step-by-step explanation:

A.x(−3x^3+2x−4) then simplify −3x^4+2x^2−4x

B. -2p^3(-2p^4+5p^2-p) then simplify 4p^7−10p^5+2p^4

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Unknown interior for 65° and 45°

fomenos

To find the answer, use this equation:
65 + 45 + x = 180
Simplify:
110 + x = 180
Subtract:
x = 70
Your unknown interior is 70

Hope this helps!! :)

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3 years ago

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Phoenix [80]

18 percent is the answer

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2 years ago

Find three consecutive integers such that three times The largest increased by two is equal to five times the smallest increased

rosijanka [135]

$a,a+1,a+2-\ three\ consecutve\ integers\\\\
3*(a+2)+2=5*(a+3)+3(a+1)\\\\
3a+6+2=5a+15+3a+3\\\\
3a+8=8a+18\ \ \ | subtract\ 3a\\\\
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a=-2\\\\Numbers\ are:\ \ -2,-1,0
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3 years ago

What is the Domain and Range of the function f(x)<img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx-7%7D%20%2B9" id="TexFormula1" tit

omeli [17]

For the given function, the domain is D : { x ≥ 7} and the range is R: { y ≥ 9}

<h3>How to get the domain and range?</h3>

Here we have a square root, remember that the argument of the square root must be equal or larger than zero, so the domain is such that:

x - 7 ≥ 0.

Solving for x we get:

x ≥ 0 + 7

Then the domain is:

x ≥ 7

To get the range, we evaluate in the minimum of the domain:

f(7) = √(7 - 7) + 9 = 9

Then the range is the set of all values larger than 9, because the function is increasing.

So the range is R: y ≥ 9.

If you want to learn more about domain and range:

brainly.com/question/10197594

#SPJ1

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1 year ago

Find the value of x² - x+1 when x=√3 -1, in the form a+b√3 where a and b are integers

lidiya [134]

Answer:

$6-3\sqrt{3}$

If you compare this to $a+b\sqrt{3}$ , then $a=6$ and $b=-3$ .

Step-by-step explanation:

We are given $x=\sqrt{3}-1$ and we are asked to find the value of $x^2-x+1$ .

So we will need to find $x^2$ .

$x=\sqrt{3}-1$

Square both sides:

$x^2=(\sqrt{3}-1)^2$

Expand using identity: $(x-a)^2=x^2-2ax+a^2$ .

$x^2=(\sqrt{3})^2-2(\sqrt{3})(1)+1^2$

$x^2=3-2\sqrt{3}+1$

$x^2=4-2\sqrt{3}$

Let's go back to the full $x^2-x+1$ .

$x^2-x+1$

$(4-2\sqrt{3})-(\sqrt{3}-1)+1$

$4-2\sqrt{3}-\sqrt{3}+1+1$

$6-3\sqrt{3}$

If you compare this to $a+b\sqrt{3}$ , then $a=6$ and $b=-3$ .

5 0

3 years ago