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3 years ago

14

Your answer should be a polynomial in standard form. ( − 8 p 3 + 9 p 2 q − 5 p q 2 ) + ( 8 p 3 − p q + 5 q 2 ) = (−8p 3 +9p

2 q−5pq 2 )+(8p 3 −pq+5q 2 )

1 answer:

gavmur [86]3 years ago

5 0

9p^2-6p5q^2 in standard form it would be: 9p^2+5q^2-6p

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I WILL MARK BRAINLIEST

Korvikt [17]

Answer:

I believe the answer is A. 15-(3xy)

Step-by-step explanation:

The expression you are given is 15-(yx3) and option A is 15-(3xy). All they did was flip the values in the parentheses. It doesen't matter what value is first because the negative sign will be distributed to everything in the parentheses.

Hope this helps!!

Mark me brainliest please!

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2 years ago

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KATRIN_1 [288]

The diameter of the plate is 12, and the formula for finding circumference is diameter times pie. so you multiply pie (or 3.14 if you aren't using a calculator) times 12 and it should give you 37.6991. rounded to the nearest tenth, that's 37.7, hope I helped!

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3 years ago

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RUDIKE [14]

The perimeter, by definition, is the outside measure of that figure. MN and LM are the same length and LK and NK are the same length....we just need to find the lengths! Use the distance formula to find the distance between the 2 points:
$\sqrt{( x_{2} - x_{1} ) ^{2} +( y_{2}- y_{1} ) ^{2} }$
For the segment MN, use the coordinates of M as your x1, y1, and use the coordinates of N for x2, y2:
$\sqrt{(3-2) ^{2}+(4-3) ^{2} }$
which simplifies to
$\sqrt{(1 )^{2}+(1) ^{2} }$ which is $\sqrt{2}$
So that is the length of both MN and LM. So far our perimeter is $\sqrt{2} + \sqrt{2}=2 \sqrt{2}$
Now let's use the same formula to find out the length of one of the longer segments:
$\sqrt{(5-3) ^{2} +(3-2) ^{2} }$
which simplifies down to
$\sqrt{(2) ^{2} +(1) ^{2} }$
which is of course $\sqrt{5}$
Since we have 2 of those lengths, $\sqrt{5} + \sqrt{5}=2 \sqrt{5}$
So our perimeter is, in the end, $2 \sqrt{2}+2 \sqrt{5}$
That's the third choice down

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3 years ago

Read 2 more answers

Solve 2x^2 + x - 4 = 0 <br> X2 +

damaskus [11]

Answer:

$\large \boxed{\sf \ \ x = -\dfrac{\sqrt{33}+1}{4} \ \ or \ \ x = \dfrac{\sqrt{33}-1}{4} \ \ }$

Step-by-step explanation:

Hello, please find below my work.

$2x^2+x-4=0\\\\\text{*** divide by 2 both sides ***}\\\\x^2+\dfrac{1}{2}x-2=0\\\\\text{*** complete the square ***}\\\\x^2+\dfrac{1}{2}x-2=(x+\dfrac{1}{4})^2-\dfrac{1^2}{4^2}-2=0\\\\\text{*** simplify ***}\\\\(x+\dfrac{1}{4})^2-\dfrac{1+16*2}{16}=(x+\dfrac{1}{4})^2-\dfrac{33}{16}=0$

$\text{*** add } \dfrac{33}{16} \text{ to both sides ***}\\\\(x+\dfrac{1}{4})^2=\dfrac{33}{16}\\\\\text{**** take the root ***}\\\\x+\dfrac{1}{4}=\pm \dfrac{\sqrt{33}}{4}\\\\\text{*** subtract } \dfrac{1}{4} \text{ from both sides ***}\\\\x = -\dfrac{1}{4} -\dfrac{\sqrt{33}}{4} \ \ or \ \ x = -\dfrac{1}{4} +\dfrac{\sqrt{33}}{4}$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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3 years ago