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Amanda [17]
3 years ago
5

I need help plzzzz!!!!!!!

Mathematics
1 answer:
svp [43]3 years ago
3 0

Answer:

-8x² - 6

Step-by-step explanation:

-1 + 2 will give us "+1"

+1 - 7 will give us "-6"

The answers given already are correct as they are properly matched, so we are left with the following to complete:

Thus,

-4y³ + 4y³ will give us "0".

We would be left with:

-3x² - 6 - 5x²

Collect like terms again

-3x² - 5x² - 6

-3x² - 5x² will give us "-8x²"

Final result would be:

-8x² - 6

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If AB vector =(4 5) and PQ vector=(2p 10) are parallel to each other, find the value of p.​
Sedaia [141]

Answer:

Answer:

They both have q+3/2p, so that means that 2PQ=CB and that means they are parallel to each other

Step-by-step explanation:

PQ=PA+QA

PQ=1/2(2q-p)+2/5*5p=q-1/2p+2p=q+3/2p

CB=2q+3p=2(q+3/2p)

Other explanation: It should be written like this PQ=q+3/2p and CB=2q+3p=2(q+3/2p) they are parallel bcs CB=2*PQ.

8 0
3 years ago
Determine the value of r so that a line through the points (r,3) and (5,-4) has a slope of -1/2
CaHeK987 [17]

\bf (\stackrel{x_1}{r}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{-4}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-4-3}{5-r}=\stackrel{\stackrel{slope}{\downarrow }}{-\cfrac{1}{2}}\implies \cfrac{-7}{5-r}=\cfrac{1}{-2} \\\\\\ 14=5-r\implies 9=r

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3 0
3 years ago
A company claims its video game players have a defective rate of 4%. What is the probability that you and a friend both receive
alex41 [277]
Since you're looking for the chance that the defective player occurs twice, you need to find the chance your friend receives a defective player given that you also receive one. The chance you receive a defective player is 4%, or 0.04. If you friend also receives a defective player, then the chance of both occurring is 4% of 4%, or 0.04 * 0.04, which equals 0.0016. So the probability that you can a friend both receive a defective player is 0.16%.
5 0
4 years ago
Let p=x^2+6
stepan [7]
That's a bit of a nasty. Where ever you see (x^2 + 6) you put in p.
p^2  - 21 = 4(x^2 + 6)
p^2 - 21 = 4p
p^2 - 4p - 21 = 0

A <<<< answer. 

8 0
4 years ago
Ex 2.8<br> 3. find the maximum value of y for the curve y=x^5 -3 for -2≤x≤1
harkovskaia [24]
y=x^5-3\\ y'=5x^4\\\\ 5x^4=0\\ x=0\\ 0\in [-2,1]\\\\ y''=20x^3\\\\&#10;y''(0)=20\cdot0^3=0

The value of the second derivative for x=0 is neither positive nor negative, so you can't tell whether this point is a minimum or a maximum. You need to check the values of the first derivative around the point.
But the value of 5x^4 is always positive for x\in\mathbb{R}\setminus \{0\}. That means at x=0 there's neither minimum nor maximum.
The maximum must be then at either of the endpoints of the interval [-2,1].
The function y is increasing in its entire domain, so the maximum value is at the right endpoint of the interval.

y_{max}=y(1)=1^5-3=-2
4 0
3 years ago
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