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3 years ago

Let p = x2 + 6. Which equation is equivalent to (22 + 6)2 – 21 = 4x2 + 24 in terms of p?

Mathematics

1 answer:

Alona [7]3 years ago

4 0

Answer:

$\frac{35}{4} =p$

Step-by-step explanation:

p= $x^{2} +6$ ,

(22+6)2-21= $4x^{2} +24$

44+12-21= $4x^{2} +24$

35= $4x^{2} +24$ ,

If you factor out the equation, you get that 35=4p(if you factor it out)

Therefore, $\frac{35}{4} =p$

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Answer:

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Step-by-step explanation:

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2 years ago

Solve these two equations using the Addition Method (also known as the elimination method).

Brums [2.3K]

Answer:

(-2, 3)

Step-by-step explanation:

5/2(3/4x + 1/3y = -1/2)

1/2x - 5/6y = -7/2

15/8x + 5/6y = -5/4

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4 0

3 years ago

Which number is a prime 49 51 53 55

shtirl [24]

53 is a prime number !!

3 0

3 years ago

Read 2 more answers

can anyone help me its urgent. i have to get it done with the right answer by 3:25. can anybody help me and dont post saying dum

solmaris [256]

Answer:

Step-by-step explanation:

All pieces of data for seventh grade classes are larger than that of any kindergarten class on the table. This also proves with the mean provided.

The MAD, mean absolute deviation, helps you to identify the variation. The MAD for seventh grade was significantly larger than that of the kinder classes. The range for both sets of data is two times one another, further proving that 7th grade classes varies more.

C and D are irrelevant, and B says "varies less" which is the opposite of what's going on.

Hope that helps!

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3 years ago

Read 2 more answers

I really really really need help!!!!

Yakvenalex [24]

For $f$ to be continuous at $x=1$ , you need to have the limit from either side as $x\to1$ to be the same.

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2$
$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b$

If $a=2$ and $b=3$ , then the limit from the right would be $2+3=5\neq2$ , so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation $a+b=2$ .

Part (c) is done similarly to part (1). This time, you need to limits from either side as $x\to2$ to match. You have

$\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b$
$\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0$

So, $a$ and $b$ have to satisfy the relation $4a+2b=0$ , or $2a+b=0$ .

Part (4) is done by solving the system of equations above for $a$ and $b$ . I'll leave that to you, as well as part (5) since that's just drawing your findings.

8 0

2 years ago