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jasenka [17]
3 years ago
15

Complete the standard form of the equation -4y=3-2y.

Mathematics
1 answer:
timama [110]3 years ago
8 0
-4y = 3 - 2y
+2y = +2y
----------------
-2y = 3
---- ---
-2 -2

Y= -1.5

So your goal is to get x all alone
To achieve this you would have to add 2y to both sides to become -2y = 3
Then you would have to divide both sides by -2 (so y is all alone)
When you divide it it comes out to be y = -1.5
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If b=p+5 and a=p^2+10p+1 find an expression that equals 2b+3a in standard form
olya-2409 [2.1K]

Answer:

3p² + 32p + 13

Step-by-step explanation:

Okay, so lets first solve for 2b. 2b = 2(p + 5), which is equal to 2p + 10. Now, let's solve for 3a. 3a = 3(p² + 10p +1), simplifying to 3p² + 30p +3. After adding 2b and 3a, we are able to get 2p + 10 + 3p² + 30p + 3 = 3p² + 32p + 13

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2 years ago
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Step-by-step explanation:

2 + 3 × 6 + 4 × 3

Here, Parantheses should be put,

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2 years ago
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8 0
3 years ago
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The dance committee at Jefferson High School decides to charge students different prices for dance tickets depending on whether
valentina_108 [34]

Answer:

30

Step-by-step explanation:

4 0
3 years ago
Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
kherson [118]

Answer:

The answer is

f(x) = {\displaystyle  8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }

Step-by-step explanation:

Remember that Taylor says that

f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }

For this case

f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18

f(x) = {\displaystyle  8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }

5 0
3 years ago
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