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Dmitrij [34]
3 years ago
12

Simplify (5x^3 - 7x^2 + 3x - 4) - (8x^3 + 2x^2 + 3x - 7)

Mathematics
1 answer:
muminat3 years ago
6 0

Answer:

-3x^4 - 14x^3 + 3x^2 - 10

Step-by-step explanation:

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Good x sold for $40 in 1945. the Cpi in 1945 was 18.0 and the cpi in 1999 was 166.6. what was the price of good x in 1999 dollar
andrey2020 [161]

Answer:

the price of good x in 1999 dollars is 370.89 dollars

Step-by-step explanation:

Given that good x sold for $40 in 1945. the Cpi in 1945 was 18.0 and the cpi in 1999 was 166.6.

We have cpi and sale price have direct variation

In other words S = kC where C = CPi and S = sales price

In 1945, 40 = 18k or K = 20/9

Using this we can say

Sales price in 1999 would be k (166.6)

=\frac{20}{9} (166.6)\\= 370.89

the price of good x in 1999 dollars is 370.89 dollars

5 0
4 years ago
Please help me online school isnt fit for me lol
worty [1.4K]

Answer:

The answer is A!!

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E3_t%7B%7Cx%2B2%7C%7D%20%5C%2C%20dx%20" id="TexFormula1" title=" \int\limi
photoshop1234 [79]
\displaystyle\int_{-3}^3|x+2|\,\mathrm dx

Recall the definition of absolute value:

|x|=\begin{cases}x&\text{for }x\ge0\\-x&\text{for }x

So we can split up the integration interval at x=-2 and apply this definition to rewrite the integral as

\displaystyle\int_{-3}^{-2}(-(x+2))\,\mathrm dx+\int_{-2}^3(x+2)\,\mathrm dx
=\displaystyle-\int_{-3}^{-2}(x+2)\,\mathrm dx+\int_{-2}^3(x+2)\,\mathrm dx
=-\left(\dfrac12x^2+2x\right)\bigg|_{x=-3}^{x=-2}+\left(\dfrac12x^2+2x\right)\bigg|_{x=-2}^{x=3}
=\dfrac12+\dfrac{25}2=13
7 0
3 years ago
It's my last question just nee help because I'm dumb<br>​
inna [77]
Answer: C

4x7*2-25

4x49-25

196-25

= 171

3 0
4 years ago
Read 2 more answers
What is the value of ( h o k ) (2)<br><br>a -5<br><br>b 1<br><br>c -2<br><br>d -3
Bond [772]

Answer:

-3

Step-by-step explanation:

h(k(2)) = h(k(x=2)) = h(-2) = h(x= -2)= 3/( -2+1) = 3/-1= -3

sorry if I'm wrong.

3 0
3 years ago
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