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dusya [7]
3 years ago
6

Add. (1+7x3−x2)+(x3+2+2x2) Express the answer in standard form. Enter your answer in the box.

Mathematics
2 answers:
Elis [28]3 years ago
8 0

Answer:

8x^3+x^2+3

Step-by-step explanation:

In order to solve this you just have to put together the numbers witht the same base and exponent and then you just add and withdraw, and you can do it like this:

""7x^3-x^2+1\\+x^3+2x^2+2\\\\------\\8x^3+ x^2+3

So the answer would be :8x^3+x^2+3

Juli2301 [7.4K]3 years ago
6 0
The answer to the question

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Beatrice is planning a personal budget. Which part of her budget will be easiest to change?
Lerok [7]

Answer: It’s C

Step-by-step explanation: It is easiest to change what wants you want to spend your money on. It is hard to ask for a raise or income change. And spending on needs might not change, same goes for savings.

4 0
3 years ago
I have 5 digits my 9 is worth 9 times 10,000 my 2 is worth 2 thousand One of my 7s is worth 70 the other one is worth 10 times a
omeli [17]
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5 0
3 years ago
Two functions are shown below
photoshop1234 [79]
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3 years ago
The expression (x+j)(x-j)(x-k) can be rewritten as x^3-5x^2-4x+t, where j,k, and t are constants. Which of the following is the
kirza4 [7]

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<u><em>Explanation</em></u>

The expression (x+j)(x-j)(x-k) can be rewritten as x^3-5x^2-4x+t

That means:   (x+j)(x-j)(x-k)= x^3-5x^2-4x+t

Now simplifying the left side, we will get....

(x+j)(x-j)(x-k)= x^3-5x^2-4x+t\\ \\ (x^2-j^2)(x-k)= x^3-5x^2-4x+t\\ \\ x^3-kx^2-j^2x+j^2k= x^3-5x^2-4x+t

Now <u>comparing the co-effcients of like terms in left and right side</u>, we will get.....

k=5, j^2=4 \\ \\ and\\ \\ t=j^2k\\ \\ So, t= (4)(5) =20

Thus, the value of 't' will be 20.


8 0
3 years ago
Please help to find the HCF and LCM of this question ​
poizon [28]

Step-by-step explanation:

Explanation is in the attachment

<em>H</em><em>ope </em><em>it</em><em> is</em><em> helpful</em><em> to</em><em> you</em>

<em> </em><em> </em><em>✌️</em><em>✌️</em><em>✌️</em><em>✌️</em><em>✌️</em><em>✌️</em><em>✌️</em>

4 0
3 years ago
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