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diamong [38]
2 years ago
12

Can someone pls help me

Mathematics
2 answers:
Novosadov [1.4K]2 years ago
8 0
9(2n + 1)

Distribute 9 to the terms in the brackets.

18n + 9

Add = 0 to solve for x

18n + 9 = 0

Move 9 to the other side by subtracting

18n = -9

Divide by 18 on both sides

n = -9/18

Simplify

n = -1/2

In decimal form the answer can also be -0.5 depending on your question.

Hope this helped!


anygoal [31]2 years ago
6 0

Answer:

Sorry that I cant help.

But have a good rest of your dayFollow if you liked this.

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Gnom [1K]

That would be 10ft²

Because LxW=A

so we did 5x2=10

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Find a cubic function with the given zeros. (1 point) Square root of seven. , - Square root of seven. , -4
Debora [2.8K]

Given :

Three roots , \sqrt{7},\ -\sqrt{7},\ -4 .

To Find :

A cubic function with the given zeros.

Solution :

Equation of polynomial of 3 zeroes is given by :

(x-a_1)(x-a_2)(x-a_3)=0\\\\(x-\sqrt{7})(x+\sqrt{7})(x+4)=0\\\\( x^2-7)(x+4)=0\\\\x^3+4x^2-7x-28=0

Therefore, the cubic function of given zeroes is x^3+4x^2-7x-28=0.

Hence, this is the required solution.

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3 years ago
Find the value of each variable.<br> 105°<br> 86°<br> 106°/2°
forsale [732]

Answer:

z = 74 degrees

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5 0
2 years ago
ANSWER FAST!!!! What is the surface area of the following solid figure?
Charra [1.4K]
42in.^2 should be the answer that you are looking for. Hope this helps
6 0
3 years ago
Read 2 more answers
2. Determine the sum of the first 400 ODD numbers.<br><br>​
il63 [147K]

Odd numbers take the form 2n-1, where n\ge1 is an integer. When n=400, the last odd number would be 799. So we're adding

S=1+3+5+\cdots+795+797+799

By reversing the order of terms, we have

S=799+797+795+\cdots+5+3+1

and we can pair up terms in both sums at the same position to write

2S=(1+799)+(3+797)+(5+795)+\cdots(795+5)+(797+3)+(799+1)

so that we are basically adding 400 copies of 800, and from there we can find the value of the sum right away:

2S=400\cdot800\implies S=160,000

###

We could also make use of the formulas,

\displaystyle\sum_{i=1}^n1=n

\displaystyle\sum_{i=1}^ni=\dfrac{n(n+1)}2

We have

S=\displaystyle\sum_{i=1}^{400}(2i-1)=2\sum_{i=1}^{400}i-\sum_{i=1}^{400}1=400(400+1)-400=400^2=160,000

3 0
3 years ago
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