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Harrizon [31]
3 years ago
8

tex-formula">
Mathematics
2 answers:
Leviafan [203]3 years ago
8 0
N = 14
I think this is what you wanted. I hope this helps :)
enyata [817]3 years ago
5 0

Answer:

n=14

Step-by-step explanation:

12.5=n-1.5

-n=-1.5-12.5

-n=-14

n=14

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Graph the functions on the same coordinate plane.
beks73 [17]

Answer:

select -3 and 2

Step-by-step explanation:

f(x)=g(x) means set them equal and solve for x

x^2+2x-8=-x^2+4

Add x^2 on both sides and subtract 4 on both sides.

2x^2+2x-12=0

Divide both sides by 2

x^2+x-6=0

Think of two numbers that multiply to be -6 and add up to be 1

that is 3 and -2

so The factored form is (x+3)(x-2)=0

So x=-3 or x=2

7 0
4 years ago
HELPPPP <br> this is due today <br> WITH THE EQUATION
timofeeve [1]
60% more than regular size. 230.4.
5 0
3 years ago
SOMEONE PLEASE HELP ME!!!!! URGENT
IgorLugansk [536]

Answer:

AB=14

Step-by-step explanation:

Since there are 360° in a circle we can do 140°/360°=7/18

To find AB we can do (7/18)*(6^2)=(7/18)*(36)=14

AB=14

8 0
4 years ago
Using principle of mathematical induction prove that 6^-1 divisble by 5 .​
salantis [7]

I suppose the claim is 5 \mid 6^n - 1 for n\in\Bbb N.

When n=1, we have 6^1 - 1 = 6 - 1 = 5, and of course 5 divides 5.

Assume the claim holds for n=k, that 5 \mid 6^k - 1. We want to use this to show it holds for n=k+1, that 5 \mid 6^{k+1} - 1.

We have

6^{k+1} - 1 = \left(6^{k+1} - 6\right) + \left(6 - 1) = 6\left(6^k - 1\right) + 5

Since 5 \mid 6^k - 1, we can write 6^k - 1 = 5\ell for some integer \ell. Then

6^{k+1} - 1 = 6\cdot5\ell + 5 = 5(6\ell + 1)

which is clearly divisible by 5. QED

5 0
2 years ago
Please help.........
anygoal [31]
-\dfrac{ax}{b}\geq c-d\ \ \ |\text{change the signs}\\\\\dfrac{ax}{b}\leq d-c\ \ \ |\cdot b\\\\ax\leq bd-bc\ \ \ \ |:a\\\\\boxed{x\leq\dfrac{bd-bc}{a}}
3 0
3 years ago
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