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nalin [4]
3 years ago
8

What is the constant of proportionality in the equation y = 5x? ​

Mathematics
1 answer:
Yuki888 [10]3 years ago
4 0

Answer:

no constants

Step-by-step explanation:

a constant for example is 5, when the expression is 5+2x=20

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How many one degree ngles are in a 70 degree angle
velikii [3]

Answer:

um, seventy one-degree-angles would be in a seventy-degree angle...

Step-by-step explanation:

Are you being serious about this question?

I assume the answer to be pretty self-explanatory.

8 0
2 years ago
What's 0.003 1/10 of
andrew-mc [135]
Simple. 
Just multiply .003 by 10.

0.003 x 10 = 0.00030, or 0.0003.

0.0003 is 1/10 of 0.003
3 0
3 years ago
Find the value of x. Round to the nearest tenth.
emmasim [6.3K]

Answer:

x = 8.5

Step-by-step explanation:

\frac{sin28^{o} }{x} =\frac{sin83^{o} }{18}

x=\frac{18sin28^{o} }{sin83^{o} } =8.513

Rounded x = 8.5

Hope this helps

5 0
2 years ago
Andre is making paper cranes to decorate for a party. He plans to make one large papercrane for a centerpiece and several smalle
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3 0
3 years ago
If x^2 +x-12 is a factor of x^3+ax^2-10x-b then find the values of a and b
Marysya12 [62]

Answer:

a=3

b=24

Step-by-step explanation:

If x^2+x-12 is a factor of x^3+ax^2-10x-b, then the factors of x^2+x-12 must also be factors of x^3+ax^2-10x-b.

So what are the factors of x^2+x-12?  Well the cool thing here is the coefficient of x^2[tex] is 1 so all we have to look for are two numbers that multiply to be -12 and add to be positive 1 which in this case is 4 and -3.-12=4(-3) while 1=4+(-3).So the factored form of [tex]x^2+x-12 is (x+4)(x-3).

The zeros of x^2+x-12 are therefore x=-4 and x=3.  We know those are zeros of x^2+x-12 by the factor theorem.  

So x=-4 and x=3 are also zeros of x^3+ax^2-10x-b because we were told that x^2+x-12 was a factor of it.

This means that when we plug in -4, the result will be 0. It also means when we plug in 3, the result will be 0.

Let's do that.

(-4)^3+a(-4)^2-10(-4)-b=0  Equation 1.

(3)^3+a(3)^2-10(3)-b=0  Equation 2.

Let's simplify Equation 1 a little bit:

(-4)^3+a(-4)^2-10(-4)-b=0

-64+16a+40-b=0

-24+16a-b=0

16a-b=24

Let's simplify Equation 2 a little bit:

(3)^3+a(3)^2-10(3)-b=0

27+9a-30-b=0

-3+9a-b=0

9a-b=3

So we have a system of equations to solve:

16a-b=24

9a-b=3

---------- This is setup for elimination because the b's are the same. Let's subtract the equations.

16a-b=24

9a-b=   3

------------------Subtracting now!

7a    =21

Divide both sides by 7:

 a   =3

Now use one the equations with a=3 to find b.

How about 9a-b=3 with a=3.

So plug in 3 for a.

9a-b=3

9(3)-b=3

27-b=3

Subtract 27 on both sides:

   -b=-24

Multiply both sides by -1:

    b=24

So a=3 and b=24

7 0
4 years ago
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