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1 year ago

6

PLEASE HELP WILL GIVE BRANILIEST

1 answer:

masya89 [10]1 year ago

8 0

Answer:

It is a function because the x values don't repeat

Step-by-step explanation:

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What is 6900g converted to kg???

stich3 [128]

1000 g=1kg
to convert to kg
1kg/1000g=conversion factor
multiply
6900g times 1kg/1000g
grams cancel
6.9 kg

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2 years ago

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Which of the following is not true for the equation y= x+4?

Natalka [10]

The answer is B.
Hope that helped! :)

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2 years ago

Use the elimination method to solve the system of equations. What is the correct ordered pair. Explain.

natta225 [31]

3y = x - 1...I am going to re-arrange this.....-x + 3y = -1

-x + 3y = -1
x - 2y = 2
---------------add...u will notice that the x's cancel
y = 1

3y = x - 1
3(1) = x - 1
3 = x - 1
3 + 1 = x
4 = x

solution is (4,1)

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2 years ago

Solving Rational equations. LCD method. Show work. Image attached.

posledela

$(k-2)(k-6)=k^2-2k-6k+12=k^2-8k+12$

So in order to get all the fractions to have a common denominator, we need to multiply $\dfrac k{k-2}$ by $\dfrac{k-6}{k-6}$ , and $\dfrac1{k-6}$ by $\dfrac{k-2}{k-2}$ :

$\dfrac k{k-2}\cdot\dfrac{k-6}{k-6}=\dfrac{k(k-6)}{(k-2)(k-6)}=\dfrac{k^2-6k}{k^2-8k+12}$

$\dfrac1{k-6}\cdot\dfrac{k-2}{k-2}=\dfrac{k-2}{(k-2)(k-6)}=\dfrac{k-2}{k^2-8k+12}$

Now,

$\dfrac4{k^2-8k+12}=\dfrac{(k^2-6k)+(k-2)}{k^2-8k+12}$

As long as $k\neq2$ and $k\neq6$ (which we can't have because otherwise $k^2-8k+12=0$ ), we can cancel $k^2-8k+12$ in the denominators on both sides:

$4=(k^2-6k)+(k-2)$

$4=k^2-5k-2$

$0=k^2-5k-6$

We can factorize the right side:

$0=(k-6)(k+1)$

which tells us that $k=6$ and $k=-1$ are solutions.

4 0

3 years ago

What is x2+2x-8=0 by factoring ?

Vaselesa [24]

Answer:

(x+4)(x-2)

Step-by-step explanation:

We need to find two numbers that multiply to -8 and add up to 2. These are 4,-2.

Rewrite the polynomial:

x^2-2x+4x-8

and factor each pair:

x(x-2)+4(x-2) = (x+4)(x-2)

7 0

3 years ago