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2 years ago

What is the vertex of the graph of y=2(x+6)^2-2

Mathematics

2 answers:

daser333 [38]2 years ago

6 0

Answer:

the vertex is 4

Step-by-step explanation:

nexus9112 [7]2 years ago

6 0

Answer:

(-6, -2)

Step-by-step explanation:

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Find simple interest amount for 1000 for one year at 6%

DIA [1.3K]

Answer:

The ans is in the picture with the steps how i got it

(hope this helps can i plz have brainlist :D hehe)

Step-by-step explanation:

8 0

2 years ago

Pls help stuck on big ideas math

beks73 [17]

Answer:

x = 15

Step-by-step explanation:

Substitute: -4/5x + 7 = -5

Rearrange unknown terms to the left side of the equation: -4/5x = -5 - 7

Calculate the sum or difference: -4/5x = -12

Divide both sides of the equation by the coefficient of variable: x = -12 × (-5/4)

Remove the parentheses: x = 12 × 5/4

Cross out the common factor: x = 3 × 5

Calculate the product or quotient: x = 15

So the answer is: x = 15

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2 years ago

The stem-and-leaf plot lists the number of coupons saved per person in a coupon club.

navik [9.2K]

Answer:

Your answers 5 :)

Step-by-step explanation:

I took the test and i did it in my head ignore the other person, thanks please mark brainliest!

6 0

3 years ago

Read 2 more answers

One half of the sum of 3 and q

ivann1987 [24]

Here you go let me know if you have questions

7 0

3 years ago

Find+the+positive+value+for+α+if+the+radius+of+the+circle+3x^2+3y-6αx+12y-3α=0+is+4

Alik [6]

Answer:

$\alpha=3$

Step-by-step explanation:

<u>Equation of a Circle</u>

A circle of radius r and centered on the point (h,k) can be expressed by the equation

$(x-h)^2+(y-k)^2=r^2$

We are given the equation of a circle as

$3x^2+3y^2-6\alpha x+12y-3\alpha=0$

Note we have corrected it by adding the square to the y. Simplify by 3

$x^2+y^2-2\alpha x+4y-\alpha=0$

Complete squares and rearrange:

$x^2-2\alpha x+y^2+4y=\alpha$

$x^2-2\alpha x+\alpha^2+y^2+4y+4=\alpha+\alpha^2+4$

$(x-\alpha)^2+(y+2)^2=r^2$

We can see that, if r=4, then

$\alpha+\alpha^2+4=16$

Or, equivalently

$\alpha^2+\alpha-12=0$

There are two solutions for $\alpha$ :

$\alpha=-4,\ \alpha=3$

Keeping the positive solution, as required:

$\boxed{\alpha=3}$

8 0

2 years ago