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mart [117]
3 years ago
12

What is the vertex of y=(x-2)-7

Mathematics
1 answer:
Zina [86]3 years ago
7 0

Answer:

The vertex of this equation is (2, -7)

Step-by-step explanation:

In order to find the vertex of this equation we start with the base form of the vertex form.

y = a(x - h) + k

With this equation (h, k) is the vertex. You can see that 2 lines up with h and -7 lines up with k. This shows that (2, -7) is the vertex.

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Please help! Will give brainliest! Trig ratio!
DiKsa [7]
Good question!

First of all, you need to be aware of the following trigonometrical ratios\functions:

For angle A:

adjacent=12
opposite=35
hypotenuse=37

Hence:
cos(A)= \frac{12}{37} 
\\~\\
sin(A)= \frac{35}{37} 
\\~\\
tan(A)= \frac{35}{12}

I hope that helps!



I am with you if you faced any difficulties!


 


4 0
3 years ago
How to write y-1=-x-6 In slope intercept form
DiKsa [7]

y=-x-5. you just need to add 1 to another side to get rid of 1 from the left side

4 0
3 years ago
Read 2 more answers
Write the equation of a line in slope intercept form given;<br> (6,-3), (0, 5)
olganol [36]

Answer:

y=(-4/3)x+5

Step-by-step explanation:

so you want to start by using y=mx+b

step one: find the slope (remember m is the slope in this)

m= \frac{y2-y1}{x2-x1}

m= \frac{5+3}{0-6}

m= \frac{8}{-6}

m= - \frac{4}{3\\}

step two: insert the value we found for m into the equation and take one of the given points and insert the x and y into the equation (im using (0,5)) and solve for b.

y=mx+b

5= -4/3(0)+b

5=b

step three: rewrite y=mx+b with the values we found for m and for b

y=(-4/3)x+5

3 0
2 years ago
I need help solving systems of linear equations by substitution. <br> {4x+3y=14<br> 3x-2y=2
Sauron [17]

Answer:

A) 4x+3y=14

B) 3x-2y=2

We multiply equation A) by (2/3)

A) (8/3) x +2y = 28/3 then we add this to equation B)

B) 3x-2y=2

5 (2/3) x = 11 (1/3)

x = 2

A) 4 * 2 + 3 y = 14

A) 3y = 6

y = 2

Step-by-step explanation:

4 0
3 years ago
Help me out here pleaseeeeeeeeee
patriot [66]
Just multiply 1/4 times 6
5 0
2 years ago
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