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bija089 [108]
3 years ago
5

Question 11 and 12 please help

Mathematics
1 answer:
Maksim231197 [3]3 years ago
7 0

Answer:

11. 10

12. 13

Step-by-step explanation:

Use the distance formula.

11.

(3, 4) and (-3, -4)

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

d = \sqrt{(-3 - 3)^2 + (-4 - 4)^2}

d = \sqrt{(-6)^2 + (-8)^2}

d = \sqrt{36 + 64}

d = \sqrt{100}

d = 10

12.

(0, 0) and (-5, -12)

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

d = \sqrt{(-5 - 0)^2 + (-12 - 0)^2}

d = \sqrt{(-5)^2 + (-12)^2}

d = \sqrt{25 + 144}

d = \sqrt{169}

d = 13

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Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{x =  - 4}}}}}

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\sf{3x - 5 = 2x - 9}

Move 2x to left hand side and change it's sign

⇒\sf{3x - 2x - 5 =  - 9}

Move 5 to right hand side and change it's sign

⇒\sf{3x - 2x =  - 9 + 5}

Collect like terms

⇒\sf{x =  - 9 + 5}

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Graph f(x) = x2 + 2x - 3, label the function’s x-intercepts, y-intercept and vertex with their coordinates. Also draw in and lab
Semenov [28]
<h2>Answer with explanation:</h2>

The given function : f(x)=x^2+2x-3

Using completing the squares, we have

f(x)=x^2+2x+1-1-3        [∵ (x+1)^2=x^2+2x+1]

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Comparing (1) to the standard vertex form f(x)=(x-h)^2+k , the vertex of function is at (h,k)=(-1,-4)

For x-intercept, put f(x)=0 in (1), we get

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Square root on both sides, we get

x+1=\pm2\\\\x+1=-2\ or\ \ x+1=2\\\\=x=-3\ \ or\ x=1

∴ x intercepts : x= (-3,0) and (1,0)

For y-intercept put x=0 in (1), we get

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Please help me with this!!<br> Thank u
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Answer:

Step-by-step explanation:

Left

When a square = a linear, always expand the squared expression.

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x^2 - 2x - 3x + 1 = -5

x^2 - 5x +1 = - 5                      Add 5 to both sides

x^2 - 5x + 1 + 5 = -5 + 5

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28

Second from the Right

This one is rather long. I'll get you the equations, you can solve for a and b. Maybe not as long as I think.

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<u>17 = 12a + b         Subtract</u>

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Right

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(2x - 2)(2x - 3)     This is correct.

So the answer is D

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