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2 years ago

Find the slope of the line through the given points

1 answer:

7 0

I can’t solve all of them but maybe one and I will give you the formula for you to solve the rest.
Slope = (y2 -y1) / (x2 -x1).
m= ( 16 -(-3)) / ( -14 -(-9))
m = (16+3)/(-14+9)
m = 19/ -5

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3.) (4 x 9 - 21) \ 3+6^2

nignag [31]

Answer:

Step-by-step explanation:

Given Expression:

(4 x 9 - 21) \ 3+6^2

Multiply 4 and 9 to get 36.

36 - 21 / 3 + 6^2

Subtract 21 from 36 to get 15.

15/3 + 6^2

Divide 15 by 3 to get 5.

5+6^2

Calculate 6 to the power of 2 and get 36.

5+36

Add 5 and 36 to get 41.

3 0

3 years ago

There are 30boys in a sporting club 20of them play hockey and 15 play volley ball each boy plays atleast one of the true game.il

Fed [463]

5 boys play both hockey and volleyball (30-20-15=5)

7 0

2 years ago

Solve for y please help

ss7ja [257]

Answer:

The answer is (-16.6)

Step-by-step explanation:

y/2 ≥ -8.3

y ≥ (-8.3) × 2

y ≥ -16.6

Thus, The value of y is (-16.6)

<u>-TheUnknownScientist 72</u>

7 0

2 years ago

Graph 25 = -5y -3x using the slope and y-intercept

wlad13 [49]

Step-by-step explanation:

you move -5y to the other side.

25=-5y-3xsubtract 25 both sides

0=-5y-3x-25then add 5y both sides

5y=-3x-25divide 5 both sides

y=-3/5x-5

m=-3/5 do rise over runrise/run

b=(0,-5)plot it

3 down and 5 to right or 3 up and 5 left

8 0

3 years ago

Find the complex zeros of x^3+27. write f in factored form

aliina [53]

$x^3+27=0
\\x^3+3^3=0
\\(x+3)(x^2-3x+9)=0
\\x^2-3x+9=0
\\a=1,b=-3,c=9
\\
\\x_{1,2}= \frac{-b\pm \sqrt{b^2-4ac} }{2a} = \frac{-(-3)\pm \sqrt{(-3)^2-4\times1\times9} }{2\times1} =\frac{3\pm \sqrt{-27} }{2} =\frac{3\pm 3\sqrt{3}i }{2} 
\\
\\(x+3)(x-\frac{3+ 3\sqrt{3}i }{2})(x-\frac{3- 3\sqrt{3}i }{2})=0

$

4 0

3 years ago