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3 years ago

Five less than twice what negative number squared is 157?

1 answer:

3 0

Five less than twice what negative number squared is 157 <- put into an expression
-5 + 2(-x^2) = 157
Now simplify:
-5 + -2x^2 = 157
-2x^2 = 162
-2x^2 - 162 = 0
2x^2 + 162 = 0
x^2 + 81 = 0

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Find the slope of the graph

zhuklara [117]

Answer:

Slope = $\frac{3}{2}$

Step-by-step explanation:

Slope = $\frac{6}{4} =\frac{3}{2}$

4 0

3 years ago

If p=(-3,-2) and q=(1,6) are the endpoints of the diameter of a circle find the equation of the circle

stira [4]

Answer:

The equation of the circle (x +1) )² +(y-(2))² = (2(√5))²

 or 

The equation of the circle x² + 2 x + y² - 4 y = 15

Step-by-step explanation:

Given points end Points are p(-3,-2) and q( 1,6)

The distance of two points formula

P Q = $\sqrt{x_{2} - x_{1})^{2} + ((y_{2} -y_{1})^{2} }$

P Q = $\sqrt{1 - (-3)^{2} + ((6 -(-2))^{2} }$

P Q = $\sqrt{16+64} = \sqrt{80}$

The diameter 'd' = 2 r

2 r = √80

= $\sqrt{16 X 5}$

= $4 \sqrt{5}$

r = 2√5

Mid-point of two end points

$(\frac{x_{1} + x_{2} }{2} , \frac{y_{1} +y_{2} }{2} ) = (\frac{-3+1}{2} ,\frac{-2 +6}{2} )$

= (-1 ,2)

Mid-point of two end points = center of the circle

(h,k) = (-1 , 2)

The equation of the circle

(x -h )² +(y-k)² = r²

(x -(-1) )² +(y-(2))² = (2(√5))²

x² + 2 x + 1 + y² - 4 y + 4 = 20

x² + 2 x + y² - 4 y = 20 -5

x² + 2 x + y² - 4 y = 15

Final answer:-

The equation of the circle (x +1) )² +(y-(2))² = (2(√5))²

 or 

The equation of the circle x² + 2 x + y² - 4 y = 15

6 0

4 years ago

4c+6+2 if anyone can tell me the answer it would mean so much!

hodyreva [135]

Answer:

Just combine like terms

6+2=8

4c can’t be combines

so 4c+8

3 0

3 years ago

Read 2 more answers

Inez says you can plot 40 on a number line, but Jeremiah says you cannot. Who is correct?

Alborosie

Answer:

I am correct

Step-by-step explanation:

⊕

4 0

3 years ago

What is x and y? 2x-2y=-4 2x+y=11 Is there infinitely many solutions, 1 solution, or no solution? solve by elimination.

Kaylis [27]

Let's solve your system by substitution.

2x−2y=−4;2x+y=11

Rewrite equations:

2x+y=11;2x−2y=−4

Step: Solve2x+y=11for y:

2x+y=11

2x+y+−2x=11+−2x(Add -2x to both sides)

y=−2x+11

Step: Substitute−2x+11foryin2x−2y=−4:

2x−2y=−4

2x−2(−2x+11)=−4

6x−22=−4(Simplify both sides of the equation)

6x−22+22=−4+22(Add 22 to both sides)

6x=18

6x/6 = 18/6

(Divide both sides by 6)

x=3

Step: Substitute3forxiny=−2x+11:

y=−2x+11

y=(−2)(3)+11

y=5(Simplify both sides of the equation)

Answer:

x=3 and y=5

5 0

3 years ago