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2 years ago

8

What is the value of x? Enter your answer in the box.

2 answers:

natka813 [3]2 years ago

4 0

The answer is 80 degrees
70+30 =100
180-100=80

Katyanochek1 [597]2 years ago

3 0

The value of x would have to be 90 degrees as it is a right angle.

Thank you,

Jolly Santa

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9 = 3(5t -2) help pls

horsena [70]

Answer:

t=1

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

4 0

2 years ago

Estimate the value of √2π/√5

Luba_88 [7]

Answer: 1.12099824328 or estimated is 1

6 0

3 years ago

URGENT!! ILL GIVE<br> BRAINLIEST!!!! AND 100 POINTS!!!!!

Naddik [55]

The value of angle x is 113 degrees and y is 67 degrees.

What is angle?

In euclidean geometry, an angle is that the figure shaped by 2 rays, referred to as the perimeters of the angle, sharing a typical end point, referred to as the vertex of the angle. Angles shaped by 2 rays exist the plane that contains the rays.

Main body:

Given ;

∠C = 62°

∠A =51°

according to triangle sum property: Sum of angles of triangle is 180°

Using it,

62° +51° +∠y = 180°

113°+ ∠y =180°

∠y = 67°

∠x is equal to the sum of interior opposite angle ,so using it we get:

∠x = 61°+52°

∠x = 113°

Hence the value of ∠x is 113 degrees and y is 67 degrees.

To know more about angle , visit:

brainly.com/question/25770607

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4 0

1 year ago

Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

so

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

<em>Find the values of y</em>

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

2 years ago

Solve: 3y+5=-1 . What’s the solution?

Harman [31]

Answer:

hmmm

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers