What are the real solutions of x squared =225

Mathematics

1 answer:

Gnesinka [82]2 years ago

3 0

The real solutions the equation as given in the task content; x² = 225 are; +25 and -25.

<h3>What are the real solutions of the equation as given in the task content?</h3>

It follows from the task content that the real solutions of the equation as given in the task content can be determined as follows;

x² = 225

x = ± 15

Therefore, the real solutions of the equation are; +25 and -25.

Read more on real solutions of equations;

brainly.com/question/3122484

#SPJ1

You might be interested in

Angle G and H are vertical angles. If angle G is 45 degrees and angle H is 2x degrees, what is the value of x?

Veseljchak [2.6K]

Answer:

x = 22.5

Step-by-step explanation:

Vertical angles are congruent by the vertical angles theorem. Therefore is angle G is 45 degrees, then that means that angle H must be 45 degrees as well.

The value of x can be found by making the expression that represents angle H (2x) equal to 45 in an equation to solve for x.

2x = 45

Divide both sides by 2.

x = 22.5

6 0

3 years ago

Find the value of x+10=11

drek231 [11]

Answer:

x=1

Step-by-step explanation:

×+10=11

-10 -10

x = 1

8 0

3 years ago

Sofia hits a softball straight up. the equation h= -16t^2 + 90t models the height h, in feet, of the ball after t seconds. How l

andrezito [222]

It is in the air 5.625 seconds.

Using the quadratic formula,

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
\\
\\=\frac{-90\pm \sqrt{90^2-4(-16)(0)}}{2(-16)}
\\
\\=\frac{-90\pm \sqrt{8100-0}}{-32}
\\
\\=\frac{-90\pm \sqrt{8100}}{-32}=\frac{-90\pm90}{-32}
\\
\\=\frac{-90+90}{-32}\text{ or }\frac{-90-90}{-32}=\frac{0}{32}\text{ or }\frac{-180}{-32}
\\
\\=0\text{ or }5.625$

0 is when the ball is first launched, 5.625 is when it hits the ground again.

4 0

3 years ago

Find the volume of a cone of radius 49cm and height 10cm

den301095 [7]

Answer: