PLEASE HELP ME BRAINLIEST WILL BE GIVEN!!

Mathematics

1 answer:

hjlf3 years ago

3 0

Answer:

False

Step-by-step explanation:

The people arent random

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Helppppp!?..............

madreJ [45]

Answer:

For number 1: y=2 and x=3

4 0

3 years ago

Question 2 Not yet answered Points out of 2

nata0808 [166]

Answer:

Trapezoid

Step-by-step explanation:

perpendicular: They are perpendicular due to the fact they cross at right angles to each other. Another true instance of perpendicular lines we will see in nature is a football field. All four corners of a football field are perpendicular to every other.

In the trapezoid attached there aren't any perpendicular sides. A and B are both obtuse angles in this diagram, and C and D are both acute angles. However, now not all trapezoids appear like this. ... Definition B: A trapezoid is any quadrilateral with at the least one pair of parallel sides.

please mark me brainliest!

3 0

3 years ago

What are the solutions to the following system of equations? y=x^2+12x+30 and 8x-y=10

Ierofanga [76]

Note that the 2nd equation can be re-written as y=8x-10.
According to the second equation, y=x^2+12x+30.
Equate these two equations to eliminate y:

8x-10 = x^2+12x+30

Group all terms together on the right side. To do this, add -8x+10 to both sides. Then 0 = x^2 +4x +40. You must now solve this quadratic equation for x, if possible. I found that this equation has NO REAL SOLUTIONS, so we must conclude that the given system of equations has NO REAL SOLUTIONS.

If you have a graphing calculator, please graph 8x-10 and x^2+12x+30 on the same screen. You will see two separate graphs that do NOT intersect. This is another way in which to see / conclude that there is NO REAL SOLUTION to this system of equations.

6 0

3 years ago

What two numbers add to be 6 and multiply to be 15

yarga [219]

$\displaystyle \bf \\
x_1+x_2=6\\\\
x_1\cdot x_2=15\\\\
\Longrightarrow~~x^2-6x+15=0\\\\
x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{6\pm\sqrt{36-60}}{2}=\\\\
=\frac{6\pm\sqrt{-24}}{2}=\frac{6\pm2\sqrt{-6} }{2}=3\pm i\sqrt{6}\\\\
x_1=\boxed{3-i\sqrt{6}}\\\\
x_2=\boxed{3+i\sqrt{6}}\\\\
\texttt{Verify:}\\\\
x_1+x_2=3-\underline{i\sqrt{6}}+3+\underline{i\sqrt{6}}=3+3=6~~correct\\\\
x_1\cdot x_2=(3-i\sqrt{6})(3+i\sqrt{6})=3^2-(i\sqrt{6})^2=9+6=15~~correct$

4 0

3 years ago

Linda is planning to build a rectangular deck that is 12 feet longer than it is wide. Find its width if its area is 589 square f

BaLLatris [955]

Let the width be x, then the length is x + 12.
Area = length x width = x(x + 12) = 589
$x^{2} +12x = 589$
Solving the quadratic equation, we have that
x = 19 feet
i.e. the width is 19 feet.

6 0

3 years ago