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3 years ago

What is 1+2= because I don't really know

Mathematics

2 answers:

Black_prince [1.1K]3 years ago

4 0

1+2=3. So your answer is 3.

sweet-ann [11.9K]3 years ago

3 0

The answer is 3, and now you know

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Researchers wanted to determine if having a DVD player in the bedroom is associated with obesity. The researchers administered a

Mashutka [201]

Answer:

is c

Step-by-step explanation:

3 0

3 years ago

The height of a trapezoid is 8 in. And it's area is 96 squared inch. One base of the trapezoid is 6 inches longer than the other

MaRussiya [10]

we name bases k , k+6. so

area: 96= (k + k+6).8/2

=> 2k+6= 24 => k=9

7 0

2 years ago

If (x,y) is a solution to the system of equations shown below, what is the product of the y-coordinates of the solutions? x^2+4y

aliina [53]

Answer:

The product of the y-coordinates of the solutions is equal to 3

Step-by-step explanation:

we have

$x^{2}+4y^{2}=40$ -----> equation A

$x+2y=8$ ------> equation B

Solve by graphing

Remember that the solutions of the system of equations are the intersection point both graphs

using a graphing tool

The solutions are the points (2,3) and (6,1)

see the attached figure

The y-coordinates of the solutions are 3 and 1

therefore

The product of the y-coordinates of the solutions is equal to

(3)(1)=3

4 0

2 years ago

Please help me- AND PLEASE EXPLAIN

GuDViN [60]

Answer:

A=35, B=55, and C=25

Step-by-step explanation:

Angle A and the angle next to it are linear pairs, so you can subtract 145 from 180.

if you already know Angle A, and that the other angle is 90 degress, you can add 90+35 and subtract it from 180 to get Angle B

Angle C is paired with the angle across from it (I forgot what the term is called) so it is 25 degrees.

7 0

2 years ago

An object is launched from ground level with an initial velocity of 120 meters per second. For how long is the object at or abov

babunello [35]

Answer:

D) 14 seconds

Step-by-step explanation:

First we will plug 500 in for y:

500 = -4.9t² + 120t

We want to set this equal to 0 in order to solve it; to do this, subtract 500 from each side:

500-500 = -4.9t² + 120t - 500

0 = -4.9t²+120t-500

Our values for a, b and c are:

a = -4.9; b = 120; c = -500

We will use the quadratic formula to solve this. This will give us the two times that the object is at exactly 500 meters. The difference between these two times will tell us when the object is at or above 500 meters.

The quadratic formula is:

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

Using our values for a, b and c,

$x=\frac{-120\pm \sqrt{120^2-4(-4.9)(-500)}}{2(-4.9)}\\\\=\frac{-120\pm \sqrt{14400-9800}}{-9.8}\\\\=\frac{-120\pm \sqrt{4600}}{-9.8}\\\\=\frac{-120\pm 67.8233}{-9.8}\\\\=\frac{-120+67.8233}{-9.8}\text{ or }\frac{-120-67.8233}{-9.8}\\\\=\frac{-57.1767}{-9.8}\text{ or }\frac{-187.8233}{-9.8}\\\\=5.3242\text{ or }19.1656$

The two times the object is at exactly 500 meters above the ground are at 5 seconds and 19 seconds. This means the amount of time it is at or above 500 meters is

19-5 = 14 seconds.

6 0

2 years ago