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2 years ago

What are these questions

Mathematics

1 answer:

alina1380 [7]2 years ago

3 0

Answer:

Theres quadrants so -3,9 would be lower about III

6,-6 would be IV

0,-1 1/2 That would be near the 0 obviously so II

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When 0.5 and the number n are multiplied the result is 2.5 what is the value of n

tresset_1 [31]

first form an equation to solve the problem

<em>bring 0.5 over to the other side and divide it by 2.5</em>

<em>the answer is </em>

<em>so,</em>

hope that helps :)

8 0

3 years ago

If ef bisects angle ceb,angle cef=7x+31 and angle feb=10x-3

Nastasia [14]

Given : Angle < CEB is bisected by EF.

< CEF = 7x +31.

< FEB = 10x-3.

We need to find the values of x and measure of < FEB, < CEF and < CEB.

Solution: Angle < CEB is bisected into two angles < FEB and < CEF.

Therefore, < FEB = < CEF.

Substituting the values of < FEB and < CEF, we get

10x -3 = 7x +31

Adding 3 on both sides, we get

10x -3+3 = 7x +31+3.

10x = 7x + 34

Subtracting 7x from both sides, we get

10x-7x = 7x-7x +34.

3x = 34.

Dividing both sides by 3, we get

x= 11.33.

Plugging value of x=11.33 in < CEF = 7x +31.

We get

< CEF = 7(11.33) +31 = 79.33+31 = 110.33.

< FEB = < CEF = 110.33 approximately

< CEB = < FEB + < CEF = 110.33 +110.33 = 220.66 approximately

7 0

2 years ago

The sum of the lengths of any two sides of a triangle must be less than the length of the remaining side. True or false?

Pavlova-9 [17]

False. The Triangle Inequality Theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side.

6 0

3 years ago

Figure BCDE was transformed using the composition rx-axis ◦ T2,2. If point B on the pre-image lies at (3, 4), what are the coord

kaheart [24]

Thank you for posting your question here at brainly. If point B on the pre-image lies at (3, 4), the are the coordinates of B'' on the final is (5, -6). Below is the solutions, I hope the answer helps.

rx-axis ◦ T2,2 (3,4)
<span>= rx-axis ( T2,2 (3,4) ) </span>
<span>= rx-axis ( 5,6 ) </span>
= (5, -6)

4 0

2 years ago

Read 2 more answers

A cannonball of mass 1kg is shot vertically upward from the top of a building with an unknown velocity v_0(m/sec).v 0 (m/sec).

Bumek [7]

Taking the upward direction to be positive, the cannonball's height $y(t)$ in the air at time $t$ is given by

$y(t)=y_0+v_0 t-\dfrac g2t^2$

where $g$ is the magnitude of the acceleration due to gravity, 10 m/s^2, and $y_0$ is the height of the building from which the ball is being thrown.

At the moment the cannonball reaches its maximum height of 30 m, its velocity at that time is 0, so that

$0^2-{v_0}^2=-2g(30\,\mathrm m-y_0)\implies v_0=\sqrt{\left(20\dfrac{\rm m}{\mathrm s^2}\right)(30\,\mathrm m-y_0)}$

Substitute this into the height equation above, and let $t=2\,\mathrm s$ , for which we have $y(2\,\mathrm s)=30\,\mathrm m$ :

$30\,\mathrm m=y_0+\sqrt{\left(20\dfrac{\rm m}{\mathrm s^2}\right)(30\,\mathrm m-y_0)}(2\,\mathrm s)-\left(5\dfrac{\rm m}{\mathrm s^2}\right)(2\,\mathrm s)^2$