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3 years ago

11

please help will give BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLI

EST BRAINLIEST BRAINLIEST for correct answer!!!!!!!!!!!!!

2 answers:

o-na [289]3 years ago

8 0

Answer:

25 inches is the answer

Temka [501]3 years ago

3 0

Answer:

25 in or 20in

Step-by-step explanation:

it is not half way if it was it would be 15in its way of half way

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Look at this system of equations.<br> 2x +5 =9.5<br> 3x + 5y =10.5<br> What is the value of x+y

stiv31 [10]

Answer:

x = 2.25 y = 0.75

Step-by-step explanation:

2x = 9.5- 5

2x = 4.5

x = 4.5 ÷ 2

x = 2.25

know we now what is x so in the second one we evaluate x

3(2.25) + 5y = 10.5

6.75 + 5y = 10.5

5y = 10.5- 6.75

5y = 3.75

y = 3.75 ÷ 5

y = 0.75

that's it !

if you want to check your answer you can evaluate x and y so it will be like

2×2.25 + 5

= 4.5 + 5 = 9.5

and

3× 2.25 +5×0.75

= 6.75 + 3.75 = 10.5

6 0

3 years ago

Write the point-slope form of the line that passes through (-8, 2) and is perpendicular to a line with a slope of -8

Schach [20]

Answer:

y - 2 = (1/8)(x + 8)

Step-by-step explanation:

The slope of the given line is -8. That means the slope of any line perpendicular to the given line will be the negative reciprocal of -8, which would be (1/8).

Then the desired equation is

y - 2 = (1/8)(x + 8)

4 0

3 years ago

BRAINIEST ANSWER HERRREEE!!! :D

steposvetlana [31]

By asking students to answer un-biased, and with detail. Comparing and contrasting each others' genre preference is also useful.

4 0

3 years ago

Given: 2x + 3y= 6. When xhas a value of zero, the y value is 2. true or false<br>

SIZIF [17.4K]

The answer is going to be true

8 0

3 years ago

A brick is thrown upward from the top of a building at an angle of 250 above the horizontal, and with an initial speed of 15 m/s

mojhsa [17]

Answer:

(a) 25.08 m

(b) 2.05 m

Step-by-step explanation:

Let the height of the building be 'h'.

Given:

Angle of projection is, $\theta=25$ °

Initial speed is, $u=15\ m/s$

Time of flight is, $t=3.0\ s$

(a)

Consider the vertical motion of the brick.

Vertical component of initial velocity is given as:

$u_y=u\sin\theta\\\\u_y=15\sin(25)=6.34\ m/s$

Vertical displacement of the brick is equal to the height of the building.

So, vertical displacement = $-h$ (Negative sign implies downward motion)

Acceleration is due to gravity in the downward direction. So,

Acceleration is, $g=-9.8\ m/s^2$

Now, using the following equation of motion;

$-h=u_yt+\frac{1}{2}gt^2\\-h=6.34(3)-\frac{9.8}{2}(3)^2\\\\-h=19.02-44.1\\\\-h=-25.08\\\\h=25.08\ m$

Therefore, the building is 25.08 m tall.

(b)

Let the maximum height be 'H'.

At maximum height, the vertical component of velocity is 0 as the brick stops temporarily in the vertical direction.

So, $v_y=0\ m/s$

Now, using the following equation of motion, we have:

$v_y^2=u_y^2+2gH\\\\0=(6.34)^2-2\times 9.8\times H\\\\19.6H=40.2\\\\H=\frac{40.2}{19.6}=2.05\ m$

Therefore, the maximum height of the brick is 2.05 m.

5 0

3 years ago