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3 years ago

I NEED HELP PLEASE!!

Mathematics

2 answers:

RUDIKE [14]3 years ago

5 0

Answer:

D, -79, plz brainliestt

Step-by-step explanation:

-8(2) -3 = -16-3= -19

-19(4) -3 = -76-3 = -79

TEA [102]3 years ago

5 0

D, -79 plzzz BRAINLIEST

-8 (2) -3 = -16 - 3 = -19

-19 (4) -3 = -76 - 3 = -79

Hope this helps :)

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Can someone please help

Fynjy0 [20]

Answer:

angle C = 78

angle D = 108

Step-by-step explanation:

the measure of angle A is 102

To find the measure of angle C

let Angle C be x

102 + x = 180 (linear pair)

x = 180 -102

x = 78

angle C = 78

angle B = 78 (vertically opposite angles are equal)

angle D = 108 (vertically opposite angles are equal)

8 0

3 years ago

Can someone tell me what 2/3 divided by 5 is it’s for a quiz.

docker41 [41]

Answer:2/15 hope this helps

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Solve 2∕3 + 5∕6 and put answer in simplest form.<br> A. 3∕2<br> B. 2∕3<br> C. 9∕6<br> D. 7∕6

kaheart [24]

2/3 + 5/6
= 4/6 + 5/6
= 9/6
= 3/2

A. 3/2

6 0

4 years ago

Jamal earns $2200 each month and has a car payment of $264. What percent of his income goes toward his car payment?

finlep [7]

2200 / 264 = 8.333333-

Round the answer, which is 8.

So, your answer is A. 8%

3 0

3 years ago

A person on a runway sees a plane approaching. The angle of elevation from the runway to the plane is 11.1° . The altitude of th

Gnoma [55]

Answer:

The horizontal distance from the plane to the person on the runway is 20408.16 ft.

Step-by-step explanation:

Consider the figure below,

Where AB represent altitude of the plane is 4000 ft above the ground , C represents the runner. The angle of elevation from the runway to the plane is 11.1°

BC is the horizontal distance from the plane to the person on the runway.

We have to find distance BC,

Using trigonometric ratio,

$\tan\theta=\frac{Perpendicular}{base}$

Here, $\theta=11.1^{\circ}$ ,Perpendicular AB = 4000

$\tan\theta=\frac{AB}{BC}$

$\tan 11.1^{\circ} =\frac{4000}{BC}$

Solving for BC, we get,

$BC=\frac{4000}{\tan 11.1^{\circ} }$

$BC=\frac{4000}{0.196}$ (approx)

$BC=20408.16$ (approx)

Thus, the horizontal distance from the plane to the person on the runway is 20408.16 ft

8 0

3 years ago