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luda_lava [24]
3 years ago
11

Let s = 8.

Mathematics
jadon
3 years ago
where is the question?
2 answers:
sleet_krkn [62]3 years ago
6 0
B. 22

Because 6+2s

is the same as 6+2(8).

6+2(8)
6+16
=22
klasskru [66]3 years ago
5 0
So 2s mean 2 times s
if s=8 then pemdas
multiply
2 times 8=16
6+16=22
the answer is B, 22
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Solve the expression (7th grade math):<br><br> -0.8m+3-12.5m+95.36m
Lana71 [14]

Step-by-step explanation:

<u>Combine Like-Terms:</u>

-0.8m + 3 - 12.5m + 95.36m

-13.3m + 3 + 95.36m

82.06m + 3

82.06m + 3 is your simplified expression.

7 0
2 years ago
Given that ABDC : EFGH, what is m G?<br> 27°<br> 90°<br> 117°<br> 120°
oksian1 [2.3K]
I think it would be 117

Because H is 63
So 90+90+63=243
360-243=117
I used 360 because angles in quadrilaterals add up to 360 degrees
4 0
3 years ago
The elevation at ground level is 0 feet. An elevator starts 90 feet below ground level. After traveling for 15 seconds, the elev
sweet [91]
The answer is -4 2/3

Explanation : The elevator is at 0, and went below ground level 90 feet, that would be -90, then it went up 20 feet, which would be -90 + 20 = -70

-70 ft / 15sec = - 4 2/3

I am not sure about the ground level and negative difference but that it is what I was taught. Below ground level is negative and above is positive. Therefore the elevator traveled up and it traveled -4 2/3 per second.
It so either 4 2/3 or - 4 2/3 but pretty sure it is - 4 2/3.

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6 0
3 years ago
What is plane z in this image
denpristay [2]

Answer:

the left side of the line

Step-by-step explanation:

6 0
4 years ago
A helicopter is flying above a town. the local high school is directly to the east of the helicopter at a 20° angle of depressio
guapka [62]
Answer: 4.5 miles

Explanation:

When you draw the situation you find two triangles.

1) Triangle to the east of the helicopter

a) elevation angle from the high school to the helicopter = depression angle from the helicopter to the high school = 20°

b) hypotensue = distance between the high school and the helicopter

c) opposite-leg to angle 20° = heigth of the helicopter

d) adyacent leg to the angle 20° = horizontal distance between the high school and the helicopter = x

2) triangle to the west of the helicopter

a) elevation angle from elementary school to the helicopter = depression angle from helicopter to the elementary school = 62°

b)  distance between the helicopter and the elementary school = hypotenuse

c) opposite-leg to angle 62° = height of the helicopter

d) adyacent-leg to angle 62° = horizontal distance between the elementary school and the helicopter = 5 - x

3) tangent ratios

a) triangle with the helicpoter and the high school

tan 20° = Height / x ⇒ height = x tan 20°

b) triangle with the helicopter and the elementary school

tan 62° = Height / (5 - x) ⇒ height = (5 - x) tan 62°

c) equal the height from both triangles:

x tan 20° = (5 - x) tan 62°

x tan 20° = 5 tan 62° - x tan 62°

x tan 20° + x tan 62° = 5 tan 62°

x  (tan 20° + tan 62°) = 5 tan 62°

⇒ x = 5 tant 62° / ( tan 20° + tan 62°)

⇒ x = 4,19 miles

=> height = x tan 20° = 4,19 tan 20° = 1,525 miles

4) Calculate the hypotenuse of this triangle:

hipotenuese ² = x² + height ² = (4.19)² + (1.525)² = 19.88 miles²

hipotenuse = 4.46 miles

Rounded to the nearest tenth = 4.5 miles

That is the distance between the helicopter and the high school.
5 0
3 years ago
Read 2 more answers
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