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spayn [35]
3 years ago
14

I NEED HELP ASAP IT'S DUE IN 10 MINUTES!!! I'LL GIVE BRAINLIEST!!!

Mathematics
1 answer:
ycow [4]3 years ago
3 0
You answer is 12 :)...
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Mumz [18]
180 degress because 60 degrees + 120 degrees =180 degrees ab and cd are the same angle
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3 years ago
Hello help me please​
laila [671]

Answer:

7

Step-by-step explanation: it intersects on the y-axis

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Six men completed a task in 24 hrs and 48 min. How long, in hours and minutes, would it take four men to complete the same task?
igor_vitrenko [27]

Answer:

It would take 4 men 37 hrs and 12 mins to complete the same task.

step-by-step explanation:

Let X be the time the four men would use to complete the task.

From the question, 6 men completed a task in 24 hrs and 48 mins

But

24 \: hrs \: 48 \: mins = 24(60mins) + 48 \: mins=1488 \: mins

Using ratio and proportion

6 \: men:1488 \: mins

4  men:x\:mins

if more, less divides and if less, more divides.

4 men required more time to complete the task

Hence:

x =  \frac{6}{4} \times 1488 \: mins

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5 0
3 years ago
1. Find the sum. -5 + ( -1 ) + 6 + 4
Alex73 [517]
<span>1. Find the sum. -5 + ( -1 ) + 6 + 4

</span> -5 + ( -1 ) + 6 + 4 
= -5 -1 + 10
= 4
answer <span>B: 4

</span><span>2. Add -10 + ( -5 ) + 15
</span><span> -10 + ( -5 ) + 15
= -10 - 5 + 15
= -15 + 15
= 0

answer
</span><span>B: 0</span>
8 0
4 years ago
Read 2 more answers
Scientists have found a relationship between the temperature and the height above a distant planet's surface. , given below, is
ikadub [295]

The question is incomplete, the complete question is below:

Scientists have found a relationship between the temperature and the height above a distant planet's surface. , given below, is the temperature in Celsius at a height of kilometers above the planet's surface. The relationship is as follows: T(h) = 30.5 -2.5h.

a) Calculate T^-1(x)

b) T^-1(15)

Answer:

a) T^{-1} (x)= \frac{30.5-x}{2.5}

b) 6.2 °C

Step-by-step explanation:

a) T(h) = 30.5 - 2.5h\\2.5h = 30.5-T(h)\\h = \frac{30.5-T(h)}{2.5}\\ T^{-1} (x)= \frac{30.5-x}{2.5}

b) T^{-1} (x)= \frac{30.5-x}{2.5}\\\\T^{-1} (15)= \frac{30.5-15}{2.5}=6.2

3 0
3 years ago
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