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3 years ago

Find the missing side or angle.

Mathematics

1 answer:

bonufazy [111]3 years ago

3 0

Answer:

c = 6.1

Step-by-step explanation:

Given that: B = 18° , C = 152°, and b = 4.

The missing side, c, can be determined by applying the sine rule which states that:

$\frac{b}{Sin B}$ = $\frac{c}{Sin C}$

$\frac{4}{Sin 18^{o} }$ = $\frac{c}{Sin 152^{o} }$

c = $\frac{4* Sin 152^{o} }{Sin 18^{o} }$

= $\frac{1.8779}{0.3090}$

= 6.0773

c = 6.1

Therefore, the value of the missing side, c, is 6.1

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Solve -9-8(1+4h)=-17

Mariulka [41]

-9-8(1+4h) = -17. We need to solve for h.
First, use the distributive property for 8(1+4h):
8(1+4h) = 8*1 + 8*4h = 8 + 32h

So -9-8(1+4h) = -17
-9 - (8+32h) = -17
-9 - 8 - 32h = -17
-17 - 32h = - 17
Add 17 on both sides to have the variables on a side and the numbers on the other:
-17 - 32h + 17 = -17 + 17
-32h = 0
divide both sides by -32 to get the variable h alone and its value on the other side:
(-32h)/-32 = 0/-32
h = 0.

So -9-8(1+4h) = -17 for h = 0.

You can recheck your answer (very important):
-9 - 8(1+4h) = -9 -8(1+4*0) = -9 - 8(1+0) = -9-8*1 = -9-8 = -17
The answer has been approved.

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On Monday 4 bricklayers took 3 hours to lay a total of 4200 bricks.

lisabon 2012 [21]

<h2>Hello!</h2>

The answer is:

It will take 4.5 hours for two bricklayers to lay a total of 3150 bricks.

From the statement, we know that on Monday, bricklayers took 3 hours to lay a total of 4200 bricks.

So, we have that the rate of the four bricklayers is given by the following equation:

$Production=Rate*Time\\\\Rate=\frac{Production}{time}$

$Rate=\frac{4200bricks}{3hours}=\frac{1400bricks}{hour}$

Therefore, we have that the rate of the four bricklayers is 1400 bricks per hour, meaning that each bricklayer can lay 350 bricks per hour.

Now, to calculate how many hours it will take to the 2 bricklayers to lay a total of 3150, we need to divide the calculated rate (for bricklayers) by two, since there are only 2 bricklayers working on Tuesday.

So, calculating the time, we have:

$Production=\frac{Rate}{2}*Time$