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melomori [17]
3 years ago
10

Choices; a) 21.8° b) 35.2° c) 45.1°

Mathematics
1 answer:
kifflom [539]3 years ago
6 0

Answer:

\displaystyle a) \: {x}^{ \circ}  =  21.8 ^{ \circ}

Step-by-step explanation:

we are given opposite and adjacent

we want to figure out x°

remember that,

\displaystyle \: \tan( \theta)  =  \frac{opp}{adj}

where our opp is 5 and adj is 12.5

let \theta be x°

so

\displaystyle \: \tan(  {x}^{ \circ} )  =  \frac{5}{12.5}

\displaystyle \: {x}^{ \circ}  =  \arctan \left( \frac{5}{12.5}  \right)

by using calculator

\displaystyle \: {x}^{ \circ}  =  21.8

hence,

our answer is a

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Please help will mark branliest
Citrus2011 [14]

Answer:

RS = 17

Step-by-step explanation:

QS = 62

RS = ?

QR = 45

QS - QR = RS

62 - 45 = 17

Thus RS = 17

3 0
2 years ago
There is a mound of g pounds of gravel in a quarry. Throughout the day, 300 pounds of gravel are added to the mound. Two orders
galina1969 [7]

Answer:

g+300-700-700=1500, g=2600

Step-by-step explanation:

At the start of the day there are g pounds of gravel in the quarry.

Then it says that there were 300 pounds added to the mound, so at this point the pounds of gravel in the mound are now g+300.

Then two orders of 700 pounds are sold and removed from the mound. That means they removed 700 pounds from the mound 2 times. Therefore at this point the pounds of gravel in the mound are now g+300-700-700.

Then it says that at this point, the amount of pounds in the mound are 1500.  Hence we get

g+300-700-700=1500

To solve for g we just add 700 two times to both sides of the equation, and subtract 300 from both sides of the equation, getting:

g+300-700-700+700+700-300=1500+700+700-300

And so g=2600

4 0
3 years ago
What is the area, in square feet, of the trapezoid below?
slavikrds [6]

Answer:

68.75

Step-by-step explanation:

Measure the length of the parallel sides and add them together. Then multiply this number by the height of the area. Then multiply by ½ or divide by two to find the square footage of the trapezoid.  

13.6 + 5.9= 19.5

19.5 + 8 = 27.5

27.5 x 5 = 137.5

137.5 x 1/2 = 68.75

3 0
3 years ago
15. Explain why we need to specify 0 < b < 1 and b > 1 as valid values for the base b in the expression logb(xx).
damaskus [11]

Answer:

The base (b) has to be positive and different of 1. The logarithm is the inverse of exponential, so:

logb(a) = x ⇒ a = bˣ

So, for b = 0 ⇒ 0ˣ = a

And there is impossible, "a" only could be 0.

For b = 1 ⇒ 1ˣ = a

And the same thing would happen, the logarithming would be to be 1, and the function will be extremally restricted.

For b<0, then the expression a = bˣ will be also restricted, and will not represent all values of a.

So, 0<b<1 and b >1.

3 0
3 years ago
In what ways does adding a constant or coefficient to a quadratic equation affect the graph
lesya692 [45]
A constant can either shift the graph up or down. If coefficient is less than 1 it will make the graph wider, if it's more than 1 it will make graph more narrow
7 0
3 years ago
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