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stealth61 [152]

3 years ago

14

the courtyard for a building is in the shape of a triangle. one angle measures 25 degrees, the side opposite this angle is 14 me

ters and an adjacent side is 12 meters. find the area of the triangle.

1 answer:

Free_Kalibri [48]3 years ago

4 0

Ignore this “answer” i just need to be able to ask more questions

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Does anyone have the answers to this worksheet

fgiga [73]

I can't see paper. Maybe try to take a picture closer and more bright and clear.

3 0

3 years ago

Henry had $24.12. His friend Greg had one-

Daniel [21]

Answer: $42.21

Step-by-step explanation:

Henry; $24.12

Gregg; $24.12 / 2= $12.06

Kasey; $ 24.12 / 4= $6.03

$24.12 + $12.06 + $6.03

6 0

3 years ago

Solve this equation for y. <br> 3x - 5y = 12

Volgvan

-5y=12-3x (subtract 3x from both sides)
y = - 12-3x/5 (divide both sides by -5)
y = - 3(4-x)/5 (common term is 3)

6 0

3 years ago

X^2+14x-15=0 solve by completing square<br>

Softa [21]

Answer:

${x}^{2} + 14x - 15 = 0 \\ {x}^{2} - x + 15x -15 = 0 \\ x(x - 1) + 15(x - 1) = 0 \\ (x - 1)(x + 15) = 0 \\ \boxed{x = 1} \\ \boxed{x = - 15}$

<h3><u>x=1 and x=</u><u>-</u><u>15</u> is the right answer.</h3>

7 0

3 years ago

In △ABC, point P∈ AC with AP:PC=1:3, point Q∈AB so that AQ:QB=3:4, Find the ratios APBQ : APBC and AAQP : AABC.

weqwewe [10]

Answer: The ratios APBQ : APBC=4:21 and AAQP : AABC= 3:28

Explanation:

Here, ABC is a triangle where, P and Q are the midpoints of the edges AC and AB respectively,

Now, according to the question ,AP:PC=1:3 let AP=x and PC=3x where x is any real number. Thus, AC=AP+AC= x+3x= 4x

Similarly, AQ:QB=3:4 let AQ=3y and QB=4y where y is also any real number. Thus, AB=AQ+QB=3y+4y=7y

Since, $\frac{AP.BQ}{AP.BC} =\frac{AP.BQ}{AB.PC} =\frac{x.4y}{7y.3x} =\frac{4xy}{21xy} =\frac{4}{21}$

And, $\frac{AA.QP}{AA.BC} =\frac{AQ.AP}{AB.AC} =\frac{3y.1x}{7y.4x} =\frac{3xy}{28xy} =\frac{3}{28}$

5 0

4 years ago