3.5x-10>-3 add 10 to both sides
3.5x>7 divide both sides by 3.5
x>2
... now for the other inequality:
8x-9<39 add 9 to both sides
8x<48 divide both sides by 3
x<6
So we have x>2 and x<6, so the compound inequality is:
2<x<6 and this means that the solution set is:
x=(2, 6)
Answer:
The first step would be to subtract the two equations.
5/6 - 3/8
Rewrite the fractions to have a common denominator:
5/6 x 4 = 20/24
3/8 x 3 = 9/24
Now you have 20/24 - 9/24 = 11/24
The answer is D. 11/24
we know the segment QP is an angle bisector, namely it divides ∡SQR into two equal angles, thus ∡1 = ∡2, and ∡SQR = ∡1 + ∡2.
![\bf \begin{cases} \measuredangle SQR = \measuredangle 1 + \measuredangle 2\\\\ \measuredangle 2 = \measuredangle 1 = 5x-7 \end{cases}\qquad \qquad \stackrel{\measuredangle SQR}{7x+13} = (\stackrel{\measuredangle 1}{5x-7})+(\stackrel{\measuredangle 2}{5x-7}) \\\\\\ 7x+13 = 10x-14\implies 13=3x-14\implies 27=3x \\\\\\ \cfrac{27}{3}=x\implies 9=x \\\\[-0.35em] ~\dotfill\\\\ \measuredangle SQR = 7(9)+13\implies \measuredangle SQR = 76](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20%5Cmeasuredangle%20SQR%20%3D%20%5Cmeasuredangle%201%20%2B%20%5Cmeasuredangle%202%5C%5C%5C%5C%20%5Cmeasuredangle%202%20%3D%20%5Cmeasuredangle%201%20%3D%205x-7%20%5Cend%7Bcases%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Cmeasuredangle%20SQR%7D%7B7x%2B13%7D%20%3D%20%28%5Cstackrel%7B%5Cmeasuredangle%201%7D%7B5x-7%7D%29%2B%28%5Cstackrel%7B%5Cmeasuredangle%202%7D%7B5x-7%7D%29%20%5C%5C%5C%5C%5C%5C%207x%2B13%20%3D%2010x-14%5Cimplies%2013%3D3x-14%5Cimplies%2027%3D3x%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B27%7D%7B3%7D%3Dx%5Cimplies%209%3Dx%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cmeasuredangle%20SQR%20%3D%207%289%29%2B13%5Cimplies%20%5Cmeasuredangle%20SQR%20%3D%2076)
Answer:
∠B ≅ ∠Y △ABC ~ △ZYX by the SAS similarity theorem.
Step-by-step explanation:
1.
units
units
units
units, then

2.
and
are right angles - given
3.
two right angles are always congruent.
4.
by SAS similarity theorem.
SAS Similarity Theorem states that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar.