All categories

3 years ago

What are RQS and TQS?

Mathematics

1 answer:

yaroslaw [1]3 years ago

8 0

Answer:

m∠RQS = 72°

m∠TQS = 83°

Step-by-step explanation:

m∠RQS +m ∠TQS = m∠RQT

The two angles combine to make a larger angle

m∠RQS = (4x - 20)

m∠TQS = (3x + 14)

(4x - 20) + (3x + 14) = 155

Group the Xs and the constants

4x + 3x - 20 + 14 = 155

Combine like terms

7x - 6 = 155

Add 6 to both sides

7x = 161

Divide by 7 on both sides

x = 23

Check:

4(23) - 20 + 3(23) + 14 = 155

92 - 20 + 69 + 14 = 155

155 = 155

But we need to find m∠RQS and m∠TQS. So plug in x = 23 to the values.

m∠RQS = 4(23) - 20 = 72°

m∠TQS = 3(23) + 14 = 83°

Checking:

72 + 83 = 155

You might be interested in

Find the value of x, & y.<br> 3x-y+8x-7=x+2y+5x+3y

Advocard [28]

Please rewrite this equation(s) using parantheses than needed

thank you

5 0

3 years ago

What is the value of p?

aksik [14]

Answer:

P = 18

Step-by-step explanation:

:) that the answer.

3 0

3 years ago

Read 2 more answers

Quadrilateral ABCD is translated to get quadrilateral A′B′C′D′. Vertex A is at (-5, 2), and vertex A′ is at (2, -2).

fgiga [73]

Then vertex B’ is at (1,1)
proof: the translation from A to A’ is (7,-4) therefore applying the same conditions to B would give you the outcome of (1,1)

4 0

3 years ago

13 POINTS- please help me

allsm [11]

Answer:

See explanation

Step-by-step explanation:

16. Two parallel lines are cut by transversal. Angles with measures $(6x+20)^{\circ}$ and $(x+100)^{\circ}$ are alternate exterior angles. By alternate exterior angles, the measures of alternate exterior angles are the same:

$6x+20=x+100\\ \\6x-x=100-20\\ \\5x=80\\ \\x=16$

Then

$(6x+20)^{\circ}=(6\cdot 16+20)^{\circ}=116^{\circ}\\ \\(x+100)^{\circ}=(16+100)^{\circ}=116^{\circ}$

17. Two parallel lines are cut by transversal. Angles with measures $(2x+12)^{\circ}$ and $(3x-22)^{\circ}$ are alternate interior angles. By alternate interior angles, the measures of alternate interior angles are the same:

$2x+12=3x-22\\ \\2x-3x=-22-12\\ \\-x=-34\\ \\x=34$

Then

$(2x+12)^{\circ}=(2\cdot 34+12)^{\circ}=80^{\circ}\\ \\(3x-22)^{\circ}=(3\cdot 34-22)^{\circ}=80^{\circ}$

18. Two parallel lines are cut by transversal. Angles with measures $(6x-7)^{\circ}$ and $(5x+10)^{\circ}$ are alternate exterior angles. By alternate interior angles, the measures of alternate exterior angles are the same:

$6x-7=5x+10\\ \\6x-5x=10+7\\ \\x=17$

Then

$(6x-7)^{\circ}=(6\cdot 17-7)^{\circ}=95^{\circ}\\ \\(5x+10)^{\circ}=(5\cdot 17+10)^{\circ}=95^{\circ}$

19. The diagram shows two complementary angles with measures $2x^{\circ}$ and $56^{\circ}$ . The measures of complementary angles add up to $90^{\circ},$ then

$2x+56=90\\ \\2x=90-56\\ \\2x=34\\ \\x=17$

Hence,

$2x^{\circ}=2\cdot 17^{\circ}=34^{\circ}$

Check:

$34^{\circ}+56^{\circ}=90^{\circ}$

20. Angles $\angle 1$ and $\angle 2$ are vertical angles. By vertical angles theorem, vertical angles are congruent, so

$m\angle 1=m\angle 2\\ \\5x+7=3x+15\\ \\5x-3x=15-7\\ \\2x=8\\ \\x=4$

Hence,

$m\angle 1=(5x+7)^{\circ}=(5\cdot 4+7)^{\circ}=27^{\circ}\\ \\m\angle 2=(3x+15)^{\circ}=(3\cdot 4+15)^{\circ}=27^{\circ}$

21. $\angle 5$ and $\angle 8$ are supplementary. The measures of supplementary angles add up to $180^{\circ},$ so

$m\angle 5+m\angle 8=180^{\circ}\\ \\3x-40+7x-120=180\\ \\10x-160=180\\ \\10x=180+160\\ \\10x=340\\ \\x=34$

Therefore,

$m\angle 5=(3x-40)^{\circ}=(3\cdot 34-40)^{\circ}=62^{\circ}\\ \\m\angle 8=(7x-120)^{\circ}=(7\cdot 34-120)^{\circ}=118^{\circ}\\ \\62^{\circ}+118^{\circ}=180^{\circ}$

7 0

3 years ago

14 is 56% of what number

Sedbober [7]

7.84 <span> 56% off 14 is equal to (56 x 56) / 100 </span>

8 0

3 years ago

Read 2 more answers