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Marta_Voda [28]
3 years ago
13

Need to help to find the slope pls helppp

Mathematics
2 answers:
kodGreya [7K]3 years ago
7 0

Answer:

it is 2

Step-by-step explanation:

rise over run

it goes up 2 points and

moves right 1 point

2/1 =2

AlexFokin [52]3 years ago
5 0

Answer:

Step-by-step explanation:

easy

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If QS bisects PQT, SQT=(8x- 25),PQT=(9x+34), and SQR=112, find each measure.
Goryan [66]

The values of the angles are; x = 12°, m∠PQS = 71°, m∠PQT = 142° and m∠TQR = 41°

<h3>What are the measure of the each angle?</h3>

The required angles; x, m∠PQS, m∠PQT, and m∠TQR

The given parameters are bisects ∠PQT

m∠SQT = (8·x - 25)°

m∠PQT = (9·x + 34)°

m∠SQR = 112°

We have;

m∠PQT = m∠SQT + m∠PQS (Angle addition postulate)

m∠SQT ≅ m∠PQS (Angles formed by angle bisector are congruent)

m∠SQT = m∠PQS  by Definition of congruency

m∠PQT = 2 × m∠SQT

Therefore;

(9·x + 34)° = 2 × (8·x - 25)° = (16·x - 50)°

Collecting like terms gives:

(34 + 50)° = 16·x - 9·x = 7·x

7·x = 84°

x = 84°/7 = 12°

x = 12°

m∠SQT = (8·x - 25)°

Therefore;

m∠SQT = (8 × 12 - 25)° = 71°

m∠SQT = 71°

m∠PQT = 2 × m∠SQT

∴ m∠PQT = 2 × 71° = 142°

m∠PQT = 142°

m∠PQS = m∠SQT (Angles formed by the same bisector )

∴ m∠PQS = m∠SQT = 71°

m∠PQS = 71°

m∠SQR = m∠SQT + m∠TQR (Angle addition postulate)

m∠SQT = 71°

∴  m∠SQR = 112° = 71° + m∠TQR

m∠TQR = 112° - 71° = 41°

m∠TQR = 41°

Learn more about angles here:

brainly.com/question/2882938

#SPJ1

4 0
1 year ago
You notice that the client's bedroom is also proportional to a larger office that you have already completed. The diagram below
9966 [12]

Answer:

D. 75

Step-by-step explanation:

8 0
3 years ago
What is the equivalent decimal to this fraction? 10/45 0.2 0.222222222... 0.232323232323... 0.181818181818...​
Ivan

Factorize the numerator and denominator, then simplify:

10/45 = (2×5)/(9×5) = 2/9

Now,

1/9 = 1/10 + 1/90

1/90 = 1/10 × 1/9 = 1/10 × (1/10 + 1/90) = 1/100 + 1/900

1/900 = 1/100 × 1/9 = 1/100 × (1/10 + 1/90) = 1/1000 + 1/9000

and so on, which is to say

1/9 = 1/10 + 1/100 + 1/1000 + …

or

1/9 = 0.111…

so that multiplying both sides by 2 gives

2/9 = 0.222…

5 0
2 years ago
Wht is the solutioin to the system of equations y = 4x – 10 y = 2
gtnhenbr [62]

Answer:

(3,2)

Step-by-step explanation:

I graphed the two lines using a graphing tool. The lines intercept at (3,2). Therefore, the solution to the system is (3,2).

7 0
3 years ago
The sum of the diagonals of a rhombus is 5√2.
Alex
Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC
= \frac{1}{2} BD (AO + OC)
\frac{1}{2} BD \times AC

So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}
BD \times AC = 8
Now, AC + \: BD = 5 \sqrt{2}

Squaring both sides, we get
AC {}^{2} + BD {}^{2} + 2 AC.BD =50
AC {}^{2} + BD {}^{2} + 2 \times 8 = 50
AC {}^{2} + BD {}^{2} = 50 - 16
AC {}^{2} + BD {}^{2} = 34

In △AOB, we have
OA {}^{2} + OB {}^{2} =AB {}^{2}
( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}
\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}
AC {}^{2} + BD {}^{2} = 4 AB {}^{2}
34 = 4 AB {}^{2}

Square rooting both sides
\sqrt{34} = 2 AB
Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units.

Hope it helps!

7 0
3 years ago
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