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Dafna1 [17]
2 years ago
12

A total of 585 tickets were sold for the school play. They were either adult tickets or student tickets. There were 65 fewer stu

dent tickets sold than adult tickets. How many adult tickets were sold?
Mathematics
1 answer:
natulia [17]2 years ago
3 0

Answer:

325

Step-by-step explanation:

Total Tickets = 585

Let the adult tickets be = x

If 65 fewer student tickets were sold than adult tickets then Student tickets = x-65

Student Tickets + Adult Tickets = 585

x + (x-65) = 585

x + x - 65 = 585

2x - 65 = 585

2x = 585 + 65

2x = 650

2x/2 = 650/2

x = 325

Answered by Gauthmath

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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
A local hamburger shop sold a combined total of 518 hamburgers and cheeseburgers on Tuesday. There were 68 more cheeseburgers so
Nonamiya [84]
518-68=450. . . . . . . . . 450/2 = 225 hamburgers
7 0
2 years ago
Limit   
STatiana [176]

Rationalize the numerator:

\dfrac{\sqrt{x+4}-2}x\cdot\dfrac{\sqrt{x+4}+2}{\sqrt{x+4}+2}=\dfrac{(\sqrt{x+4})^2-2^2}{x(\sqrt{x+4}+2)}=\dfrac x{x(\sqrt{x+4}+2)}=\dfrac1{\sqrt{x+4}+2}

This is continuous at x=0, so we can evaluate the limit directly by substitution:

\displaystyle\lim_{x\to0}\frac{\sqrt{x+4}-2}x=\lim_{x\to0}\frac1{\sqrt{x+4}+2}=\frac1{\sqrt4+2}=\frac14

5 0
3 years ago
What is the horizontal distance between the two points (-7,-4) and (12,-4)? Remember that distance is always positive! Hint: The
RSB [31]
<h3>Answer: 19</h3>

==================================================

Explanation:

Draw out a number line. Plot -7 and 12 on the number line. Draw in the tickmarks between them. You should find the distance from -7 to 12 is 19 units since you need to go 19 spaces from either -7 to 12, or vice versa.

You can use subtraction to get

-7-12 = -19

or

12-(-7) = 12+7 = 19

The final result is made positive since negative distance does not make sense.

So you'd have |-7-12| = |-19| = 19 or |12-(-7)| = |19| = 19.

All of this only works because the two y coordinates are the same, which makes a horizontal line through the two given points.

7 0
3 years ago
If sally got 28 questions correct and received 70% on the exam, how many questions were on the exam
Furkat [3]

Answer: 40

Step-by-step explanation:

Number of correct questions Sally got = 28 questions

Percentage of correct questions answered= 70% on the exam,

Total number of questions in the exam = Unknown

Let the total questions in the exam be represented as y.

Since Sally got 70% correctly, this will be:

70% of y = 28

70/100 × y = 28

0.7 × y = 28

0.7y = 28

Divide both side by 0.7

0.7y/0.7 = 28/0.7

y = 40

There are 40 questions in the exam.

7 0
3 years ago
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