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Musya8 [376]
3 years ago
14

Find the area of triangle PQR.

Mathematics
1 answer:
Alex3 years ago
4 0

Answer:

70°

Step-by-step explanation:

Since  PTS shape is equal to QTR

and  PTQ shape is equal to STR

and Since QTP = 10

and Since  PTS = 60

Add both up 70

Therefore PQR = 70°

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Latoya buys candy that costs $6 per pound. She will spend more than $54 on candy. What are the possible numbers of pounds she wi
kotegsom [21]

Answer:

11 pounds

Step-by-step explanation:

4 0
2 years ago
Th SEDelco
Delvig [45]

<u>Corrected Question</u>

Matt makes $80 a month mowing lawns. His friend James also mows lawns, but he makes x dollars less per  month than Matt does. James makes a total of $816 per year. How much less each month does James make than Matt? Solve the linear equation to find the value for x

. Remember that x represents how much less per month James makes than Matt

Equation: 12(80 – x) = 816

Answer:

x=$12

Step-by-step explanation:

To solve the given equation

12(80 – x) = 816

First, we distribute:

960-12x=816

Collect like terms

12x=960-816

12x=144

Divide both sides by 12

x=12

Therefore, James makes $12 less than Matt every Month.

6 0
3 years ago
Read 2 more answers
Write a linear function f with the values f (0) =4 and f(3)=-8
Darya [45]

Answer:

Answer:

f(x) = - 4x + 4

Step-by-step explanation:

Step-by-step explanation:

A linear function has the form

f(x) = ax + b, thus

f(0) = a(0) + b = 4, that is

0 + b = 4 , hence b = 4

f(3) = 3a + b = - 8 ← substitute b = 4

3a + 4 = - 8 ( subtract 4 from both sides )

3a = - 12 ( divide both sides by 3 )

a = - 4, thus

f(x) = - 4x + 4

8 0
3 years ago
Find the exact solutions of the problem algebraically: tan 2x-cot x=0
Eva8 [605]
Tan 2x - cot x = 0
2tan x / (1 - tan^2 x) - 1/tan x = 0
2tan^2 x - (1 - tan^2 x) = 0
2tan^2 x - 1 + tan^2 x = 0
3tan^2 x - 1 = 0
tan^2 x = 1/3
tan x = + or - 1/√3
5 0
3 years ago
Algebra need some assistance
d1i1m1o1n [39]
X=0,2,3/8
That is what x equal i don’t know the question but I hope this help
5 0
2 years ago
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