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V125BC [204]
3 years ago
13

BRAINLIST AND A THANK YOU AND 5 stars WILL BE REWARDED PLS ANSER

Mathematics
1 answer:
vichka [17]3 years ago
3 0

Answer:

The first picture's answer would be (6, 21)

Step-by-step explanation:

You have to find the points on the 8th and the 9th day, and then you would add them together, and then divide by two finding the average, which would be 24 and 18, so when added, you get 42, divided by 2 you get 21. You look on the graph for the point with 21, and you find it is on 6.

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Melinda goes bowling on Saturday afternoon, she bowls three games and pays for shoe rental. Define a variable and write an expre
BabaBlast [244]
Fisrt let us define the variableslet y be the total cost melinda paysx be the number of games she playedb be the cost of shoe rentalm is the cost of one game
y = mx + b
y = 4(3) + by = 12 + b 
7 0
2 years ago
Read 2 more answers
A video game has five stages. If Will beat 4/5 of the stages, what percent of the game has he finished
fiasKO [112]

Answer:

75%

Step-by-step explanation:

1 = 10%

2 = 25%

3 = 50%

4 = 75%

5 = 100%

5 0
2 years ago
8600⋅0.0395x=21000<br> ok this time there is an x<br> what is x
11Alexandr11 [23.1K]

Answer: x = 61.81925228

Step-by-step explanation:

6 0
2 years ago
Find the average rate of change for f(x) = x2 + 9x + 18 from x = −10 to x = 10. A) 3 B) 7 C) 9 D) 11
Sergeeva-Olga [200]

Answer:

C

Step-by-step explanation:

The average rate of change of f(x) in the closed interval [ a, b ] is

\frac{f(b)-f(a)}{b-a}

Here [ a, b ] = [ - 10, 10 ], thus

f(b) = f(10) = 10² + 9(10) + 18 = 100 + 90 + 18 = 208

f(a) = f(- 10) = (- 10)² + 9(- 10) + 18 = 100 - 90 + 18 = 28, thus

average rate of change = \frac{208-28}{10-(-10)} = \frac{180}{20} = 9

3 0
3 years ago
Solve for m -mk-90&gt;85
loris [4]
First add 90 to both sides) -mk>85+90simplify 85+90 to 175) -mk>175divide both sides by K) -m>175/kmultiply both sides by -1) m<-175/k
 YOUR ANSWER IS
M<-175/K 
hope l helped today l was stuck on the same question doing part 2 of alg exam , and l came through your question l hope it's not too late :)
5 0
3 years ago
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