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Anika [276]
3 years ago
13

HELP PLZ ASPA 100 POINTS WILL GIVE BRAINLIEST!!!

Mathematics
2 answers:
Naddika [18.5K]3 years ago
8 0
(x10,y-2) y=1

That is the answer 100 percent sure
Varvara68 [4.7K]3 years ago
6 0

Step-by-step explanation:

(x10,y-2), reflection over y=1

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100 PONITS PLEASE HELP ME!!
Dennis_Churaev [7]

Answer:

Your business partner is correct.

General Formulas and Concepts:

<u>Stats - Correlations</u>

  • Positive means that it is trending upwards
  • Negative means that is is trending downwards

Step-by-step explanation:

If we were to draw a best line of fit into the scatterplot, we would see that we would get a positive slope and a positive line.

That means the scatterplot will be increasing calories as we increase fat.

Therefore, we would have a positive correlation. The business partner is correct.

5 0
3 years ago
Read 2 more answers
Question 15 of 20
ArbitrLikvidat [17]

72

Very simple Its pretty Blantant Its 72

6 0
2 years ago
Mac's age is 3 years less than twice Joe's age. The sum of their ages is 30. Find their ages.
k0ka [10]

Answer:

Mac is 19 years old, Joe is 11 years old

Step-by-step explanation:

2J - 3 = M

J + M = 30

M = 30 - J

2J - 3 = 30 - J

3J = 33

J = 11

M = 30 - 11

M = 19

4 0
3 years ago
Will a triangle with side length 3 cm, 4 cm and 1 cm exist
sveticcg [70]

No, because sum of any 2 sides of the triangle must be greater than the remaining side, but we have 1+3=4 => The triangle does not exist

3 0
3 years ago
Pls help w this question
luda_lava [24]

Answer:

f(x) = -2x + 1

Step-by-step explanation:

The given expression is \frac{64^x}{4^{5x-1}}

By solving the given expression further,

\frac{64^x}{4^{5x-1}} = \frac{[(4)^{3}]^x}{(4)^{5x-1}} [Since 64 = 4³]

        = \frac{4^{3x}}{4^{5x-1}}

        = 4^{3x}\times 4^{-(5x-1)} [Since \frac{1}{a}=a^{-1}]

        = 4^{3x-5x+1} [Since a^x\times a^y=a^{(x+y)}]

        = 4^{(-2x+1)}

By comparing the result with 4^{\text{f(x)}}

f(x) = -2x + 1

Therefore, f(x) = (-2x + 1) will be the answer.

3 0
3 years ago
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