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Valentin [98]
2 years ago
10

Suppose that G(X) = F(x+ 9). Which statement best compares the graph of

Mathematics
1 answer:
podryga [215]2 years ago
6 0

Answer:

Suppose that G(X) = F(x+ 9). Which statement best compares the graph of

G(X) with the graph of F(x)?

O A. The graph of G(X) is the graph of F(x) shifted 9 units try the left.

B. The graph of G(x) is the graph of F(X) shifted 9 units down.

C. The graph of G(x) is the graph of F(x) shifted 9 units up.

D. The graph of G(X) is the graph of F(x) shifted 9 units to the right.

Step-by-step explanation:

Suppose that G(X) = F(x+ 9). Which statement best compares the graph of

G(X) with the graph of F(x)?

O A. The graph of G(X) is the graph of F(x) shifted 9 units try the left.

B. The graph of G(x) is the graph of F(X) shifted 9 units down.

C. The graph of G(x) is the graph of F(x) shifted 9 units up.

D. The graph of G(X) is the graph of F(x) shifted 9 units to the right.

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Help me with this please I will give you Brainliest
oksian1 [2.3K]
Answer

h = 8.12 cm

Explanation

Area of a parallelogram = b • h

We know the area is 20.3 cm^2 and that the base is 2.5cm.

Therefore 20.3 cm^2 = 2.5h

20.3/2.5 = 8.12 cm

h = 8.12 cm
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Step-by-step explanation:

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Step-by-step explanation:

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2 years ago
Need Help! I don't get it
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Well hmmm let's say you take the car and go in the city for 60 miles with it, well, the car can do 60 miles per gallon, since you just drove it for 60 miles, you only spent 1 gallon of gasoline then.

that only happens if you drive it for 60 miles, what if you drive it for more, let's do a quick table on that,

\bf \begin{array}{ccll}&#10;miles&cost\\&#10;\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\&#10;60&3.6(1)\\\\&#10;&3.6\left( \frac{60}{60} \right)\\\\&#10;120&3.6(2)\\\\&#10;&3.6\left( \frac{120}{60} \right)\\\\&#10;180&3.6(3)\\\\&#10;&3.6\left( \frac{180}{60} \right)&#10;\end{array}

and so on, now let's check if you less than 60 miles,

\bf \begin{array}{ccll}&#10;miles&cost\\&#10;\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\&#10;40&3.6\left( \frac{40}{60} \right)\\\\&#10;20&3.6\left( \frac{20}{60} \right)\\\\&#10;10&3.6\left( \frac{10}{60} \right)&#10;\end{array}

so, if you divide the amount of miles driven, by 60, when you have driven it for 120 miles, 120/60 is just 2, and the cost is for 2 gallons, or 3.6 * 2, which is 7.2 bucks, for 180 miles is 180/60 or 3 gallons for 3.6 * 3 bucks, and so on.

now, what if you drive it instead for "m" miles?

\bf \begin{array}{ccll}&#10;miles&cost\\&#10;\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\&#10;m&3.6\left( \frac{m}{60} \right)\\\\&#10;\end{array}\implies c=3.6\left( \cfrac{m}{60} \right)\implies c=\cfrac{3.6m}{60}&#10;\\\\\\&#10;c=\cfrac{3.6}{60}m\implies c=0.06m
3 0
3 years ago
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