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grandymaker [24]
4 years ago
7

(02.03)Figure ABC is reflected about the x-axis to obtain figure A'B'C':

Mathematics
1 answer:
faust18 [17]4 years ago
5 0

Answer:

B.(–2, –7)

Step-by-step explanation:

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10. is f 11. is a I dont know for 12 though
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What is the product of two numbers is 48?; What is the formula for finding product of two numbers?; What two numbers have a prod
Nadusha1986 [10]

The two numbers that have a product of 48 are 6 and 8 or 12 and 4, etc. They can be multiplied together to get 48, which is the product of the two numbers.

The formula for finding the product of two numbers is to multiply them together. For example, if you have two numbers, a and b, the product of these two numbers is a multiplied by b or an a x b.

The two numbers that have a product of 48 and a sum of 16 are 4 and 12. You can multiply 4 and 12 together to get 48, and you can add them together to get 16.

When the product of two numbers is one of each number, the other number is 1. This means that when the product of two numbers is the same as one of the numbers, the other number must be 1, as 1 multiplied by any number is always equal to itself.

Learn more about the product of two numbers with an example:

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1 year ago
Which is the correct equation for the graph of f(x), a transformation of the graph of g(x) = log2x? A. f(x) = -log2(x + 2) B. f(
krok68 [10]

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Step-by-step explanation:

8 0
3 years ago
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masha68 [24]

there are 14 green grapes in a bowl of fruits

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2 years ago
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cos theta = 4/5, 0˚< theta < 90˚ use the information given to find sin2theta, cos2theta, and tan 2theta
NikAS [45]
First calculate \sin\theta.

\sin^2\theta+\cos^2\theta=1\\\\\sin^2\theta=1-\cos^2\theta\\\\\sin^2\theta=1-\left(\dfrac{4}{5}\right)^2\\\\\\\sin^2\theta=1-\dfrac{16}{25}\\\\\\\sin^2\theta=\dfrac{9}{25}\quad|\sqrt{(\dots)}\\\\\\\sin\theta=\sqrt{\dfrac{9}{25}}\\\\\\\boxed{\sin\theta=\dfrac{3}{5}}

We take positive value of sinθ because 0° < θ < 90°
Now we could calculate sin2θ, cos2θ and tan2θ:

\sin2\theta=2\sin\theta\cos\theta=2\cdot\dfrac{3}{5}\cdot\dfrac{4}{5}=\dfrac{2\cdot3\cdot4}{5\cdot5}=\dfrac{24}{25}\\\\\\\cos2\theta=\cos^2\theta-\sin^2\theta=\left(\dfrac{4}{5}\right)^2-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}-\dfrac{9}{25}=\dfrac{7}{25}\\\\\\&#10;\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=\dfrac{\frac{24}{25}}{\frac{7}{25}}=\dfrac{24\cdot25}{7\cdot25}=\dfrac{24}{7}=3\dfrac{3}{7}
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3 years ago
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