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3 years ago

12

The graph of g is a vertical stretch by a factor of 4 and a reflection in the x-axis, followed by a translation 2 units up of th

e graph of f(x)=x2. Write a rule for g. Then identify the vertex.

1 answer:

Eddi Din [679]3 years ago

7 0

Answer:

f(x)=x^2

g(x)= -4(x^2)+2

(0,2)

Step-by-step explanation:

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docker41 [41]

Answer:

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2 years ago

How should the decimal point in 34.05 be moved to determine the product 34.05×106 ?

r-ruslan [8.4K]

Answer: Six places to the right

Step-by-step explanation:

For this exercise it is important to remember that. by definition, the exponent of a number indicates the number of times you must use the same factor to multiply.

Given:

$b^n$

"b" is the base and "n" is the exponent.

In this case, you have the following multiplication provided in the exercise:

$34.05*10^6$

Notice that the base 10 has an exponent 6. This indicates the following:

$10^6=10*10*10*10*10*10=1,000,000$

By definition, moving the decimal point 6 places to the right (because there are six zeros), is the same as multiplying the decimal number 34.05 by 1,000,000.

Therefore, based on the explained, when you move the decimal point six places to the right, you get the following product.

$34.05*10^6=34,050,000$

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3 years ago

Need help with question 2

Alex777 [14]

The answer is y= -x-1

8 0

3 years ago

I WILL MARKED BRAINLIEST IF YOU COULD ANSWER THIS!

tester [92]

Answer:

s = -2400t + 17400

Step-by-step explanation:

Let's say t is the x value on a coordinate plane, and s is the y. Then, we have the points (0, 17400) and (6, 3000). The slope of these is 14400/-6 or -2400.

Now we just have the equation y = -2400x + b, and from the point (0, 17400) we can find that b is 17400. So, we have y = -2400x + 17400. Convert these back into t and s and you get your answer, s = -2400t + 17400.

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3 years ago

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ASHA 777 [7]

$\bf tan\left( \frac{x}{2} \right)+\cfrac{1}{tan\left( \frac{x}{2} \right)}\\\\
-----------------------------\\\\
tan\left(\cfrac{{{ \theta}}}{2}\right)=
\begin{cases}
\pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}}
\\ \quad \\

\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}
\\ \quad \\

\boxed{\cfrac{1-cos({{ \theta}})}{sin({{ \theta}})}}
\end{cases}\\\\$

$\bf -----------------------------\\\\
\cfrac{1-cos(x)}{sin(x)}+\cfrac{1}{\frac{1-cos(x)}{sin(x)}}\implies \cfrac{1-cos(x)}{sin(x)}+\cfrac{sin(x)}{1-cos(x)}
\\\\\\
\cfrac{[1-cos(x)]^2+sin^2(x)}{sin(x)[1-cos(x)]}\implies 
\cfrac{1-2cos(x)+\boxed{cos^2(x)+sin^2(x)}}{sin(x)[1-cos(x)]}
\\\\\\
\cfrac{1-2cos(x)+\boxed{1}}{sin(x)[1-cos(x)]}\implies \cfrac{2-2cos(x)}{sin(x)[1-cos(x)]}
\\\\\\
\cfrac{2[1-cos(x)]}{sin(x)[1-cos(x)]}\implies \cfrac{2}{sin(x)}\implies 2\cdot \cfrac{1}{sin(x)}\implies 2csc(x)$

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3 years ago