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3 years ago

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The scale of a map is 5 mm = 2 feet. What is the map distance if the actual distance is 50 feet. BE SURE TO LABEL!

2 answers:

Vikentia [17]3 years ago

7 0

It would probably be 20 feet if I’m not wrong

Murljashka [212]3 years ago

5 0

It should be 20 feet!!!!!!!!!!!

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Assume the parent function has been shifted up

mars1129 [50]

Answer:

If $y=(x-2)^2+4$ is the transformed function, then the parent function $y=x^2$ has been shifted up vertically by 4 units and horizontally right by 2 units

Step-by-step explanation:

Assume the function is $y=f(x)$ is the parent function.

A vertical shift,b units up will transform the function to $y=f(x)+b$ .

A horizontal shift $a$ units.

$y=f(x-a)+b$

For instance, if $y=x^2$ , b=4, and a=2 units.

Then the transformed function becomes:

$y=(x-2)^2+4$

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3 years ago

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valentina_108 [34]

She can travel <span>665.4375 miles in 68.25 min.</span>

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3 years ago

What is longer: a mile or a kilometer?

kotykmax [81]

A miler is longer than a kilometer.

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3 years ago

The question is this;

zheka24 [161]

Step-by-step explanation:

X is 24

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Express the function as the sum of a power series by first using partial fractions. f(x) = 8 x2 − 4x − 12 f(x) = ∞ n = 0 find th

alexandr1967 [171]

I'm guessing the function is

$f(x)=\dfrac8{x^2-4x-12}=\dfrac8{(x-6)(x+2)}$

which, split into partial fractions, is equivalent to

$\dfrac1{x-6}-\dfrac1{x+2}$

Recall that for $|x|$ we have

$\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n$

With some rearranging, we find

$\dfrac1{x-6}=-\dfrac16\dfrac1{1-\frac x6}=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n$

valid for $\left|\dfrac x6\right|$ , or $|x|$ , and

$\dfrac1{x+2}=\dfrac12\dfrac1{1-\left(-\frac x2\right)}=\displaystyle\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

valid for $\left|-\dfrac x2\right|$ , or $|x|$ .

So we have

$f(x)=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n-\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

$f(x)=\displaystyle-\sum_{n=0}^\infty\left(\frac{x^n}{6^{n+1}}+\frac{(-x)^n}{2^{n+1}}\right)$

$f(x)=\displaystyle-\sum_{n=0}^\infty\frac{1+3(-3)^n}{6^{n+1}}x^n$

Taken together, the power series for $f(x)$ can only converge for $|x|$ , or $-2$ .

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3 years ago