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trasher [3.6K]
3 years ago
7

Imagine if you wrapped a rope tightly around the earth at the equator. how much longer would you have to make the rope if you wa

nted it to be exactly one foot above the surface all the way around?
Mathematics
1 answer:
jasenka [17]3 years ago
6 0
Answer: Use the formula: 2(pi)(r + 1) - 2(pi)r

In your problem, you didn't give any measurements to use, so we can't determine an exact answer. However, the given formula above would work. 

The first term calculates the extended length of rope. The second term calculates the original length of rope. All you need to do is subtract them.

Online, you can find the radius of the Earth if you need an exact value. 3,959 miles
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The graph of g(x) = -x^2 is shifted 3 units left and 1 unit up. If this new graph is f(x), then what is the value of f(-1.5)
kondor19780726 [428]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\end{array}\\\\
--------------------\\\\

\bf \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{  D}}\\
\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\
\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind, let's see.

to the left by 3 units, C = +3
up by 1  unit, D = 1

\bf g(x)=-x^2\\\\\\ g(x)=-1(1x+\stackrel{C}{0})^2+\stackrel{D}{0}\implies g(x)=-1(1x+3)^2+1
\\\\\\
g(x)=-(x+3)^2+1\impliedby f(x)\qquad \qquad f(-1.5)=-[(-1.5)+3]^2+1

and surely you know how much that is.
5 0
4 years ago
# 15. Q. Find the area of the trapezoid
harina [27]
Answer:
219.96 yd²

Explanation:
The area of the trapezoid can be calculated using the following rule:
Area of trapezoid = \frac{b1 + b2}{2} * h
where:
b1 is the length of the first base
b2 is the length of the second base
h is the height

Now, in the given we have:
b1 = 15.8 yd
b2 = 21.8 yd
h = 11.7 yd

Substitute with the givens in the above equation to get the area as follows:
area = \frac{15.8 + 21.8}{2}  * 11.7 = 219.96 yd^2

Hope this helps :)
7 0
3 years ago
Write the equation of the line that has a slope of m=3 and passes through (4 , 2)
mina [271]

Answer:

y=3x-10

Step-by-step explanation:

For this, you would use the point slope formula, which is (y-y1)=m(x-x1). x1 and y1 are the coordinates from the ordered pair.

(y-y1)=m(x-x1)

(y-2)=3(x-4)

y-2=3x-12

y=3x-10

To check, you can plug the coordinates into the equation.

y=3x-10

(2)=3(4)-10

2=12-10

2=2

3 0
3 years ago
What is the value of a and b?
prohojiy [21]
A = 34 x 2 = 68
b = 56/2 = 28

hope it helps
3 0
3 years ago
Read 2 more answers
At noon, the temperature was 59º. The temperature rose an average of 2° per hour for the next 6 hours. What was the temperature
saw5 [17]

Answer:

The temperature was 71 degrees at 6pm.

5 0
3 years ago
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