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Angelina_Jolie [31]
3 years ago
7

Work out the Real distance 6cm represents. Give your answer in kilometres

Mathematics
1 answer:
lilavasa [31]3 years ago
3 0
Check the picture below.

keep in mind that Bearings on the North-South line are angles that start off at the North line and go clockwise.

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PLEASE HELP DOUBLE POINTS!!!!
kow [346]

Answer:

I'm pretty sure it's d

Step-by-step explanation:

20 percent is the discount. Everything has 100%, so if you take away 20% of the 68$ you will get only the amount of the discount; however, we are looking for the price after the discount. Since 20% is the discount, then perhaps 80% is the cost after the discount is used; therefore, if you multiply .8 which is equal to 80% to 68, you will get the cost after the discount. The answer would be .8(68)=T

4 0
3 years ago
Consider the graph of the linear function h(x) = –x + 5. Which could you change to move the graph down 3 units?
Feliz [49]
\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

\bf h(x)=-x+5\implies h(x)=\stackrel{A}{-1}(\stackrel{B}{1}x\stackrel{C}{+0})\stackrel{D}{+5}

a shift down by 3 units, means a vertical shift downwards, so D needs to drop by 3 units.

\bf h(x)=\stackrel{A}{-1}(\stackrel{B}{1}x\stackrel{C}{+0})\boxed{\stackrel{D}{+5-3}}\implies h(x)=-1(1x+0)+2
\\\\\\
h(x)=-x+2
4 0
3 years ago
Daniel is 5 feet tall. Every year he grows 5% taller. What will be his height next year?
forsale [732]
I'm pretty sure he would be 5 ft 3 in. tall.
8 0
3 years ago
Read 2 more answers
Describe the prime factorization of 180
Sindrei [870]

Answer:

<h2>5 x 2 x 2 x 3 x 3 = 180</h2>

Step-by-step explanation:

HOPE THIS HELPS’

PLZZ MARK BRAINLIEST

7 0
3 years ago
The most confused that i have been
Trava [24]

Answer:

(-5, -3)

Step-by-step explanation:

To graph the two equations, you can look at two things, the slope and the y-intercept.

The two equations are in slope-intercept form, y= mx + b. m is the slope and b is the y-intercept. The slope is rise/run and the y-intercept is where the line crosses the y-axis.

In the first equation, the slope is -1 and the y-intercept is -8. So, to graph you would start at (0, 8) and move up one unit and then left one unit to get the next point.

For the second equation, the slope is 2/5 and the y-intercept is -1. So, to graph you would do the same thing. Start at (0, -1) and then go up 2 units and right 5 units to get the next point.

I've attached a graph below if you need it.

6 0
3 years ago
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