Ask question

Ask question

All categories

3 years ago

15

Ted likes to run long distances. He can run 20 \text{ km}20 km20, start text, space, k, m, end text in 959595 minutes. He wants

to know how many kilometers (k)(k)left parenthesis, k, right parenthesis he will go if he runs at the same pace for 285285285 minutes.

1 answer:

Ghella [55]3 years ago

5 0

Answer:

im looking for the anwser

Step-by-step explanation:

You might be interested in

Javier throws two dice and does not care about the order but only that a 6 and a 5 fall, what is the probability that this will

FinnZ [79.3K]

Not very much because things don't always happen the way we want them too

7 0

3 years ago

Please help with this

KonstantinChe [14]

I believe the answer is 32

4 0

3 years ago

The mass of a bar of chocolate is 1800g.

Umnica [9.8K]

Hello,

<span>Density = Mass/Volume

9 = 1800/volume
9volume = 1800
volume = 200 cm^3</span><span>
</span>
The answers is 200 cm^3

Hope that helped :)

5 0

3 years ago

Read 2 more answers

HEEEEEEEEEEEEELLLLLLLLLLPPPPPPPPPP

Vika [28.1K]

The answer is B, 5/6, because when you multiply 4 by 5 it is 20, and when you multiply 5 by 6, you get 30, so 20/30. And, 20/30=2/3. I hope I answered your question!

4 0

2 years ago

Write an equation that has a graph with the shape of y = x^2, but shifted left 3 units.

svetoff [14.1K]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf % template detailing
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

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

now, with that template in mind, let's see

$\bf y=x^2\implies 
\begin{array}{llll}
y=(&1x&-0)^2\\
&B&C
\end{array}$

so, just change C to +3, thus C/B is 3/1

6 0

2 years ago