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Kipish [7]
2 years ago
6

What is the distance between the points (21 , 11) and (4 , 11) in the coordinate plane?

Mathematics
1 answer:
mash [69]2 years ago
4 0
The different is that 21 is going ALL the way down to 21 while 4 is only going to 4..
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Convert the following equation<br> into standard form.<br> y = 2x – 4<br> -2x + y = [?]
zlopas [31]

Answer:

-4

Step-by-step explanation:

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Please help solve -2|5y-1|=-10
sweet-ann [11.9K]
Alright, so we can start by dividing -2 from both sides, getting |5y-1|=20. Then, since 5y-1 is in an absolute value, 5y-1 is either 20 or -20.
In 5y-1=20, we can add one to both sides, getting 5y=21 and y=21/5. In
5y-1=-20, we can add one again, getting 5y=-19 and y=-19/5.

If you have any more equations, make sure to plug both numbers in to check, but otherwise y has two answers , which are -19/5 and 21/5.
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3 years ago
The patch of the blade has a circumference of about 12.5 inches. What is the radius of the model.
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3 years ago
PLEASE HELP find the arch shown in red leave your answer in terms of pi
Norma-Jean [14]

Answer:

Answer:1296π

Step-by-step explanation:

As we know,

Area of cicle=πr^2

= π(36)^2

=1296π

6 0
2 years ago
Let h(x)=20e^kx where k ɛ R (Picture attached. Thank you so much!)
zloy xaker [14]

Answer:

A)

k=0

B)

\displaystyle \begin{aligned} 2k + 1& = 2\ln 20 + 1 \\ &\approx 2.3863\end{aligned}

C)

\displaystyle \begin{aligned} k - 3&= \ln \frac{1}{2} - 3 \\ &\approx-3.6931 \end{aligned}

Step-by-step explanation:

We are given the function:

\displaystyle h(x) = 20e^{kx} \text{ where } k \in \mathbb{R}

A)

Given that h(1) = 20, we want to find <em>k</em>.

h(1) = 20 means that <em>h</em>(x) = 20 when <em>x</em> = 1. Substitute:

\displaystyle (20) = 20e^{k(1)}

Simplify:

1= e^k

Anything raised to zero (except for zero) is one. Therefore:

k=0

B)

Given that h(1) = 40, we want to find 2<em>k</em> + 1.

Likewise, this means that <em>h</em>(x) = 40 when <em>x</em> = 1. Substitute:

\displaystyle (40) = 20e^{k(1)}

Simplify:

\displaystyle 2 = e^{k}

We can take the natural log of both sides:

\displaystyle \ln 2 = \underbrace{k\ln e}_{\ln a^b = b\ln a}

By definition, ln(e) = 1. Hence:

\displaystyle k = \ln 2

Therefore:

2k+1 = 2\ln 2+ 1 \approx 2.3863

C)

Given that h(1) = 10, we want to find <em>k</em> - 3.

Again, this meas that <em>h</em>(x) = 10 when <em>x</em> = 1. Substitute:

\displaystyle (10) = 20e^{k(1)}

Simplfy:

\displaystyle \frac{1}{2} = e^k

Take the natural log of both sides:

\displaystyle \ln \frac{1}{2} = k\ln e

Therefore:

\displaystyle k = \ln \frac{1}{2}

Therefore:

\displaystyle k - 3 = \ln\frac{1}{2} - 3\approx-3.6931

3 0
2 years ago
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