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ss7ja [257]
2 years ago
9

Solve 70=-2.5x (one-step equation)

Mathematics
1 answer:
Lapatulllka [165]2 years ago
7 0

Answer:

-28

Step-by-step explanation

70/-2.5=-2.5x/-2.5      divide

-28=x  

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Answer:the median is 20 and the lower quartile is 16.5

Step-by-step explanation:

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Find the slope of 9x=3y+5
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<u>9x</u> = <u>3y + 5</u>
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x = 1/3y + 5/9
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A football team lost 4 yards on each of 2 plays, gained 14 yards on the third play, and lost 5 yards on the forth play. Write an
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3 years ago
The table of values represents the function t(x) and the graph shows the function g(x).
IrinaK [193]

Answer:

  The maximum value of the table t(x) has a greater maximum value that the graph g(x)

Step-by-step explanation:

The table shows t(x) has two (2) x-intercepts: t(-3) = t(5) = 0. The graph shows g(x) has two (2) x-intercepts: g(1) = g(5) = 0. Neither function has fewer x-intercepts than the other.

The table shows the y-intercept of t(x) to be t(0) = 3. The graph shows the y-intercept of g(x) to be g(0) = -1. The y-intercepts are not the same, and that of t(x) is greater than that of g(x).

The table shows the maximum value of t(x) to be t(1) = 4. The graph shows the maximum value of g(x) to be g(3) = 2. Thus ...

  the maximum value of t(x) is greater than the maximum value of g(x)

8 0
2 years ago
Use the exponential decay​ model, Upper A equals Upper A 0 e Superscript kt​, to solve the following. The​ half-life of a certai
Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

A=A_0 e^{kt}

Where,

A_0 = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = \frac{A_0}{2},

i.e.

\frac{A_0}{2}=A_0 e^{19k}

\frac{1}{2} = e^{19k}

0.5 = e^{19k}

Taking ln both sides,

\ln(0.5) = \ln(e^{19k})

\ln(0.5) = 19k

\implies k = \frac{\ln(0.5)}{19}\approx -0.03648

Now, if the substance to decay to 78​% of its original​ amount,

Then A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0

0.78 A_0=A_0 e^{-0.03648t}

0.78 = e^{-0.03648t}

Again taking ln both sides,

\ln(0.78) = -0.03648t

-0.24846=-0.03648t

\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0
2 years ago
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