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Deffense [45]
3 years ago
8

After learning how to fly, wendy reduced her daily commute time by 75%, percent. Previously, her commute took m minutes.what is

a expression that could represent her commute time in minutes after she learned how to fly?
Mathematics
2 answers:
Basile [38]3 years ago
5 0
The answer is:
Commute = m-m(0.75)
Tanya [424]3 years ago
4 0
That the answer. will be 0.7 because you put it ib similarly
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Find equation of a line through 5 -3 that is parallel to y=1/2x+3
xxTIMURxx [149]

Answer:

The equation of a line through (5 -3) that is parallel to y = 1/2 x+3 is

y = - 2 x  + 7

Step-by-step explanation:

Let us assume the slope of the line whose equation we need to find is m 1.

The line parallel to the needed line  is:   y=1/2x+3

Comparing it with the general form: y = m x + C

we get m 2 = 1/2

Now, as Line 1 is Perpendicular to Line 2.

⇒ m 1 x m 2  = -1

⇒ m 1 x ( 1/2)  = -1

⇒ m 1  = - 2

Also, the point son the line 1 is given as: (x,y)  = (5,-3)

Put the value of point and Slope in y = m x + C to find the value of Y- INTERCEPT.

we get: -3 = (-2) (5) +  C

or, C = -3 + 10 = 7

⇒ C = 7

The general line equation is given as:  y = m x + C

Substituting  the values of C and m, we get:

y = - 2 x  + 7

Hence, the equation of a line through 5 -3 that is parallel to y = 1/2 x+3 is

y = - 2 x  + 7

8 0
3 years ago
An oil tank is leaking into a lake at a rate of 0.1 m3/day. The oil slick forms a semicircular disk whose center is at the leak
gregori [183]

Answer:

17.9  m/s

Step-by-step explanation:

Volume of the slick =  0.5 x π r² h--------------------------------- (1)

Where r = radius of slick

           h = thickness of slick, 10⁻⁶m

If 0.5m³ of oil leaked, then the radius of the semicircular slick can be calculated from equation (1)

V  =  0.5 x π r² h

0.5=  0.5 x  π x  r² x 10⁻⁶

r²   =  10⁶/ π

r =     10³/√π

dV/dt =   πrh dr/dt  + 0.5π r² dh/dt----------------------------------- (2)

Asumming the film thickness is constant , equation (2) becomes

dV/dt =   πrh dr/dt-------------------------------- (3)

dV/dt = 0.1m³/day

r=  10³/√π

dr/dt= rate of expansion of the slick

Substituting  into (3);

0.1 =  π x 10³/√π x 10⁻⁶ x dr/dt

dr/dt = 0.1  x 10⁶/ ( π x 10³/√π)

       =  17.9479 m/s

       ≅  17.9  m/s

8 0
3 years ago
Which ordered pairs is a solution to −5x + 3y > 12?
In-s [12.5K]

<u>Answer </u><u>:</u><u>-</u>

Given inequality ,

  • -5x +3y > 12
  • 3y > 5x + 12
  • y > 5/3x + 12/3
  • y > 5/3x + 4

From the options ,

<u>When </u><u>x </u><u>is </u><u>3</u><u> </u><u>and </u><u>y </u><u>=</u><u> </u><u>9</u>

  • y >1* 5 + 4
  • y > 5 +4
  • 9> 9 ( Not possible )

<u>When</u><u> </u><u>x </u><u>is </u><u>-</u><u>5</u><u> </u><u>y</u><u> </u><u>is </u><u>5</u><u> </u>

  • y > 5/3*-5 + 4
  • y > -8.3 + 4
  • 5 > -3.7 ( Possible )

<u>When </u><u>x </u><u>is </u><u>3</u><u> </u><u>y </u><u>is </u><u>-</u><u>6</u><u> </u>

  • y > 5 + 4
  • -6 > 9 ( Not possible )

<u>When </u><u>x </u><u>is </u><u>-</u><u>2</u><u> </u><u>y </u><u>is </u><u>-</u><u>5</u><u> </u>

  • y > -3.33 + 4
  • -5 > -0.67 ( Possible )

<u>When </u><u>x </u><u>is </u><u>2</u><u> </u><u>y </u><u>is </u><u>8</u><u> </u>

  • y > 3.33 + 4
  • 8 > 7.77 ( possible )

<u>When</u><u> </u><u>x </u><u>is </u><u>-</u><u>6</u><u> </u><u>y </u><u>is </u><u>0</u><u> </u>

  • y > -10 +4
  • 0 > -6 ( possible )
7 0
3 years ago
Read 2 more answers
Determine whether the graph table or equation represents a linear or nonlinear function explain
Anna [14]
<h3>Answer: Linear</h3>

Each time x goes up by 3, y goes up by 2. The rate of change is constant.

slope = (change in y)/(change in x) = 2/3

y intercept = 6 since this is where x = 0

The equation of this line is y = (2/3)x+6

5 0
3 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
3 years ago
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