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Ann [662]
3 years ago
13

Alex ran 12.8 km in the morning. After​ lunch, he continued running. When he finished the​ run, he had covered 31.4 km in all. W

rite an equation that can be used to find how far Alex ran after lunch.
Let r be the distance in kilometres that Alex ran after lunch.
Mathematics
1 answer:
Ulleksa [173]3 years ago
3 0

Answer:

12.8 km + r = 31.4 km

Step-by-step explanation:

First, Alex ran 12.8 km, then he ran an r amount of km. Added together they equal 31.4 km.

If you need to solve the equation you subtract 12.8 from 31.4 and will give you the amount of miles he ran after lunch.

Ex.: r = 31.4 km - 12.8 km

      r = 18.6 km

Alex ran 18.6 km after lunch

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5 grams of chucknorrisidium created, decays at a half-life of 16,000,000 Need 1 gram for experiments. How long before the is 1 g
devlian [24]

Answer:

the time taken for the element to remain 1 gram is 38,400,000 unit.

Step-by-step explanation:

Given;

initial mass, m = 5 g

half life, t = 16,000,000 unit

The time taken for the element to remain 1 gram is calculated as;

          mass remaining                                      time

             5 g                                                        0

             2.5 g                                                    16,000,000

             1.25 g                                                   32,000,000

            0.625 g                                                 48,000,000

Interpolate to obtain the time at 1 gram;

1.25 g ----------------------------------   32,000,000

1 g --------------------------------------------- x

0.625 g --------------------------------------- 48,000,000

\frac{1.25 - 1}{1.25 - 0.625} = \frac{32,000,000 - x}{32,000,000 - 48,000,000} \\\\\frac{0.25}{0.625} = \frac{32,000,000 - x}{-16,000,000} \\\\-4,000,000= 20,000,000 - 0.625x\\\\0.625x = 20,000,000 + 4,000,000\\\\0.625x =  24,000,000\\\\x = \frac{24,000,000}{0.625} \\\\x = 38,400,000

Therefore, the time taken for the element to remain 1 gram is 38,400,000 unit.

         

8 0
3 years ago
Bob can body paint 12 cars in 3 weeks. How many cars can Bob body paint in 8 weeks?
Oksana_A [137]

Answer:

32 cars

Step-by-step explanation:

we know that he can paint 12 cars in 3 weeks.

that means that he can paint 4 cars per week.

4cars x 8weeks = 32 cars in 8 weeks

3 0
3 years ago
Read 2 more answers
You can text 44 words in 1 minute. Write a proportion to help you find
Natalka [10]

Answer:

176 words

Step-by-step explanation:

1) well we want to find the amount of words we'd be able to type in 4 minutes. so, what i did was:

<u>44 words</u> = <u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>x</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

1 minutes = 4 minutes

2) now look at how i set up the proportion, notice how i left a variable of x for the value i didn't know.

3) we also know that 1 times 4 is 4, so it'd also make sense to also multiply the numerator by 4 too!!

4) multiplying 44 by 4 would get us our answer which is 176

im not the best at math but i this explanation helped <33

5 0
3 years ago
Find the point on the parabola y^2 = 4x that is closest to the point (2, 8).
guapka [62]

Answer:

(4, 4)

Step-by-step explanation:

There are a couple of ways to go at this:

  1. Write an expression for the distance from a point on the parabola to the given point, then differentiate that and set the derivative to zero.
  2. Find the equation of a normal line to the parabola that goes through the given point.

1. The distance formula tells us for some point (x, y) on the parabola, the distance d satisfies ...

... d² = (x -2)² +(y -8)² . . . . . . . the y in this equation is a function of x

Differentiating with respect to x and setting dd/dx=0, we have ...

... 2d(dd/dx) = 0 = 2(x -2) +2(y -8)(dy/dx)

We can factor 2 from this to get

... 0 = x -2 +(y -8)(dy/dx)

Differentiating the parabola's equation, we find ...

... 2y(dy/dx) = 4

... dy/dx = 2/y

Substituting for x (=y²/4) and dy/dx into our derivative equation above, we get

... 0 = y²/4 -2 +(y -8)(2/y) = y²/4 -16/y

... 64 = y³ . . . . . . multiply by 4y, add 64

... 4 = y . . . . . . . . cube root

... y²/4 = 16/4 = x = 4

_____

2. The derivative above tells us the slope at point (x, y) on the parabola is ...

... dy/dx = 2/y

Then the slope of the normal line at that point is ...

... -1/(dy/dx) = -y/2

The normal line through the point (2, 8) will have equation (in point-slope form) ...

... y - 8 = (-y/2)(x -2)

Substituting for x using the equation of the parabola, we get

... y - 8 = (-y/2)(y²/4 -2)

Multiplying by 8 gives ...

... 8y -64 = -y³ +8y

... y³ = 64 . . . . subtract 8y, multiply by -1

... y = 4 . . . . . . cube root

... x = y²/4 = 4

The point on the parabola that is closest to the point (2, 8) is (4, 4).

4 0
3 years ago
write an equation in slope intercept form for the line that passes through (4, -4) and is parallel to 3x+4x=2y-9
photoshop1234 [79]

Answer:

\large\boxed{y=\dfrac{7}{2}x-18}

Step-by-step explanation:

The slope-intercept form of an equation of a line:

y=mx+b

Convert the equation of a line 3x + 4x = 2y - 9 to the slope-intercept form:

3x+4x=2y-9

7x=2y-9             <em>add 9 to both sides</em>

7x+9=2y       <em>divide both sides by 2</em>

\dfrac{7}{2}x+\dfrac{9}{2}=y\to y=\dfrac{7}{2}x+\dfrac{9}{2}

Parallel lines have the same slope. Therefore we have the equation:

y=\dfrac{7}{2}x+b

Put the coordinates of the point (4, -4) to the equation:

-4=\dfrac{7}{2}(4)+b

-4=7(2)+b

-4=14+b       <em>subtract 14 from both sides</em>

-18=b\to b=-18

Finally we have the equation:

y=\dfrac{7}{2}x-18

8 0
3 years ago
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