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SVEN [57.7K]
3 years ago
6

A student on the track team runs 45 minutes each day as a part of her training. She

Mathematics
1 answer:
wel3 years ago
3 0

Answer:D

Step-by-step explanation:

The basic equation to solve this would be D = RT, which is

D is Distance

R is Rate (speed)

T is time

To find total distance D, we can find individual distances for two legs of the whole.

First Leg:

R = 8

T = a

D = 8a

Step-by-step explanation:

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Two equivalent ratios for 12/60
Marat540 [252]
If you would like to know two equivalent ratios for 12/60, you can calculate this using the following steps:

12/60 = 12/2 / 60/2 = 6/30
6/30 = 6/2 / 30/2 = 3/15
3/15 = 3/3 / 15/3 = 1/5

The result would be 6/30 or 3/15 or 1/5.
8 0
3 years ago
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What is the area of a triangle whose vertices are P(−1, 4), Q(3, 2), and R(3, −4)?
sp2606 [1]
What kind of triangle is it?
7 0
3 years ago
In 1985, the cost of clothing for a certain familiy was $620. In 1995, 10 years later, the cost of clothing for this famility wa
Sholpan [36]

Answer:

$848

Step-by-step explanation:

Calculation for the cost of this familiy's clothing in 1991

First step is to calculate the amount that is increase per year

Increase per year= ($1000-$620)/(1995-1985)

Increase per year= 380/10

Increase per year= $38

Now let y be :38*x + $620 and let x be 6 years (1991-1985)

Second step is to calculate the cost of the clothing in 1991

y = 38*6 + 620

y=228+620

y = $848

Therefore the cost of this familiy's clothing in 1991 will be $848

6 0
3 years ago
Find the function value, if possible. (If an answer is undefined, enter UNDEFINED.)
ddd [48]

Answer:

f(x + 2) = 3x + 2

Step-by-step explanation:

Simply replace wherever you find x in the function f(x) with (x + 2), like so:

f(x + 2) = 3(x + 2) - 4

f(x + 2) = 3x + 6 - 4

f(x + 2) = 3x + 2

7 0
3 years ago
Help me pls.........!
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If each square represents a square foot half of it would be .5 so then add all of them. The formula for Area is A=l•w and the Area formula for a triangle is A=hbb/2. Hope this helps :)
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3 years ago
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