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Andreyy89
3 years ago
6

Val walks 2 3/5 miles each day.Bill runs 10 miles once every 4 days. In 4 days, who covers the greatest distance?

Mathematics
2 answers:
lana [24]3 years ago
5 0

Answer:

Val

Step-by-step explanation:

2 3/5 = 10.4

bill only runs 10 miles so Val runs 0.4 more than Bill

Alik [6]3 years ago
3 0

Answer:

Val

Step-by-step explanation:

10mi/4days=2.5mi/day

2 3/5=2.6

2.6>2.5

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Evaluate this expression for x=-5 and y=-5​
Talja [164]

Answer:

x - 2y=5

Step-by-step explanation:

Given

x = -5; y=-5

x - 2y --- missing from question

Required

Evaluate

We have:

x - 2y

Substitute for x and y

x - 2y=-5 - 2*-5

x - 2y=-5 +10

x - 2y=5

3 0
3 years ago
Use the equation below to find c, if a = 43 and b = 49. c=180-a-) ?​
Luden [163]

Answer:

88

Step-by-step explanation:

43 + 49= 92.

Then 180 - 92= 88

C= 88

7 0
3 years ago
Each square below represents one whole.<br> What percent is represented by the shaded area?
KATRIN_1 [288]

9514 1404 393

Answer:

  110%

Step-by-step explanation:

Each stripe is 1/10 of a whole, so represents 100%/10 = 10% of the whole.

There are 11 shaded stripes, representing 11×10% = 110%.

4 0
3 years ago
What is the solution of the system of equations below?
Nostrana [21]
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2x+3=7 \ \ |-3\\ \\ 2x+3-3=7-3\\ \\2x=4 \ \ /:2 \\ \\x=2 \\ \\ \begin{cases} x=2 \\ y=1 \end{cases}


3 0
3 years ago
Read 2 more answers
A car was valued at $39,000 in the year 1995. The value depreciated to $11,000 by the year 2003.
denis23 [38]

Answer:

14.6328% , $5836.03

Step-by-step explanation:

Here we are going to use the formula

A_{0}(1-r)^n = A_{n}

A_{0} = 39000

r=?

A_{8} = 11000

n=8

Hence

39000(1-r)^8 = 11000

(1-r)^8 = \frac{11000}{39000}

(1-r)^8 = 0.2820

(1-r) = 0.2820^{\frac{1}{8}

(1-r) = 0.2820^{0.125}

(1-r) = 0.8536

(1-0.8536=r

r = 0.1463

Hence r= 0.1463

In percentage form r = 14.63%

Now let us see calculate the value of car in 2003 that is after 12 years

we use the main formula again

A_{0}(1-r)^n = A_{n}

A_{0} = 39000

r=0.1463

A_{12} = ?

n=12

39000(1-0.14634)^{(12} = A_{12}

39000(0.8536)^{12} = A_{12}

39000*0.1497 = A_{12}

A_{12}=5840.34

Hence the car's value will be depreciated to $5840.34 (approx) by 2003.

5 0
3 years ago
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