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emmasim [6.3K]
3 years ago
9

How much dirt is there in a hole 3 feet deep, 6 ft long and 4 ft wide?

Mathematics
2 answers:
jasenka [17]3 years ago
4 0
Ok its a hole :) hope this helped
borishaifa [10]3 years ago
3 0
None. There can't be any dirt in it because it's a hole.
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If you can walk 3 1/3 miles in an hour, how many miles can you walk in 30 minutes, given the
Vedmedyk [2.9K]

Answer:

d = 1.67 miles

Step-by-step explanation:

Given that,

Speed of a person,  s=3\dfrac{1}{3}\ mph

We need to find how many miles can you walk in 30 minutes, given the  same pace.

30 minutes = 0.5 h

3\dfrac{1}{3}\ mph = \dfrac{10}{3}\ mph

Speed = distance/time

d=v\times t\\\\d=\dfrac{10}{3}\ mph\times 0.5\ h\\\\d=1.67\ miles

So, the person will cover a distance of 1.67 miles

8 0
3 years ago
I only need 6-10 pleaseeeeswws
Genrish500 [490]
6) 381/100
7) 231/1000
8) -303//25
9) 9/1000
10) -227/50
3 0
2 years ago
Nick bought a new car.
ira [324]

Using an exponential function, it is found that it takes 5.42 years for the car to halve in value.

<h3>What is an exponential function?</h3>

A decaying exponential function is modeled by:

A(t) = A(0)(1 - r)^t

In which:

  • A(0) is the initial value.
  • r is the decay rate, as a decimal.

In this problem, the car depreciates 12% a year in value, hence r = 0.12 and the equation is given by:

A(t) = A(0)(0.88)^t.

It halves in value at t years, for which A(t) = 0.5A(0), hence:

A(t) = A(0)(0.88)^t

0.5A(0) = A(0)(0.88)^t

(0.88)^t = 0.5

\log{(0.88)^t} = \log{0.5}

t\log{0.88} = \log{0.5}

t = \frac{\log{0.5}}{\log{0.88}}

t = 5.42.

It takes 5.42 years for the car to halve in value.

More can be learned about exponential functions at brainly.com/question/25537936

#SPJ1

9 0
1 year ago
1/10( x + 130) = -2(3-X)<br><br> I need help
Artist 52 [7]

Answer:

x = 10

Step-by-step explanation:

1/10(x + 130) = -2(3 - x)

x + 130 = -20(3-x)

x + 130 = -60 + 20x

x - 20x = -60 - 130

-19x = -190

x = 10

5 0
2 years ago
Question
kozerog [31]
So to find the width (w)
26=2(7+w). W=6
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2 years ago
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