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Lostsunrise [7]
3 years ago
7

Charles can drive 345 miles on a full tank of gas. If he has driven 134 miles so far, how many miles can he drive before he runs

out of gas?
Mathematics
2 answers:
Nookie1986 [14]3 years ago
7 0
Charles can drive 211 miles before he runs out of gas
Paul [167]3 years ago
7 0
If you subtract the two numbers you will get 211

The answer is 211 miles until he runs out of gas
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Choose the best estimate for the division problem below.
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Answer:

10

Step-by-step explanation:

68.362/7.12

Multiply the top and bottom by 1000

68362/7120

Rounding to the nearest 1000

70000/7000

10

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Three times the difference of a number and 9 is -3. Find the number
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3(n-9)=-3
3n-27=-3
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Evaluate each function following the specification.
telo118 [61]

Answer:

#3 a. g(-1) = 2, g(0) = 3, and g(1) = 2

b. No restrictions for all real numbers

\#4 \  a. \ h(-1) = \dfrac{1}{3}, \  h(0) =0, \  h(2) =  \infty

b. Yes, <em>x ≠ 2</em>

Step-by-step explanation:

#3 The function is given as g(x) = -x² + 3

a. From the given function, by plugging in the value of 'x' in the bracket, we have;

g(-1) = -(-1)² + 3 = -1 + 3 = 2

g(-1) = 2

g(0) = -0² + 3 = 3

g(0) = 3

g(1) = -1² + 3 = -1 + 3 = 2

g(1) = 2

g(-1) = 2, g(0) = 3, and g(1) = 2

b. The given function g(x) = -x² + 3 for finding the value of <em>g</em> can take any value of <em>x</em> which is a real number

Therefore, therefore, there are no restrictions

#4 a. The given function is given as follows;

h(x) = \dfrac{x}{x - 2}

By substitution, we get;

h(-1) = \dfrac{-1}{(-1) - 2} = \dfrac{-1}{-3} = \dfrac{1}{3}

\therefore h(-1) = \dfrac{1}{3}

h(0) = \dfrac{0}{0 - 2} = \dfrac{0}{-2} =0

\therefore h(0) =0

h(2) = \dfrac{2}{2 - 2} = \dfrac{2}{0} = \infty

\therefore h(2) =  \infty

h(-1) = \dfrac{1}{3}, \  h(0) =0, \  h(2) =  \infty

b. From the values of the function, we have that h(x) is not defined at x = 2

Therefore, there is a restriction for <em>x</em> in the function, which is <em>x ≠ 2</em>

3 0
3 years ago
A building has two sizes of apartments Smart in regular the ratio of small apartments to regular apartments is 18 to 7 what perc
tino4ka555 [31]

Answer:

72 percent of the total houses are small

Step-by-step explanation:

The first step is to know the total apartments from the ratio: This is got by adding 18 + 7 = 25

The ratio of small apartments to regular ones is 18:7

To get the percentage that is small, we have to express the number of small houses when compared to the total available apartments. (<em>Note: not the regular ones) </em>as a percentage.

This will be (18/25) X 100 = 72%

Therefore, 72 percent of the total houses are small

3 0
3 years ago
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