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2 years ago

Select the correct answer from the multiple choice below.

2 answers:

7 0

28 mi/gallon
In the beginning Sully has 16 gallons of fuel, and can drive
16 gallons*28 mi/gallons=448 mi
At the x=0, y=448 mi - this is y-intercept
28 mi/h is a slope ( for one h - 28 mi, for 2 h - 28*2 and so on)

The function is <span>y=448-28x.
Answer is C.
</span>

gtnhenbr [62]2 years ago

5 0

(Could easily be wrong, Math is not my strongsuit) I think its D, because you are suppose to convert the fuel tank capacity to the miles per gallon.

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Find the equation whose graph is shown below. Write your answer in standard form.

vladimir2022 [97]

Answer:

it as a liner function that is why we should choose to point in order to find the liner equation

y=2/3x+1

5 0

2 years ago

Helppppppppppppppppppppppppp

Wewaii [24]

Answer:

Radius: 8

Circumference: 50.27

πr^2 = Area of circle

201.06 = πr^2

divide by pi

201.06 / π = 63.9993857161 = 64

64 = r^2

Square root both sides to get radius.

$\sqrt{64}$ = 8

r = 8

Circumference = π x diameter

Diameter = 2 x radius

Circumference = π x 16 = 50.2654824574 = 50.27

8 0

3 years ago

If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t

sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, $6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296$

This implies that $6^{4}$ exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, $6^{3}=216$ and $216\times 5=1080, 216\times 4=864$ . This implies that $216\times 5$ exceeds 1052 and thus there can be at most 4 groups of $6^{3}$ .

Then,

$1052-4\times6^{3}$

$1052-4\times216$

$1052-864$

$188$

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, $6^{2}=36$ and $36\times 5=180, 36\times 6=216$ . This implies that $36\times 6$ exceeds 188 and thus there can be at most 5 groups of $6^{2}$ .

Then,

$188-5\times6^{2}$

$188-5\times36$

$188-180$

$8$

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, $6^{1}=6$ and $6\times 1=6, 6\times 2=12$ . This implies that $6\times 2$ exceeds 8 and thus there can be at most 1 group of $6^{1}$ .

Then,

$8-1\times6^{1}$

$8-1\times6$

$8-6$

$2$

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, $6^{0}=1$ and $1\times 2=2$ . This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

Then,

$1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}$