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andriy [413]
3 years ago
9

Ashley is taking a road trip from California to Washington. Every 4 hours, she travels between 204 and 272 miles. Twelve hours a

go, Ashley had traveled 118 miles from her starting point in California. Which is a reasonable measure of Ashley's distance from her starting point in California now? A. 238 miles B. 596 miles C. 832 miles D. 356 miles
Mathematics
1 answer:
nikitadnepr [17]3 years ago
3 0

Answer:

Option C. 832 miles

Step-by-step explanation:

Ashley travels between 204 and 272 miles in every 4 hours.

That means she travels between 51 (\frac{204}{4}miles) and 68 (\frac{272}{4}miles) per hour.

Therefore, in 12 hours she will travel between 612 (51×12 miles) miles and 816 (68×12 miles).

Since she has already traveled 118 miles before 12 hours then the distance traveled by her till now will be between 730 (612+118 miles) and 934 (816+118 miles).

From the given options, Option C. 832 miles is the reasonable measure of Ashley's distance from the starting point.

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4 times a number is greater than -48.what are the possible values for the number ?
Keith_Richards [23]

The answer might be - 12

4 x -12 = 48

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4 years ago
A certain math teachers likes to keep a box of prizes to distribute to students. The math teacher likes to keep at least twice a
Law Incorporation [45]

Option C

Math teacher would need to buy 130 prizes

<em><u>Solution:</u></em>

Given that,

Math teacher currently has 109 students and the box has 88 prizes in it

The math teacher likes to keep at least twice as many prizes in the box as she has students

So, she wants the number of prizes to be twice the number of students

Therefore,

number of prizes = 2 x 109 students

number of prizes = 2 x 109 = 218 prizes

The box has 88 prizes in it

Therefore, number of prizes she would need to buy is:

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Thus she would need to buy 130 prizes

3 0
3 years ago
Matching Functions
Ainat [17]

Answer:

Graph A → Equation A → Table B → Property D

Graph D → Equation B → Table A → Property B

Graph B → Equation C → Table C → Property E

Graph C → Equation D → Table E → Property A

Graph E → Equation E → Table D → Property C

Step-by-step explanation:

Graph A corresponds to equation A, x·y = 2, and Table B → Property D

xy = 2 has a vertical asymptote at x = 0, and a horizontal asymptote at y = 0

y = 2/x, when x = 0, y = ∞

Graph D corresponds to equation B, -\left |  x\right | + 1 which corresponds to table A → Property B

The domain extends from (-∞, ∞), and [1, -∞)

Graph B corresponds to equation C, -(x + 2)² which gives the values in Table C and Property E wich is a horizontal shift of 2 units left and a reflection across the x-axis

Graph C corresponds to equation D, (1/3)·x - 2, which gives the values on Table E and Property A

The y-intercept of (1/3)·x - 2 = (0, -2), and the x-intercept = (6, 0)

Graph E corresponds to equation E, x³ = y, which gives the values on Table D  Property C, y is x cubed

6 0
3 years ago
Find the gradient of the line segment between the points (-2,3) and (-5,7)
Rudiy27

Answer:

\frac{4}{ - 3}

Procedures:

\frac {y2 - y1}{x2 - x1}

\frac{7 - 3}{ - 5 -  - 2}

\frac{7 - 3}{ - 5 + 2}

\frac{4}{ - 3}

7 0
3 years ago
I need help on this question
olga2289 [7]

Hey there!

-1 ¼ • 9

= - (4+1)/4 • 9

= -5/4 • 9

= -(5 × 9)/4

= <u>-</u><u> </u><u>4</u><u>5/4</u>

= <u>- 11.25</u>

Hope it helps ya!

8 0
3 years ago
Read 2 more answers
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