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3 years ago

Nicole runs 6 miles in 44 minutes. At the same rate, how many minutes would she take to run 9 miles?

Mathematics

1 answer:

HACTEHA [7]3 years ago

6 0

66 minutes is the correct answer

You might be interested in

A number times six plus the same number times two

Travka [436]

Answer:

6x +2x

Step-by-step explanation:

when the number= x, you can just form the equation as per usual. 6x = a number times 6; 2x = a number times 2.

x is used bc the number is an unknown value. only x can be used bc it is stated the same number. hsing different alphabets such as y or z indicates the unknown value ia different.

6 0

3 years ago

Read 2 more answers

A bookstore marks up the price of a book by 25% of the cost from the publisher. Therefore, the

lesya692 [45]

Answer:

(A)

P(60) = 78.98
When the bookstore spends $60 on a textbook, the student pays $78.98

Step-by-step explanation:

Given:

P(x) = 1.053(x+0.25x)

P(60) = 1.053(60+0.25*60)

=1.053(60+15)

=1.053(75)

P(60)=$78.98

Since x is the cost from the publisher, when the bookstore spends $60 on a textbook, the student pays $78.98.

<u>The correct option is A.</u>

3 0

3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago

A point is reflected in the y axis the reflected point is(5.75,0) what is the original point. And what's the distance between th

vlada-n [284]

A point , (x,y) is reflected across the y axis. the resulting point is (-x,y)
so

if (5.75,0) is the reflected point, then (-5.75,0) is the original
the distance is the double the absolute value of the x value (basically a horizontal line beteen the 2 points)
so
2|-5.75|=2(5.75)=11.5
distance is 11.5 units

6 0

3 years ago

Which pairs of matrices are inverses?

goblinko [34]

Answer:

B is the answer

Step-by-step explanation:

5 0

3 years ago