All categories

3 years ago

Terry drove 310 miles in 5 hours at a constant speed. How long would it take him to drive 403 miles at the same speed?

2 answers:

7 0

Terry drove 310 miles in 5 hours

How long would it take him to drive 403 miles at the same speed?

First, you divide 310 by 5 which is 62

Get 403 and divide by 62 which is 6.5

or simply put:

310÷5=62

403÷62=6.5

Answer: It would take Terry 6.5 hours to drive 403 miles

NeX [460]3 years ago

6 0

Well, first, find-out how fast he was going the first time, by dividing 310 by 5. This gives you 62 mph. Now, divide 403 miles by 62 miles per hour in order to find-out how long it would take him to drive 403 miles, if he is constantly doing 62 mph. Your final answer is six-and-a-half hours.

You might be interested in

if the area of a rectangle is 48 square feet and the width of the rectangle is 6 feet what is the perimeter of the rectangle

Over [174]

Answer:

28 ft

Step-by-step explanation:

the perimeter of the rectangle is 28 feet

3 0

2 years ago

Is 2.356... rational or irrational

777dan777 [17]

Answer:

Rational

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

WILL MARK BRAINLIST!! SOS MATH. Look at picture

Aleks [24]

Answer:

Yes, the transformation is a 270° clockwise rotation

Step-by-step explanation:

(-3, 4) ,( -4, 7) and (-2,7) transformed to (-4, - 3), (-7, -4) and (-7, -2).

Rule for 270° clockwise rotation:

(x, y) --> (- y, x)

A transformation that doesn't change the size or shape of an object.

So answer is:

Yes, the transformation is a 270° clockwise rotation

5 0

2 years ago

The line $y = 3$ intersects the graph of $y = 4x^2 + x - 1$ at the points $A$ and $B$. The distance between $A$ and $B$ can be w

raketka [301]

Answer:

Step-by-step explanation:

Let's find the points $A$ and $B$ .

We know that the $y$ -coordinates of both are $3$ .

So let's first solve:

$3=4x^2+x-1$

Subtract 3 on both sides:

$0=4x^2+x-1-3$

Simplify:

$0=4x^2+x-4$

I'm going to use the quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , to solve.

We must first compare to the quadratic equation, $ax^2+bx+c=0$ .

$a=4$

$b=1$

$c=-4$

$\frac{-1 \pm \sqrt{1^2-4(4)(-4)}}{2(4)}$

$\frac{-1 \pm \sqrt{1+64}}{8}$

$\frac{-1 \pm \sqrt{65}}{8}$

Since the distance between the points $A$ and $B$ is horizontal. We know this because they share the same $y-coordinate$ .This means we just need to find the positive difference between the $x$ -values we found for the points of $A$ and $B$ .