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3 years ago

A bus gets off the freeway, stops at a rest stop for passengers to get lunch, and then gets back on the freeway.

Mathematics

2 answers:

vladimir1956 [14]3 years ago

8 0

Answer-

The first graph correctly describes the situation.

Solution-

First scenario-

As the bus gets off the freeway, that means the driver pushed the brake. That is why the speed decreases constantly to zero. So the slope is negative here.

Second scenario-

As the bus stops at the rest stop for passengers to get lunch, the vehicle is still. As the bus is in still position, so its speed is zero as long it is still at the rest stop. So the slope is zero here.

Third scenario-

As the bus now starts moving towards freeway, its speed increases constantly. So the slope is positive in this case.

lozanna [386]3 years ago

5 0

I believe the first graph in the shape of the \_/ is the correct graph because when he gets off the freeway his speed slows then he stops and is no longer moving for a period of time. Then he gets back on and his speed is going up.

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1. An outdoor art is a metal cube with edge length 39 feet. What is the volume of the cube in feet? In cubic yard? ( Can someone

Law Incorporation [45]

The volume of a cube is just the side length cubed...

V=s^3 in this case:

V=39^3=59319 ft^3

and in cubic yards it is:

59319 ft^3*(yd^3/27 ft^3)=2197 yd^3

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3 years ago

Consider the angle. Which TWO statements are true?

svetoff [14.1K]

Answer:

C, E

Step-by-step explanation:

A. INCORRECT

A is wrong because a reflection across the x-axis DOES move the position of the figure (as it is flipped, so the position changes), but it DOES NOT change the angle (since a shift in position doesn't equal to a change in angle measure)

B. INCORRECT

Although a reflection across the x-axis does change the position of the angle, it DOES NOT change the measure of the angle.

C. CORRECT

A reflection across the x-axis does in fact move the position of the figure and does not change the angle measure. Reflections only deal with flipping a figure, not changing it's shape/distorting it so that the angle will change.

D. INCORRECT

A translation right will change the position of the figure but will not change the measure of the angle.

E. CORRECT

Yes, a translation right WILL change the position of the figure but will NOT change the measure of the angle. This is because a translation is simply moving a figure up and down; it has nothing to do with changing the shape of the figure/distorting it so that the angle is different.

8 0

3 years ago

Find a and b such that gcd(a, b) = 14, a > 2000, b > 2000, and the only prime divisors of a and b are 2 and 7.

choli [55]

Answer:

$a = 2^\alpha.14 \ ,\alpha\geq 8\\b = 7^\delta.14 \ , \delta\geq 3$

Step-by-step explanation:

As 2 and 7 are the only prime divisors of both $a$ and $b$ we know that both can be written as:

$a = 2^\alpha . 7^\beta \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta \ , \gamma ,\delta\ \in \mathbb{N}_{0}\\$

Where $\mathbb{N}_{0}$ is the set of all natural numbers adding the zero (careful because this part is important as I'll explain next).

We also know that 14 divides both numbers and that is actually the greatest common divisor between them. So we can rewrite a and b as follows:

$a = 2^\alpha . 7^\beta.14 \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta.14 \ , \gamma ,\delta\ \in \mathbb{N}_{0}\\$

Why do I write them like this? Because this way is easier to observe that if $\alpha$ and $\gamma$ were both greater than zero, then 28 would divide both hence 14 wouldn't be their g.c.d.. Likewise, if $\beta$ and $\delta$ were both greater than zero, then 98 would divide both and once again, 14 wouldn't be their g.c.d.

So either of them has to be equal to zero. And then we have that

$a = 2^\alpha.14 \\b = 7^\delta.14$

All we have left to do is find the possible values for $\alpha$ and $\delta$ so that $a>2000 \ , b>2000$ and that only happens if $\alpha\geq 8$ and $\delta\geq 3$