26 = 8 + v 2) 3 + p = 8

The table shows the distance Allison drove on one day of her vacation. Is the relationship between the distance and the time a p

Anuta_ua [19.1K]

Answer:

we conclude that the relationship between distance and time is NOT proportional.

Hence, she did not drive at a constant speed.

Step-by-step explanation:

We know that when 'y' varies directly with 'x', we get the equation

y ∝ x

y = kx

k = y/x

where 'k' is called the constant of proportionality.

In our case, the table shows the distance Allison drove on one day of her vacation.

Time (h) 1 2 3 4 5

Distance (mi) 55 100 165 280 250

using the equation

k = y/x

susbtitute y = 55, x = 1

k = 55/1 = 55

substitute y = 100, x = 2

k = y/x

k = 100 / 2 = 50

substitute y = 165, x = 3

k = y/x

k = 165 / 3 = 55

substitute y = 280, x = 4

k = y/x

k = 280 / 4 = 70

substitute y = 250, x = 5

k = y/x

k = 250 / 5 = 50

It is clear that the value of 'k' does not remain constant.

Therefore, we conclude that the relationship between distance and time is NOT proportional.

Hence, she did not drive at a constant speed.

6 0

3 years ago

Person A can make a cupcake in 5 minutes, Person B can make a cupcake in 9 minutes. In total they made 168 cupcakes. How many cu

Stolb23 [73]

Answer:

Person A made 108 cupcakes and person B made 60 cupcakes.

Step-by-step explanation:

Person A makes 1 cupcake in 5 minutes.

Say in x minutes A makes : $\frac{x}{5}$ cupcakes

Person B makes 1 cupcake in 9 minutes.

Say in x minutes B makes : $\frac{x}{9}$ cupcakes

Total cupcakes prepared by A and B in x minutes = 168

$\frac{x}{5}+\frac{x}{9}=168$

$\frac{14x}{45}=168$

$x=\frac{168\times 45}{14}=540$

So, cupcakes made by A = $\frac{540}{5}=108$

Cupcakes made by B = $\frac{540}{9}=60$

Person A made 108 cupcakes and person B made 60 cupcakes.

5 0

3 years ago

How do you turn 5/2 into 10/4?

o-na [289]

That’s easy. To turn 5/2 into 10/4 you multiply by 2. :D Hope this helps!

3 0

3 years ago

Read 2 more answers

How many elements does A contains if it has 64 subsets

Amanda [17]

<h3>Answer: 6</h3>

===========================================================

Explanation:

Rule: If a set has n elements in it, then it will have 2^n subsets.

For example, there are n = 3 elements in the set {a,b,c}. This means there are 2^n = 2^3 = 8 subsets. The eight subsets are listed below.

{a,b,c} .... any set is a subset of itself
{a,b}
{a,c}
{b,c}
{a}
{b}
{c}
{ } ..... the empty set

Subsets 2 through 4 are subsets with exactly 2 elements. Subsets 5 through 7 are singletons (aka sets with 1 element). The last subset is the empty set which is a subset of any set. You could use the special symbol $\varnothing$ to indicate the empty set.

For more information, check out concepts relating to the power set.

-------------------

The problem is asking what value of n will make 2^n = 64 true.

You could guess-and-check your way to see that 2^n = 64 has the solution n = 6.

Another approach is to follow these steps.

$2^n = 64\\\\2^n = 2^6\\\\n = 6$

Which is fairly trivial.

Or you can use logarithms to solve for the exponent.

$2^n = 64\\\\\text{Log}\left(2^n\right)=\text{Log}\left(64\right)\\\\n*\text{Log}\left(2\right)=\text{Log}\left(64\right)\\\\n=\frac{\text{Log}\left(64\right)}{\text{Log}\left(2\right)}\\\\n\approx\frac{1.80617997398389} {0.30102999566399} \ \text{ ... using base 10 logs}\\\\n\approx5.99999999999983\\\\$

Due to rounding error, we don't land exactly on 6 even though we should.

5 0

2 years ago

A factory manager is packing engine components into wooden crates to be shipped on a small truck. The truck is designed to hold

Lemur [1.5K]

Answer: 63lbs

Step-by-step explanation:

The truck can only hold 16crates and each crate weigh 12lbs, the total weight of the 16 crates will be 12×16= 192lbs

This shows that the truck can only contain extra load of 1200lbs - 192lbs = 1008lbs (excluding weight of crates).

To get the shipment weigh close to the total of 1200lbs, the truck must be loaded with engine components not more than 1008lbs.

Since we have 16 crates to fill with engine components not more than 1008lb, each crates will therefore must not exceed 1008/16 pounds of engine components which is equivalent to 63lbs.

Therefore, the manager should instruct the workers to put 63lbs of machine components in "each crate" in order to get the shipment weight as close as possible to 1200 lbs.

6 0

3 years ago