All categories

2 years ago

4. If a data set is graphed, what type of graph would form if the data shows a direct variation?

1 answer:

3 0

A function represents a direct variation if the ratio between the output values and input values is constant. If two quantities vary directly, the points on a graph form a straight line, and the line runs through the origin.

Please leave a Thanks! It helps!

You might be interested in

The area of the Indian Ocean is what percent of the area of the Pacific ocean? ROund to the nearest whole?show your work

Bumek [7]

Pacific
64/100
Indian
25/100
64/100-25/100=39/100
39/100=39%
I'm not sure but I think 39% is the answer. Better to google it ;)

4 0

2 years ago

What should the balance be in Diane's register?<br><br> Will give BRAINLIEST

MArishka [77]

Answer:

The balance of Dianne's register should be $359.41

Step-by-step explanation:

We can begin at the ending balance on the bank statement at $578.30

Then a check is written for $219.25, so a debit

Next a deposit is made for $140.36, so a credit

Then a withdrawal is made for $140.00, so a debit

This means we can make the equation

$578.30-219.25+140.36-140.00=359.41$

3 0

2 years ago

Help someone please help me on tis

Angelina_Jolie [31]

Answer:

I think its b)1.5 seconds sorry if I am wrong

Step-by-step explanation:

16t²=time

16t²=16²seconds

16² seconds=256 seconds

16t²+64=distance

256+64=320

320 feet=distance

256 seconds/320feet

=1.5seconds

5 0

3 years ago

Rosie spent £2.36. Suzy spent 56p more then Rosie. How much did Suzy spend?

svetoff [14.1K]

Soory o dont no this

4 0

3 years ago

Mr. Jones' other friend Ms. O'Neill has "x" meters of fencing, what is the maximum area in square meters she can have?

Alexus [3.1K]

Answer:

$Max\ Area = \frac{x^2}{16}$

Step-by-step explanation:

Given

$P = x$ ---- the perimeter of fencing

Required

The maximum area

Let

$L \to Length$

$W \to Width$

So, we have:

$P = 2(L + W)$

This gives:

$2(L + W) = x$

Divide by 2

$L + W = \frac{x}{2}$

Make L the subject

$L = \frac{x}{2} - W$

The area (A) of the fence is:

$A = L * W$

Substitute $L = \frac{x}{2} - W$

$A = (\frac{x}{2} - W) * W$

Open bracket

$A = \frac{x}{2}W - W^2$

Differentiate with respect to W

$A' = \frac{x}{2} - 2W$

Set to 0

$\frac{x}{2} - 2W = 0$

Solve for 2W

$2W = \frac{x}{2}$

Solve for W

$W = \frac{x}{4}$

Recall that:

$L = \frac{x}{2} - W$

$L = \frac{x}{2} - \frac{x}{4}$

$L = \frac{2x- x}{4}$

$L = \frac{x}{4}$

So, the maximum area is:

$A = L * W$

$A = \frac{x}{4}*\frac{x}{4}$

$Max\ Area = \frac{x^2}{16}$

8 0

2 years ago