Ask question

Ask question

All categories

2 years ago

13

Can someone help me out with this problem

1 answer:

ohaa [14]2 years ago

3 0

Answer:

$\fbox {1) A}\\\fbox {2) A}$

Step-by-step explanation:

<u>Q1</u>

⇒ Graph A is most accurate as the situation is linear and the dots indicate per cup as you can't have any rational part (such as 0.5 cups, etc.)

<u>Q2</u>

⇒ A

⇒ The graph is discrete because the sellers have limited the purchaser to buy a distinct whole number of cups

You might be interested in

WORTH 25 POINTS PLEASE HELP!!!

GalinKa [24]

Just subtract the smaller volume from the larger:

10 × 10 ×15 - 7 × 9 × 12

First you would multiply 10 × 10 × 15 = 1500 and then multiply 7 × 9 × 12 = 756.

After you would subtract 1500 - 756 which would give you 744

I hope that helped!

7 0

4 years ago

Read 2 more answers

Question 4 (Fill-In-The-Blank Worth 6 points)

Marrrta [24]

The formula for density of an object is Density= $\frac{mass}{volume}$ First you plug in the information you already know. 163.5= $\frac{mass}{0.492}$ Now you have to solve for the mass so you multiply 0.492 by 163.5 $0.492*163.5=80.442$ Lastly, you simplify and you will get 80.44. Hope this helped :)

5 0

3 years ago

-2=k. 2 = k·1=k<br> Solve the equation.<br> Enter your answer in the box.<br> k=

Gekata [30.6K]

Answer:

k=-1

Step-by-step explanation:

-2=k. 2

Diving whole equation by 2

-2/2=k

k=-1

5 0

4 years ago

What number is 66% of 29.5

lana66690 [7]

19.47 is the answer. 29.5 * .66 or 66% is 19.47

3 0

3 years ago

Express the function as the sum of a power series by first using partial fractions. f(x) = 8 x2 − 4x − 12 f(x) = ∞ n = 0 find th

alexandr1967 [171]

I'm guessing the function is

$f(x)=\dfrac8{x^2-4x-12}=\dfrac8{(x-6)(x+2)}$

which, split into partial fractions, is equivalent to

$\dfrac1{x-6}-\dfrac1{x+2}$

Recall that for $|x|$ we have

$\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n$

With some rearranging, we find

$\dfrac1{x-6}=-\dfrac16\dfrac1{1-\frac x6}=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n$

valid for $\left|\dfrac x6\right|$ , or $|x|$ , and

$\dfrac1{x+2}=\dfrac12\dfrac1{1-\left(-\frac x2\right)}=\displaystyle\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

valid for $\left|-\dfrac x2\right|$ , or $|x|$ .

So we have

$f(x)=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n-\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

$f(x)=\displaystyle-\sum_{n=0}^\infty\left(\frac{x^n}{6^{n+1}}+\frac{(-x)^n}{2^{n+1}}\right)$

$f(x)=\displaystyle-\sum_{n=0}^\infty\frac{1+3(-3)^n}{6^{n+1}}x^n$

Taken together, the power series for $f(x)$ can only converge for $|x|$ , or $-2$ .

6 0

3 years ago