Ask question

Ask question

All categories

1 year ago

12

Classify each graph as increasing, decreasing, constant

1 answer:

posledela1 year ago

8 0

Answer:

3. increasing

4. decreasing

Step-by-step explanation:

well for graph 3 the values are increasing

and for graph 4 the values are decreasing

You might be interested in

Write the expression using rational exponents. Then simplify and convert back to radical notation.

ioda

Answer:

The radical notation is $3x\sqrt[3]{y^2z}$

Step-by-step explanation:

Given

$\sqrt[3]{27 x^{3} y^{2} z}$

Step 1 of 1

Write the expression using rational exponents.

$\sqrt[n]{a^{m}}=\left(a^{m}\right)^{\frac{1}{n}}$

$=a^{\frac{m}{n}}:\left({27 x^{3} y^{2} z})^{\frac{1}{3}}$

$(a \cdot b)^{r}=a^{r} \cdot b^{r}:(27)^{\frac{1}{3}}\left(x^{3}\right)^{\frac{1}{3}} \cdot\left(y^{2}\right)^{\frac{1}{3}} \cdot(z)^{\frac{1}{3}}$

$=$(3^3)^{\frac{1}{3}}\left(x^{3}\right)^{\frac{1}{3}} \cdot\left(y^{2}\right)^{\frac{1}{3}} \cdot(z)^{\frac{1}{3}}$$

$=\left(3\right)\left(x}\right)} \cdot\left(y}\right)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

$=3x \cdot(y)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

Simplify $3 x \cdot(y)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

$=3 x \sqrt[3]{y^{2} z}$

Learn more about radical notation, refer :

brainly.com/question/15678734

4 0

2 years ago

Principal, $3000; Annual interest rate, 5.4%; time, 4 years<br> The Balance after 4 years is?

Neporo4naja [7]

Answer: 777600

Step-by-step explanation:

you times 12 by 4 then you times 3000 by 5.4 and add the two answers

3 0

3 years ago

Chris is painting a wall with a length of 3 meters and a width of 1.6 meters

neonofarm [45]

Ok, so...what is the question?...

4 0

2 years ago

Which number line shows the solution set for |y-9| <12

RUDIKE [14]

Answer:

Option c

$y\geq-3$ or $y \leq 21$

Step-by-step explanation:

The absolute value is a function that transforms any value x into a positive number.

Therefore, for the function $f(x) = |x|$ x> 0 for all real numbers.

Then the inequation:

$|y-9| \leq 12$ has two cases

$(y-9)$ if $y \geq 9$ (i)

$-(y-9)$ if $y < 9$ (ii)

We solve the case (i)

$y \leq 9 + 12\\y\leq21$

We solve the case (ii)

$-y +9 \leq 12\\-y \leq 12-9\\y \geq -3$

Then the solution is:

$y\geq-3$ or $y \leq 21$

8 0

3 years ago

What is 457×(783×635)

Hoochie [10]

Answer:

227222685 i cant tell if u want me to multiply them all or solve for X

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers