Jenna is asked about her age. She answers as follows. "My age is 6 more than three times the age of

horrorfan [7]

<span>Z=15+2(x+y)
Use distributive property
Z= 15 + 2x + 2y
Subtract 2y from both sides
Z - 2y= 15 + 2x
Subtract 15 from both sides
Z - 2y - 15= 2x
Divide 2 on both sides
Final Answer: Z - 2y - 15(All over 2)= x</span>

8 0

3 years ago

Read 2 more answers

The work of a student to find the dimensions of a rectangle of area 8 + 12x and width 4 is shown below:

iogann1982 [59]

Answer:

Step 2: 4(2) + 4(3x)

Step-by-step explanation:

By definition, the area of a rectangle is given by:

Area = Length × Width

In this case, we know that:

Area = 8 + 12x

Width = 4

Therefore:

Step 1: 8 + 12x

Step 2: 4(2) + (4)(3x)

Step 3: 4(2 + 3x)

Therefore, the dimensions of the rectangle are 4 and 2 + 3x.

The mistake was made in STEP 2. Instead of 4(2) + 4(x2) it should be 4(2) + 4(3x). Which is the second option.

5 0

3 years ago

What if the answer is wrong

stealth61 [152]

If the answer is wrong then the answer isn't right but it also isn't left hence making it wrong.

7 0

3 years ago

A basketball-player scores a free-throw with probability 0.9. What is the probability that her first miss occurs on the 6th shot

-BARSIC- [3]

Answer:

$Pr = 0.059049$

Step-by-step explanation:

Given

$p = 0.9$ --- probability of scoring

Required

Probability that his first miss is his 6th shot

Let q represent the event that he did not score.

Using complement rule:

$q = 1 - p = 1 - 0.9 = 0.1$

The event that his first miss is his 6th is represented as:

p p p p p q ---- That he scoress the first 5 attempts

So, the probability is:

$Pr = p^5 * q$

$Pr = 0.9^5 * 0.1$

$Pr = 0.059049$

7 0

3 years ago

Math question

strojnjashka [21]

Answer:

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

Step-by-step explanation:

The volume ( $V$ ), in cubic centimeters, and surface area ( $A_{s}$ ), in square centimeters, formulas for the candle are described below:

$V = \pi\cdot r^{2}\cdot h$ (1)

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$ (2)

Where:

$r$ - Radius, in centimeters.

$h$ - Height, in centimeters.

By (1) we have an expression of the height in terms of the volume and the radius of the candle:

$h = \frac{V}{\pi\cdot r^{2}}$

By substitution in (2) we get the following formula:

$A_{s} = 2\pi \cdot r^{2} + 2\pi\cdot r\cdot \left(\frac{V}{\pi\cdot r^{2}} \right)$

$A_{s} = 2\pi \cdot r^{2} +\frac{2\cdot V}{r}$

Then, we derive the formulas for the First and Second Derivative Tests:

First Derivative Test

$4\pi\cdot r -\frac{2\cdot V}{r^{2}} = 0$

$4\pi\cdot r^{3} - 2\cdot V = 0$

$2\pi\cdot r^{3} = V$

$r = \sqrt[3]{\frac{V}{2\pi} }$

There is just one result, since volume is a positive variable.

Second Derivative Test

$A_{s}'' = 4\pi + \frac{4\cdot V}{r^{3}}$

If $\left(r = \sqrt[3]{\frac{V}{2\pi}}\right)$ :

$A_{s} = 4\pi + \frac{4\cdot V}{\frac{V}{2\pi} }$

$A_{s} = 12\pi$ (which means that the critical value leads to a minimum)

If we know that $V = 3217\,cm^{3}$ , then the dimensions for the minimum amount of plastic are:

$r = \sqrt[3]{\frac{V}{2\pi} }$

$r = \sqrt[3]{\frac{3217\,cm^{3}}{2\pi}}$

$r = 8\,cm$

$h = \frac{V}{\pi\cdot r^{2}}$

$h = \frac{3217\,cm^{3}}{\pi\cdot (8\,cm)^{2}}$

$h = 16\,cm$

And the amount of plastic needed to cover the outside of the candle for packaging is:

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$

$A_{s} = 2\pi\cdot (8\,cm)^{2} + 2\pi\cdot (8\,cm)\cdot (16\,cm)$

$A_{s} \approx 1206.372\,cm^{2}$

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

3 0

3 years ago