Solve y=4x+8x for x

eimsori [14]

Use the FOIL method (First, Outside, Inside, Last)

6r(-8r) = -48r²
6r(-3) = -18r
-1(-8r) = 8r (note: two negatives multiplied together = positive answer)
-1(-3) = 3

-48r² - 18r + 8r + 3

Combine like terms:

-48r² - 18r + 8r + 3

-48r² - 10r + 3

-48r² - 10r + 3 is your answer

hope this helps

6 0

1 year ago

Your promotional budget for the year is $300 for personal selling, $300 for sales promotion, \$1,500 for advertising, and $500 f

aleksandr82 [10.1K]

Answer:

to get the total promotion budget for the year we will add $300 for personal selling plus $300 for sales promotion plus $1500 for advertising plus $500 for image promotion to the a total of 2600 dollars

8 0

1 year ago

According to the Rational Root Theorem, the following are potential fox) 2x2 +2x 24.roots of -4, -3, 2, 3, 4Which are actual roo

aksik [14]

Correct Answer: First Option

Explanation:

There are two ways to find the actual roots:

a) Either solve the given quadratic equation to find the actual roots

b) Or substitute the value of Possible Rational Roots one by one to find out which satisfies the given equation.

Method a is more convenient and less time consuming, so I'll be solving the given equation by factorization to find its actual roots. To find the actual roots set the given equation equal to zero and solve for x as given below:

$2x^{2} +2x-24=0\\ \\ 2(x^{2} +x-12)=0\\ \\ x^{2} +x-12=0\\ \\ x^{2} +4x-3x-12=0\\ \\ x(x+4)-3(x+4)=0\\ \\ (x-3)(x+4)=0\\ \\ x-3=0, x=3\\ \\ or\\\\x+4=0, x=-4$

This means the actual roots of the given equation are 3 and -4. So first option gives the correct answer.

7 0

1 year ago

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A conical pile of road salt has a diameter of 112 feet and a slant height of 65 feet. After a storm, the linear dimensions of th

QveST [7]

we know that

the volume of a cone is equal to

$V= \frac{1}{3} \pi r^{2}h$

in this problem

the radius is equal to

$r= \frac{112}{2}= 56ft$

1) Find the height of the cone before the storm

Applying the Pythagorean Theorem find the height

$h^{2} = l^{2}-r^{2}$

$l=65 ft$

$h^{2} = 65^{2}-56^{2}$

$h^{2} = 1,089$

$h=33 ft$

2) Find the volume before the storm

$V= \frac{1}{3}*\pi* 56^{2}*33$

$V=34,496\pi\ ft^{3}$

3) Find the volume after the storm

After a storm, the linear dimensions of the pile are 1/3 of the original dimensions

so

r=(56/3) ft

h=(33/3)=11 ft

$V= \frac{1}{3}*\pi* (56/3)^{2}*11$

$V= 1,277.63\pi\ ft^{3}$

4) Find how this change affect the volume of the pile

Divide the volume after the storm by the volume before the storm

$\frac{1,277.63 \pi }{34,496 \pi } = \frac{1}{27}$

therefore

the answer part a) is

The volume of the pile after the storm is $\frac{1}{27}$ times the original volume

Part b) Estimate the number of lane miles that were covered with salt

5) Find the amount of salt that was used during the storm

$=34,496 \pi - 1,277.63 \pi \\= 33.218.37 \pi \\= 104,358.59\ ft^{3}$

6) Find the pounds of road salt used

$104,358.59*80=8,348,687.2\ pounds$

7) Find the number of lane miles that were covered with salt

$8,348,687.2/350=23,853.39 \ lane\ miles$

therefore

the answer part b) is

the number of lane miles that were covered with salt is $23,853.39 \ lane\ miles$

Part c) How many lane miles can be covered with the remaining salt? Round your answer to the nearest lane mile

the remaining salt is equal to $1,277.63\pi\ ft^{3}$

$1,277.63\pi\ ft^{3}=4,013.79\ ft^{3}$

8) Find the pounds of road salt

$4,013.79*80=321,103.20\ pounds$

9) Find the number of lane miles

$321,103.20/350=917.44 \ lane\ miles$

therefore

the answer part c) is

the number of lane miles is $917 \ lane\ miles$

7 0

2 years ago

Find the GCF of 44j5k4 and 121j2k6.

BartSMP [9]

Answer:

D. $11j^2k^4$

Step-by-step explanation:

We are asked to find the GCF of $44j^5k^4\text{ and }121j^2k^6$ .

Since we know that GCF of two numbers is the greatest number that is a factor of both of them.

First of all we will GCF of 44 and 121.

Factors of 44 are: 1, 2, 4, 11, 22, 44.

Factors of 121 are: 1, 11, 11, 121.

We can see that greatest common factor of 44 and 121 is 11.

Now let us find GCF of $j^5\text{ and }j^2$ .

Factors of $j^5$ are: $j*j*j*j*j$

Factors of $j^2$ are: $j*j$

We can see that greatest common factor of $j^5\text{ and }j^2$ is $j*j=j^2$ .

Now let us find GCF of $k^4\text{ and }k^6$ .

Factors of $k^4$ are: $k*k*k*k$

Factors of $k^6$ are: $k*k*k*k*k*k$

We can see that greatest common factor of $k^4\text{ and }k^6$ is $k*k*k*k=k^4$ .

Upon combining our all GCFs we will get,

$11j^2k^4$

Therefore, GCF of $44j^5k^4\text{ and }121j^2k^6$ is $11j^2k^4$ and option D is the correct choice.

3 0

1 year ago

Read 2 more answers