What is the least common factor of 8x^2, 6x, and 12x^3

PLEASEEEE HELP I NEED A ANSWER ASAP!!!!!!!

Licemer1 [7]

Answer:

520 - 303.93 - (10.99 * 4) - 25.25 - 73.43x ≥ 0

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1) Parentheses

520 - 303.93 - 43.96 - 25.25 - 73.43x ≥ 0

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2) Combine like terms

146.86 - 73.43x ≥ 0

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3) Get the variable term alone

-73.43x ≥ -146.86

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4) Divide to solve

x ≤ 2

** dividing by a negative number, the inequality sign flips **

ANSWER :

x ≤ 2

8 0

1 year ago

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I need help now, plz and thank you

MrRa [10]

Answer:

24

Step-by-step explanation:

3 Suit types, 2 Wig types, and 4 Talents.

3 X 2 X 4 = 24

7 0

2 years ago

A certain compound has a half-life of four days. Write and use an exponential decay function to find the amount of compound rema

dimaraw [331]

Answer:

1.97 ounces

Step-by-step explanation:

-The exponential decay function is given by the formula:

$P_t=P_oe^{-rt}\\\\$

Given that the half-life is 4 days, we can substitute in the formula to solve the rate of decay for a single half-life:

$P_t=P_oe^{rt}\\\\P_t=0.5P_o\\\\\therefore 0.5=e^{-4r}\\\\-4r=In(0.5)\\\\r=0.17329$

#Since, 1 week has 7 days, 3 weeks is a 21-day period, thus, t=21.

#We substitute and solve for the amount remaining after 21days of decay:

$P_t=P_oe^{-rt}\\\\=75e^{-21\times 0.17329}\\\\=1.97072\\\\\approx1.97 \ oz$

Hence, the amount remaining after 3 weeks of decays is approximately 1.97 ounces

4 0

2 years ago

“An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and wor

Artist 52 [7]

Answer:

Part 1) The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Part 2) The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

Step-by-step explanation:

Step 1

Find the radius of each sphere

we know that

The circumference of a circle is equal to

$C=2\pi r$

Find the radius of the sphere with a 5.50-meter circumference

For $C=5.50\ m$

assume

$\pi =3.14$

substitute and solve for r

$5.50=2(3.14)r$

$r=5.50/[2(3.14)]=0.88\ m$

Find the radius of the sphere with a 7.85-meter circumference

For $C=7.85\ m$

assume

$\pi =3.14$

substitute and solve for r

$7.85=2(3.14)r$

$r=7.85/[2(3.14)]=1.25\ m$

step 2

Find the surface area of each sphere

The surface area of sphere is equal to

$SA=4\pi r^{2}$

Find the surface area of sphere with a 5.50-meter circumference

For $r=0.88\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(0.88)^{2}$

$SA=9.73\ m^{2}$

Find the surface area of sphere with a 7.85-meter circumference

For $r=1.25\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(1.25)^{2}$

$SA=19.63\ m^{2}$

step 3

Find the cost of finishing each sphere

we know that

To find out the cost , multiply the surface area by $92 per square meter

Find the cost of sphere with a 5.50-meter circumference

$9.73*(92)=\$895.16$

therefore

The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Find the cost of sphere with a 7.85-meter circumference

$19.63*(92)=\$1,805.96$

therefore

The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

6 0

3 years ago

Pls help extra points and mark brainlist

kipiarov [429]

Answer:

x-4=19

Step-by-step explanation:

Hope that helps!

5 0

2 years ago

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