2 years ago

Whats the gcf of n^3t^2 and nt^4

Mathematics

1 answer:

melomori [17]2 years ago

4 0

$GCF-Greatest\ Common\ Factor\\\\n^3t^2=nt^2\cdot n^2\\nt^4=nt^2\cdot t^2\\\\GCF(n^3t^2;\ nt^4)=\boxed{nt^2}$

You might be interested in

Write 27/22 as a decimal rounded to the nearest tenth?

Verizon [17]

Answer:

1.2

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

For the figure above: Assume that <ABD has a mesure of 105° ANSWER ASAP WILL GIVE 15 POINTS AND BRAINLIEST

kykrilka [37]

Answer:

DBC = 75

Step-by-step explanation:

ABD and DBC are supplementary angles. They add to 180 degrees.

ABD + DBC = 180

105 + DBC = 180

Subtract 105 from each side

105-105 + DBC = 180-105

DBC = 75

4 0

2 years ago

Read 2 more answers

Evaluate f(-2) = -3x² - 2x

Delvig [45]

Answer:

-2F=7x

Step-by-step explanation:

F*-2= -2F 3x*3x=9x

-2F=9x-2x

8 0

2 years ago

Find all solutions for a triangle with A = 40°, B = 60°, and c = 20.

Ket [755]

Answer:C

Step-by-step explanation:

180-(60+40)=80°

C=80°

20/sin80. =a/sin40

a=(20sin40)÷sin80

=13.1

3 0

3 years ago

A surveyor is measuring the distance across a small lake. He has set up his transit on one side of the lake 130 feet from a pili

Dimas [21]

Answer:

x = 225.16 feet

Step-by-step explanation:

Given that,

The angle between the piling and the pier is 60°.

A surveyor has set up his transit on one side of the lake 130 feet from a piling that is directly across from a pier on the other side of the lake.

We need to find the distance between the piling and the pier. Let the distance be d. Using the trigonometry to find it as follows :

$\tan(60)=\dfrac{x}{130}\\\\x=\tan(60)\times 130\\\\x=225.16\ feet$

So, the required distance is 225.16 feet.

4 0

2 years ago