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emmainna [20.7K]
3 years ago
9

What is the volume of this cube with a side length of 6 centimeters? A cube with side lengths of 6 centimeters. Recall the formu

la Cube volume = s cubed. 12 cubic centimeters 18 cubic centimeters 36 cubic centimeters 216 cubic centimeters
Mathematics
2 answers:
Sedbober [7]3 years ago
6 0

Answer:

you would multiple length base and height together and since they're all 6 centimeters you would multiply 6 by 6 by 6 and you get216 cubic centimeters

Hope this helps!

Step-by-step explanation:

____ [38]3 years ago
3 0

Answer:36 cm cube

Step-by-step explanation:

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Identifying Possible Triangles
Masja [62]

Answer:

3rd triangle can be constructed with dimensions 2,6,7.

Step-by-step explanation:

sum of any two sides > third side.

difference of any two sides < third side

1.

8+5=13 not >14 (no triangle.)

2.

7+8=15 not >15 (no triangle)

3.

2+6=8>7

2+7=9>6

7+6=13>2

7-2=5<6

7-6=1<2

6-2=4<7

so it is a triangle.

4.

6+3=9 not >10 (not a triangle)

5 0
2 years ago
What is the product of 5/12 and 1/3
11Alexandr11 [23.1K]

Answer:

5/6

Step-by-step explanation:

6 0
2 years ago
Cos-1<br> 25²-52<br> 52 -37<br> 2625 )(37)
34kurt

Answer:

Step-by-step explanation:

i do not know

4 0
3 years ago
Can some one help me. btw the answer is not C.​
Anit [1.1K]

Answer:

A

Step-by-step explanation:

Given

x² + 6x

To complete the square

add ( half the coefficient of the x- term )² to x² + 6x

x² + 6x + (3)²

= x² + 6x + 9

= (x + 3)²

3 0
3 years ago
A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,
guajiro [1.7K]

We have been given that a huge ice glacier in the Himalayas initially covered an area of 45 square kilo-meters. The relationship between A, the area of the glacier in square kilo-meters, and t, the number of years the glacier has been melting, is modeled by the equation A=45e^{-0.05t}.

To find the time it will take for the area of the glacier to decrease to 15 square kilo-meters, we will equate A=15 and solve for t as:

15=45e^{-0.05t}

\frac{15}{45}=\frac{45e^{-0.05t}}{45}

\frac{1}{3}=e^{-0.05t}

Now we will switch sides:

e^{-0.05t}=\frac{1}{3}

Let us take natural log on both sides of equation.

\text{ln}(e^{-0.05t})=\text{ln}(\frac{1}{3})

Using natural log property \text{ln}(a^b)=b\cdot \text{ln}(a), we will get:

-0.05t\cdot \text{ln}(e)=\text{ln}(\frac{1}{3})

-0.05t\cdot (1)=\text{ln}(\frac{1}{3})

-0.05t=\text{ln}(\frac{1}{3})

t=\frac{\text{ln}(\frac{1}{3})}{-0.05}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-0.05\cdot 100}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-5}

t=-\text{ln}(\frac{1}{3})\cdot 20

t=-(\text{ln}(1)-\text{ln}(3))\cdot 20

t=-(0-\text{ln}(3))\cdot 20

t=20\text{ln}(3)

Therefore, it will take 20\text{ln}(3) years for area of the glacier to decrease to 15 square kilo-meters.

6 0
3 years ago
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