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MariettaO [177]
3 years ago
5

Three perfect cube blocks of different volumes were stacked on top of each other. The top block was 64 cubic centimeters, the mi

ddle block was 125 cubic centimeters, and the biggest block was 729 cubic centimeters. How tall was the stack of blocks?
Mathematics
1 answer:
lesya [120]3 years ago
5 0
We know that 

[volume of a cube]=b³---------> b=∛Volume
b------> is the side length of a cube

The top block was 64 cm³------> b1=∛64-------> b1=4 cm
The middle block was 125 cm³------> b2=∛125------> b2=<span>5 cm
T</span>he biggest block was 729 cm³------> b3=√729------> b3=<span>9 cm

[</span><span>the stack of blocks tall]=b1+b2+b3-------> 4+5+9-----> 18 cm
</span><span>
the answer is
</span>the stack of blocks was 18 cm tall<span>


</span>
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Mrrafil [7]
Hi again! so vertical angles are angles that are across from eachother, so it would be 5 and 7 and 6 and 8
7 0
2 years ago
Through how many radians do the minute hand and the hour hand move between 1:00 p.m. and 1:45 p.m. (on the same day)? ​
timofeeve [1]

Answer:

π/8 radians

Step-by-step explanation:

THIS IS THE COMPLETE QUESTION

In 1 h the minute hand on a clock moves through a complete circle, and the hour hand moves through 1 12 of a circle. Through how many radians do the minute hand and the hour hand move between 1:00 p.m. and 1:45 p.m. (on the same day)?

SOLUTION

✓If the minute hand on a clock moves through complete circle in 1 hour, then it means that it goes through a circle and angle of circle in radians is 2π.

Between 1:00 p.m. and 1:45pm in the same day we have 45 minutes i.e (1.45 pm -1pm)

Within the 1hour minutes, the hand can move with complete cycle of 2π radians

Then At time t= 45minutes

Angle through the circle at 45 minutes= 45/60 ×2π radians

= 3π/2 radians

And if the hour hand goes through a complete cycle 1/12 as told in the question we have 1/2 × 2π radians

For t=45 minutes

Then 1/12 × 2π ×45/60

= π/8 radians

Hence, the minute hand and the hour hand move π/8 radians between 1:00 p.m. and 1:45 p.m.

5 0
2 years ago
—13 &lt; -x+5&lt; 8 iѕ -21 <br><br> True or false?
Andrej [43]
This is True I believe
6 0
3 years ago
For the following pair of functions, find (f+g)(x) and (f-g)(x).
ch4aika [34]

Given:

The functions are

f(x)=4x^2+7x-5

g(x)=-9x^2+4x-13

To find:

The functions (f+g)(x) and (f-g)(x).

Solution:

We know that,

(f+g)(x)=f(x)+g(x)

(f+g)(x)=4x^2+7x-5-9x^2+4x-13

(f+g)(x)=(4x^2-9x^2)+(7x+4x)+(-5-13)

(f+g)(x)=-5x^2+11x-18

And,

(f-g)(x)=f(x)-g(x)

(f-g)(x)=(4x^2+7x-5)-(-9x^2+4x-13)

(f+g)(x)=4x^2+7x-5+9x^2-4x+13

(f+g)(x)=(4x^2+9x^2)+(7x-4x)+(-5+13)

(f-g)(x)=13x^2+3x+8

Therefore, the required functions are (f+g)(x)=-5x^2+11x-18

and (f-g)(x)=13x^2+3x+8.

7 0
3 years ago
Find the zeros of the function y = x^2-1.
kumpel [21]

Answer:

Step-by-step explanation:

x^2-1=0

Add  1 to both sides of the equation.

x^2= 1

Take the square root of both sides of the equation to eliminate the exponent on the left side.

x = ± √ 1

Any root of  1  is  x = ± 1

First, use the positive value of the  ±

to find the first solution.

x= 1

Next, use the negative value of the  ±

to find the second solution.

x = − 1

The complete solution is the result of both the positive and negative portions of the solution.

x = 1 , − 1

4 0
2 years ago
Read 2 more answers
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