Answer:
no
Step-by-step explanation:
11/4 plus 11/4 is 22/4
to make this a whole number and a fraction we subtract 4 from the 22 as many times as we can. we can do this 5 times, with 2 left over.
the answer is 5 2/4, which you can simplify to 5 and a half
<h2>Anser</h2>
53
Step-by-step explanation:
(6-6)+7×7+4
=0+7×7+4
=7×7+4
=49+4
=53
Answer:
Answer:
(B)
Step-by-step explanation:
We have to construct two equilateral triangles that is one with the side length of 2 inches and the another triangle with side length of 3 inches.
Now, both the triangles are equilateral triangles, thus they have same shape.
Also, both triangles have different side lengths, one having 2 inches and another having 3 inches, thus they have different size.
Hence, the true statement about the two triangles is:
The two triangles are the same shape but not the same size.
Therefore, option B is correct.
Step-by-step explanation:
Given:
The equation is,

Explanation:
Simplify the equation by using logarthimic property.

Simplify further.

Solve the quadratic equation for x.

From the above equation (x - 6) = 0 or (x - 3) = 0.
For (x - 6) = 0,

For (x - 3) = 0,

The values of x from solving the equations are x = 3 and x = 6.
Substitute the values of x in the equation to check answers are valid or not.
For x = 3,

Equation satisfy for x = 3. So x = 3 is valid value of x.
For x = 6,

Equation satifies for x = 6.
Thus values of x for equation are x = 3 and x = 6.