How to do this...Find the solution for each equation.
26) 3f-2=13
27) 2y-10=14
28) 5/7r+(-6)=29

When two the same numbers under the square root multiply, they are equal to the value under the square root, for example √2⋅√2 = 2. So then we have:

3√2⋅2√8⋅√3⋅√6 = 3⋅√2⋅2⋅√(2⋅2⋅2)⋅√3⋅√(3⋅2) = 3⋅2⋅√2⋅√2⋅√2⋅√2⋅√2⋅√3⋅√3 = 3⋅2⋅2⋅√2⋅3 = 36⋅√2

or if the bolded 2 from above is also the square root, then the solution is:

3√(2⋅2)√(2⋅2⋅2)⋅√3⋅√(3⋅2) = 54

Answer:

72√2

Step-by-step explanation:

simplify all of the radicals into their factors, then pull out integers.  If you do this, only √2 remains.  Then, multiply all the integers.

Answer:

the correct answer is C

Step-by-step explanation:

Answer:

it’s the 3rd one:)

Step-by-step explanation:

Answer:

• In this situation, (h, k) = (-2, -3)
• Radius = 8/2 = 4 units

Substitute them into the circle equation: (x - h)² + (y - k)² = r²

(x - (-2))² + (y - (-3))² = 4²

(x + 2)² + (y + 3)² = 4²

Now expand the equation:

(x + 2)² + (y + 3)² = 4²

(x + 2)(x + 2) + (y + 3)(y + 3) = 16

(x² + 2x + 2x + 4) + (y² + 3y + 3y + 9) = 16

x² + 4x + 4 + y² + 6y + 9 = 16

x² + y² + 4x + 6y + 13 - 16 = 0

x² + y² + 4x + 6y - 3 = 0

The transformations that took place are;

A: BCDE is rotated 180° counterclockwise to form JKLM.

B: BCDE is rotated 180° clockwise to form PQRS.

What is the transformation that took place?

We can see that the coordinates of Rectangle BCDE are;

B(2, 1); C(2, 2); D(4, 2); E(4, 1)

Now, for Rectangle JKLM, we see that the coordinates are;

J(-2, -1); K(-2, -2); L(-4, -2); M(-4, -1).

Now, for Rectangle PQRS, we see that the coordinates are;

P(-2, 1); Q(-2, 2); R(-4, 2); S(-4, 1).

We see the transformation from BCDE to JKLM as (x, y) → (-x, -y) and this is reflection over the x-axis which is 180° counterclockwise.

Read more about Transformation at; brainly.com/question/4289712

Answer:

actually, its 1 3 and 4 :nerd_emoji:

Step-by-step explanation: