All categories

3 years ago

The pre-image is

2 answers:

8 0

It is the last one.

A pre-image is the original before another shape is created from it, either dilated, translated, reflected, or rotated.

bezimeni [28]3 years ago

6 0

Answer

Step-by-step explanation:

It's D

You might be interested in

Which inequality matches the graph?

Alik [6]

b is the answer because it hits positive two on the x axis and -3 on the y axis

5 0

3 years ago

John collected 17 leaves to feed his caterpillar collection if he wanted to split the leaves equally amongst the four cages how

Papessa [141]

Answer:

It lies between the numbers 4 and 5

Step-by-step explanation:

If John has 17 leaves and 4 cages then he should put 17/4, or 16/4+1/4 which is 4+0.25, or 4.25 in each cage.

3 0

3 years ago

Read 2 more answers

What’s the measure of angle B?

velikii [3]

Answer:

60°

Step-by-step explanation:

∠B=(240-120)/2=120/2=60°

7 0

3 years ago

Brad bought a used bike for $8 more than half of its original price. Brad paid $40 for the bike. What was the original price?

Viefleur [7K]

$\huge\boxed{\$64}$

Start by writing a formula, where $x$ is Brad's purchase price and $y$ is the original price.

$40=8+\frac{y}{2}$

Subtract $8$ from both sides.

$32=\frac{y}{2}$

Multiply both sides by $2$ .

$\boxed{64}=y$

7 0

3 years ago

Need step-by-step. Please answer correctly. This determines what grade I get. I will mark brainliest to the first person who ans

tatyana61 [14]

Answer:

3√3

Step-by-step explanation:

For the problem shown here, your answer 3√3 is correct.

When there is a radical by itself in the denominator, you multiply numerator and denominator by a radical that results in the product being rational. For a square root, that will usually be the same square root:

$\dfrac{9}{\sqrt{3}}=\dfrac{9}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{9\sqrt{3}}{3}=\boxed{3\sqrt{3}}$

If the problem has a sum in the denominator involving a square root, then you multiply numerator and denominator by the conjugate of that sum (the sum with the sign changed). This uses the special product "difference of squares" to eliminate the radical term.

<u>Example</u>:

$\dfrac{9}{2-\sqrt{3}}=\dfrac{9}{2-\sqrt{3}}\cdot\dfrac{2+\sqrt{3}}{2+\sqrt{3}}=\dfrac{9(2+\sqrt{3})}{2^2-(\sqrt{3})^2}=\dfrac{18+9\sqrt{3}}{4-3}\\\\=\boxed{18+9\sqrt{3}}$

It is easy to demonstrate that none of the offered choices for this problem has the same value as 9/√3.

9/√3 ≈ 5.196. Offered choices have values of about 4.798, 1.732, 6.681, 23.196 -- none even close.

Please discuss this question with your teacher.

3 0

3 years ago