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4 years ago

Please help me no one even bothers to help so plzzzz i BEGGGGGGGGGGG

Mathematics

1 answer:

Rom4ik [11]4 years ago

5 0

Answer:

- The graphed is attached.

- It's an enlargement.

Step-by-step explanation:

It is important to remember that a Dilation is defined as a transformation in which the image and the pre-image have the same shape, but their sizes are different.

When the scale factor is greater than 1, the image obtained after the dilation is greater than the pre-image and it is an "Enlargement".

When the scale factor is is between 0 and 1, the image obtained after the dilation is smaller than the pre-image and it is an "Reduction".

In this case we know that the scale factor is 4.

Since:

$4>1$

It is an Enlargement.

Now we need to multiply the coordinates of the vertices of VWXY by 4, in order to find the coordinates of the vertices of V'W'X'Y'. Then:

$V'=(-1*4,-1*4)=(-4,-4)\\\\W'(-1*4,2*4)=(-4,8)\\\\X'(2*4,-1*4)=(8,-4)\\\\Y'=(2*4,2*4)=(8,8)$

Finally, we must draw the pre-image and the pre-image and label them, as you see in the picture attached.

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What is the value of -4° + (5-2)(-6)?

34kurt

Answer:

-22

Step-by-step explanation:

Due to order of operations rules, we must perform the subtraction inside parentheses first, getting:

-4 +(3)(-6)

Multiplication comes next:

-4 - 18

and the total value is thus -22.

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4 years ago

Unit 4 Test, Part 2: Congruence and Constructions<br> Help with my math please

lakkis [162]

Answer/Step-by-sep explanation:

1. Given:

∆NMK ≅ ∆TRP

$\overline{NM} = 20$

$\overline{MK} = 15$

$\overline{KN} = 25$

$\overline{TR} = 3x - 1$

a. To complete the congruent statement, thus: ∆MNK ≅ ∆RTP

b. The side that is congruent to $\overline{TR}$ is $\overline{NM}$ . Thus:

$\overline{TR}$ ≅ $\overline{NM}$

c. Since $\overline{TR}$ ≅ $\overline{NM}$ , therefore:

$\overline{TR} = \overline{NM}$

$3x - 1 = 20$ (substitution)

Add 1 to both sides

$3x = 20 + 1$

$3x = 21$

Divide both sides by 3

$x = \frac{21}{3}$

$x = 7$

2. a. Slope of LK = $\frac{rise}{run} = \frac{4}{3}$

Slope of LM = $\frac{rise}{run} = -\frac{3}{5}$

b. ✍️Length of LK is the distance between L(-7, 4) and (-4, 8):

$Lk = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$Lk = \sqrt{(-4 -(-7))^2 + (8 - 4)^2}$

$Lk = \sqrt{(3)^2 + (4)^2}$

$Lk = \sqrt{9 + 16}$

$Lk = \sqrt{25}$

$Lk = 5$

✍️Length of LM is the distance between L(-7, 4) and (-2, 1):

$LM = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$LM = \sqrt{(-2 -(-7))^2 + (1 - 4)^2}$

$LM = \sqrt{(5)^2 + (-3)^2}$

$LM = \sqrt{25 + 9}$

$LM = \sqrt{34}$

$LM = 5.8$ (nearest tenth)

∆KLM is not an isosceles ∆ because it does not has two equal side lengths. This we can see because LK and LM are not equal.

Therefore, Anthony is incorrect. Am isosceles ∆ has two equal sides.

8 0

3 years ago

What is the midpoint of a segment with These endpoints (-6,-9) and (16,5)

lora16 [44]

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