MATHEMATICS HIGH SCHOOL

What is the image of (-4,-7)(−4,−7) after a reflection over the x-axis?

Answers

Answer 1

Answer:

The image of the point (-4, -7) after a reflection over the x-axis is (-4, 7)

Calculating the image of the point after a reflection over the x-axis?

From the question, we have the following parameters that can be used in our computation:

Point = (-4, -7)

Transformation rule

A reflection over the x-axis

The rule of a reflection over the y-axis is

(x, y) = (x, -y)

Using the above as a guide, we have the following:

Image = (-4, 7)

Hence the image of the point after a reflection over the x-axis is (-4, 7)

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Which of the following is equivalent to 6x 2 + x - 12 ?A. (2x + 3)(3x - 4)
B. (2x - 3)(3x - 4)
C. (2x + 3)(3x + 4)
D. (6x + 3)(x - 12)
E. (x + 3)(2x - 12)
Please show me step by step how to solve this problem
Thanks

Answers

$6x^2+x-12=6x^2-9x+8x-12=3x(2x-3)+4(2x-3)\n\n=(2x-3)(3x+4)\n\nother\ method:\n\n6x^2+x-12\na=6;\ b=1;\ c=-12\n\Delta=b^2-4ac\to\Delta=1^2-4\cdot6\cdot(-12)=1+288=289\n\nx_1=(-b-\sqrt\Delta)/(2a);\ x_2=(-b+\sqrt\Delta)/(2a)\n\n\sqrt\Delta=√(289)=17$

$x_1=(-1-17)/(2\cdot6)=(-18)/(12)=-(3)/(2);\ x_2=(-1+17)/(2\cdot6)=(16)/(12)=(4)/(3)\n\n6x^2+x-12=6(x+(3)/(2))(x-(4)/(3))=2\cdot3(x+(3)/(2))(x-(4)/(3))\n\n=2(x+(3)/(2))\cdot3(x-(4)/(3))=(2x+3)(3x-4)$

Which of the following is not a congruence transformation? A.Translating
B.Rotating
C.Dilating

Answers

The answer is C.) Dilating
because dilating doesn't make them congruent anymore, but they change size to be either smaller or bigger than one another, but the shape itself does not change.

love, the Pineapple.
<3

Can anyone help me solve this math problem?

Answers

Step-by-step explanation:

[ Refer to the attached file ]

Used method TrigonometricRatio!!

Find the area A of polygon ABCDEFG with the given vertices. A(0,5), B(2,5), C(2,3), D(3,2), E(1,2), F(0,3), G(−4,3)

Answers

The area A of polygon ABCDEFG with the given vertices is 10 square units

How to find the area A of polygon ABCDEFG with the given vertices?

The vertices are given as

A(0,5), B(2,5), C(2,3), D(3,2), E(1,2), F(0,3), G(−4,3)

The area of the polygon is calculated using

Area = 1/2 * |(x1y2 -x2y1) +(x2y3 -x3y2) + ..... + (xny -xyn)|

So, we have

Area = 1/2 * |(0 * 5 - 5 * 2) + (2 * 3 - 5 * 2) + (2 * 2 - 3 * 3) + (3 * 2 - 2 * 1) + (1 * 3 - 2 * 0) + (0 * 3 - 3 * -4) +(-4 * 5 - 3 * 0)|

Evaluate the sum of products

So, we have

Area = 1/2 * |-20|

Remove the absolute bracket

So, we have

Area = 1/2 * 20

Evaluate the products

So, we have

Area = 10

Hence, the area A of polygon ABCDEFG with the given vertices is 10 square units

Answers

Answer:

1-7i

17i

Step-by-step explanation:

7-5i-6-2i=1-7i

1+9i-1+8i=17i

3i+14-3i=14

Jerry had an average score of 85 on his first 8 quizzes. He had an average score of 81 on his first nine quizzes. What score did he earn on his ninth quiz? side note: I got 49 the first time i tried it but when i checked it, it was wronfg

Answers

Well, let's calculate how many points in total he got so far!

He did 8 quizzes, on average 85 points: 85*8=680 points.

He scored a number of points in his ninth quiz, let's call this number "x".

If we want to calculate the average after the ninth quiz, we take the sum of points then (which is 680+x) and we divide it by 9:

(680+x)/9

we also know that this is equal to 81:

(680+x)/9=81

$((680+x))/(9)=81$

multiply both sides by 9:

$(680+x)}=729$

substracting 680 from both sides:

x=49

well, i got 49 too.

I really can't see where the mistake could be?

where did you check it? my best guess is that the place where you checked it had a typo...?

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