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3 years ago

Triangle ABC is reflected over the X axis. TRIANGLE ABC has points A(10,2) B(9,5) C(6,4). What is the C coordinate

Mathematics

1 answer:

gavmur [86]3 years ago

4 0

1. Let A(a,b) be a point in the coordinate axis. The reflection of A with respect to the x-axis is the point A'(a, -b).

2. So the reflection of C(6, 4) is C'(6, -4)

3. Remark, a picture is always helpful when dealing with reflections.

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Which statements are true of the quadrilateral she constructed? Select three options.

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3 years ago

Pls help me on this question:)

r-ruslan [8.4K]

Answer:

x= 28

Step-by-step explanation:

The angles both sum up to 180°. So, the equation would be: 39+(5x+1)= 180.

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(39+1)+5x= 180

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3 years ago

PLZ EXPLAIN. what is the area of this figure?

BlackZzzverrR [31]

The area of this figure is 532m.

First you do the square which is 18 time 18 gets you 324m.

The for the first triangle you would do 18 times 16 then divide by 2 which is 144m.

For the last triangle you would do 16•8 div by 2 which will give you 64m.

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4 years ago

HELPPPPPPPPP i have the most trouble with math what’s the answer??

lakkis [162]

Answer:

Second Option

$\frac{5\sqrt{11} +5\sqrt{3}}{8}$

Step-by-step explanation:

To simplify the expression you must multiply the numerator and the denominator of the fraction by the conjugate of the denominator.

If you have an expression of the form

$a -b$ then its conjugate will be $a + b.$

The result of that multiplication will be

$a ^ 2 -b ^ 2$

Therefore we have the expression

$\frac{5}{\sqrt{11} -\sqrt{3}}$

By multiplying by the conjugate of the denominator we have

$\frac{5}{\sqrt{11} -\sqrt{3}}*\frac{\sqrt{11} +\sqrt{3}}{\sqrt{11} +\sqrt{3}}\\\\\\\frac{5(\sqrt{11} +\sqrt{3})}{(\sqrt{11})^2 -(\sqrt{3})^2}\\\\\\\frac{5\sqrt{11} +5\sqrt{3}}{11 -3}\\\\\\\frac{5\sqrt{11} +5\sqrt{3}}{8}$

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3 years ago