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2 years ago

Khan academy how to find the domain of a function equation.

Mathematics

1 answer:

alexira [117]2 years ago

5 0

Answer:

I cant even see anything can you reupload your answer, please?

Step-by-step explanation:

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Functions quiz on khan. G(-4)=

kherson [118]

-4g is what I got if ur evaluating
If it’s something about slope I got -4

5 0

2 years ago

Please help!!

Lelechka [254]

Answer:

I made it to this and I don't know what to do for the rest sorry.

Step-by-step explanation:

y-5=-2(x-6)

+5 +5

y=3(x-6)

y=3x-18

7 0

3 years ago

Help! Urgent request

TiliK225 [7]

Answer: 6 students

What we know:

Students: 24

Students who play checkers: $\frac{2}{3}$

Students who also play sudoku: $\frac{3}{8}$ of the $\frac{2}{3}$

24 ÷ 3 = 8, so 8 × 2 = 16 (students who play checkers)

$\frac{3}{8}$ × 2 = $\frac{6}{16}$

So the answer is,

6 students play both checkers and sudoku

8 0

3 years ago

Read 2 more answers

I need helpasdasdasdasd

exis [7]

Answer: Is C

Your answer: C

8 0

3 years ago

In the figure below, AB=AE, AC=AD and AP is perpendicular to BE. Prove that BC=DE.

NISA [10]

Step-by-step explanation:

1) In the figure, as AB is equal to AE, ABE is an equilateral triangle.

As AP is perpendicular to BE

=> AP is the height of the triangle ABE.

In an equilateral triangle, the median and the height is the same, so that AP is also the median of the triangle.

=> P is the midpoint of BE

=> PE = PB

2) In the figure, as AC = AD, so that ACD is an equilateral triangle.

As AP is perpendicular to BE, so that it is perpendicular to CD as well

=> AP is the height of the triangle ACD

In an equilateral triangle, the median and the height is the same, so that AP is also the median of the triangle ACD.

=> P is the midpoint of CD

=> PC = PD

We have:

+) PB = PE

+) PC = PD

=> PB - PC = PE - PD => BC = DE

5 0

3 years ago