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3 years ago

Two children are selling lemonade. They are charging $1 what situations could the other graphs represent

Mathematics

1 answer:

lina2011 [118]3 years ago

7 0

Answer:

Step-by-step explanation:

the answer is A

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Need help!

omeli [17]

Rational numbers: B C D G
Irrational numbers: A E F
Irrational numbers are numbers that cannot be written as fractions, rational numbers are all other numbers

5 0

4 years ago

Can someone please answer. There is one problem. There's a picture. Thank you!

allochka39001 [22]

$\dfrac{60^o}{360^o}=\dfrac{1}{6}\\\\A_O=\pi\cdot4^2=16\pi\\\\Area\ of\ the\ shaded\ sector:\\\\A=\dfrac{1}{6}\cdot16\pi=\dfrac{8}{3}\pi\approx\dfrac{8}{3}\cdot3.14\approx8.4\\\\Answer:\boxed{8.4}$

8 0

3 years ago

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

case a) $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

case b) $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

case c) $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

case d) $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

the answer is

$x^{2}-2x-4=0$

5 0

3 years ago

Read 2 more answers

The product of a number and its _____ is one. Please fill in the blank.

asambeis [7]

Itself should go in the blank

4 0

3 years ago

If g(x) = -2x + 9, what is the value of x when g(x) = 3?

MissTica

Answer:

x = 3

Step-by-step explanation:

1. Plug 3 into g(x)

3 = -2x + 9

2. Subtract 9 on both sides

-6 = -2x

3. Divide by -2 on both sides to get x by itself

x = 3

I hope this helped :)

7 0

3 years ago