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3 years ago

11

A recipe for granola bars calls for 2/3 cup of oats. How much of the pats would you use to make 1/2 of the amount in the origina

l recipe?

1 answer:

Veseljchak [2.6K]3 years ago

5 0

Answer:

1/3 cups

Step-by-step explanation:

If you want to make half of the recipe that requires 2/3 cup oats, then you would use half the oats. Half of 2/3 cup of oats is 1/3 cup of oats.

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If I had a 2% solution, added 2 ounces to a gallon (128 ounces) What is the final concentration of the solution.

Ne4ueva [31]

<span>A 2% solution of chlorohexidine added to 126 oz of water. The total oz 128.

Solute: chlorohexidine, 2%=0.02
solvent: water (0%)

Concentration, C, volume V
CV=c1v1+c2v1
128C=0.02*2 + 0.00*126 = 0.04

Concentration
C=0.04/128 = 0.0003125 = 0.03125%</span>

6 0

4 years ago

Alona [7]

Answer:

The answer is C) - |x - 2| + 3 = 0

Step-by-step explanation:

a) - |x - 2| + 3 > 0

The greater than means something above the graph.

b) - |x - 2| + 3 < 0

The less than means something below the graph.

c) - |x - 2| + 3 = 0

The zero doesn't do anything.

y = - a |x - h| + k = 0

(h,k) is the origin.

h is shift left/right, h = 2 so shift right 2.

k is shift up/down, k = 3 so shift up 3.

The a, affects the y-values which cause the graph to reflect across the x-axis.

6 0

3 years ago

The radius of a circle is 3 miles. What is the circle's diameter?

Gemiola [76]

Answer: 6 miles

Step-by-step explanation:

The diameter is 2x the radius

2 x 3 = 6

4 0

3 years ago

Read 2 more answers

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.

<u>case a)</u> $x^{2}+2x+4=0$

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

<u>case b)</u> $x^{2}-2x+4=0$

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

<u>case c)</u> $x^{2}+2x-4=0$

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

<u>case d)</u> $x^{2}-2x-4=0$

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

<u>the answer is</u>

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

5 0

3 years ago

Read 2 more answers

katrin [286]

I’m pretty sure it’s added, subtracted, added !

3 0

3 years ago