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3 years ago

7

What conversion factors are used in the Currency Calamity activity

1 answer:

kodGreya [7K]3 years ago

4 0

Currency Calamity activity are the cost of euros and the cost of euros that equal to U.S dollars.

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Find the coordinates of the midpoint of VW with endpoint V(-2,-6) and W(x+2,y+3)

mariarad [96]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ V(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-6})\qquad W(\stackrel{x_2}{x+2}~,~\stackrel{y_2}{y+3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{(x+2)-2}{2}~~,~~\cfrac{(y+3)-6}{2} \right)\implies \left( \cfrac{x}{2}~~,~~\cfrac{y-3}{2} \right)$

7 0

3 years ago

What is the answer for the equation (2+3x)4

balandron [24]

Answer:

8+12x

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

For each value of y, determine whether it is a solution to 12 = 2y +8.<br> Is it a solution

grin007 [14]

Answer: 2(0) + 8 does not equal 12, not a solution.

2(2) +8 = 12 yes it is a solution

2(-3) + 8 does not equal 12, not a solution

2(5) + 8 does not equal 12, not a solution.

Step-by-step explanation:

Looks like you need to plug in each y value given and multiplied by 2 and add 8

2(0) + 8 does not equal 12, not a solution.

2(2) +8 = 12 yes it is a solution

2(-3) + 8 does not equal 12, not a solution

2(5) + 8 does not equal 12, not a solution.

3 0

3 years ago

H(x)=3x^2+7<br> h(0)<br> help!

Angelina_Jolie [31]

Answer:

h(0) = 7

Step-by-step explanation:

by saying h(0) it is asking you to plug in 0 as the x value

3(0)^2 = 0

0 + 7 = 7

8 0

2 years ago

Which are the solutions of the quadratic equation? x2 = –5x – 3 –5, 0 StartFraction negative 5 minus StartRoot 13 EndRoot Over 2

boyakko [2]

<u>Given</u>:

The quadratic equation is $x^{2}=-5 x-3$

We need to determine the solutions of the quadratic equation.

<u>Solution</u>:

Let us solve the equation to determine the value of x.

Adding both sides of the equation by 5x and 3, we get;

$x^{2}+5 x+3=0$

The solution of the equation can be determined using quadratic formula.

Thus, we get;

$x=\frac{-5 \pm \sqrt{5^{2}-4 \cdot 1 \cdot 3}}{2 \cdot 1}$

$x=\frac{-5 \pm \sqrt{25-12}}{2 }$

$x=\frac{-5 \pm \sqrt{13}}{2 }$

Thus, the two roots of the equation are $x=\frac{-5 + \sqrt{13}}{2 }$ and $x=\frac{-5- \sqrt{13}}{2 }$

Hence, the solutions of the equation are $x=\frac{-5 + \sqrt{13}}{2 }$ and $x=\frac{-5- \sqrt{13}}{2 }$

6 0

3 years ago

Read 2 more answers