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2 years ago

If x>0, what values of c and d make the equations true?

Mathematics

2 answers:

Dominik [7]2 years ago

6 0

Answer:

a=7

b=5

Step-by-step explanation:

pantera1 [17]2 years ago

5 0

Answer:

In equation A, c = 7

In equation B, d = 5

Step-by-step explanation:

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Enter your answer as a simplified fraction formatted like this: 42/53

expeople1 [14]

Answer:

$\displaystyle -0.\overline{37}=-\frac{37}{99}$

Step-by-step explanation:

<u>Convert a Repeating Decimal to a Fraction </u>

We need to convert the following decimal to a fraction:

$-0.\overline{37}$

Let's call x to the given number:

$x=-0.\overline{37}$

The number has two repeating (periodic) decimals, thus we multiply by 100:

$100x=-37.\overline{37}$

Note the decimals continue to repeat exactly like the original number.

Subtracting both expressións, the repeating decimals are simplified:

$100x - x = -37$

Operating:

99x=-37

Dividing by 99:

$\displaystyle x=-\frac{37}{99}$

The fraction cannot be simplified, thus:

$\mathbf{\displaystyle -0.\overline{37}=-\frac{37}{99}}$

8 0

2 years ago

Write as an expression " 6 times smaller than 42"

Mademuasel [1]

Answer:

42/6 or 7

Step-by-step explanation:

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2 years ago

In the graph below, Point A represents Owen's house, Point B represents David's house and Point C represents the school. Who liv

ycow [4]

Answer:

• David

• 4 miles

Explanation:

In the graph:

The given locations are:

• Owen's House, A(11,3)

• David's House, B(15,13)

• School, C(3,18)

We determine both Owen's and David's distance from the school using the distance formula.

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Owen's distance from school (AC)

$\begin{gathered} AC=\sqrt[]{(3-11)^2+(18-3)^2} \\ =\sqrt[]{(-8)^2+(15)^2} \\ =\sqrt[]{64+225} \\ =\sqrt[]{289} \\ AC=17\text{ miles} \end{gathered}$

David's distance from school (BC)

$\begin{gathered} BC=\sqrt[]{(3-15)^2+(18-13)^2} \\ =\sqrt[]{(-12)^2+(5)^2} \\ =\sqrt[]{144+25} \\ =\sqrt[]{169} \\ BC=13\text{ miles} \end{gathered}$

We see from the calculations that David lives closer to the school, and by 4 miles.

The graph below is attached for further understanding:

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10 months ago