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den301095 [7]
2 years ago
9

Use the table and equation to answer the following question.

Mathematics
1 answer:
ale4655 [162]2 years ago
8 0

go ask ur mom for help not me ty

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What is <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7B8%7D%20%20-%20%20%5Cfrac%7B1%7D%7B4%7D%20" id="TexFormula1" ti
marysya [2.9K]
Hello there, and thank you for posting your question here on brainly.

Something I noticed right away is that 4/8 can be simplified.

4/8 ===> 2/4

So now we have 2/4 - 1/4

Just subtract the numerators only.

2/4 - 1/4 = 1/4

1/4 cannot be simplified, so 1/4 is your final answer.

Hope this helped!! ☺♥
5 0
3 years ago
Which equation represents a parabola?
Evgesh-ka [11]

Answer:

(B)

Step-by-step explanation:

we know that standard equation of parabola y²= 4x or x² = 4y .

So option (B) is correct .

Hope it's helpful

5 0
3 years ago
Subtraction (must show work) 1. d - 27 = 45 2. h - 114 = 28
Arturiano [62]

Answer:

Step-by-step explanation:

1.

d - 27 = 45

add 27

d = 72

2.

h - 114 = 28

add 114

h = 142

3.

-4 + x = 15

add 4

x = 19

4.

-39 + g = 72

add 39

g = 111

7 0
2 years ago
Which choice is equivalent to the product below when x&gt;0?
V125BC [204]

Answer:

D

Step-by-step explanation:

Using the rule of radicals

\sqrt{\frac{a}{b} } = \frac{\sqrt{a} }{\sqrt{b} }

\sqrt{a\\} × \sqrt{b} ⇔ \sqrt{ab}

Given

\sqrt{\frac{6}{x} } × \sqrt{\frac{x^2}{24} }

= \frac{\sqrt{6} }{\sqrt{x} } × \frac{x^2}{24}

= \frac{\sqrt{6} }{\sqrt{x} } × \frac{x}{2 \sqrt{6\\} }

Cancel \sqrt{6} on numerator/ denominator

= \frac{1}{\sqrt{x} } × \frac{x}{2\\}

= \frac{1}{\sqrt{x} } × \frac{(\sqrt{x})^2 }{2}

Cancel \sqrt{x} on numerator/ denominator, leaving

= \frac{\sqrt{x} }{2} → D

4 0
3 years ago
Marguerite answered 79% of the questions on her test correctly. What fraction of the questions did she answer correctly?
djyliett [7]

Answer:

79 of the questions were right/coreect

Step-by-step explanation:

if you have 100 quesions and you get 79% done and correct then you did 79 answers right

3 0
2 years ago
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