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asambeis [7]
3 years ago
5

Applying GCF and LCM to fraction Operations?.. PLZ HELP

Mathematics
1 answer:
I am Lyosha [343]3 years ago
4 0
When multiplying fractions just multiply the numerator and denominators together, then simplify the fraction you get.

To solve the addition problem, first convert 2 and 1/2 to an improper fraction. It should be 5/2. Now 3/8 and 5/2 must be added together.

To add these, they must have the same denominator. Let's give 5/2 a denominator of 8, so it matches 3/8.

To give 5/2 a denominator of 8, you are starting with a denominator of 2. 2 × 4 = 8, so you must multiply 5/2 by 4.

You should get 20/8. Now you have to add 20/8 and 3/8. The only thing you add is the numerator, so you should get 23/8 as a final answer.

For the subtraction problem, do these same steps to make the fractions you must subtract have the same denominator, then subtract only the numerator to get the answer. Once you solve the subtraction problem, text me what you got so I can check it for you. :)
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-22= 10 - 8z?<br><br> What is the answer please help me!!!! I am really behind on school work.
I am Lyosha [343]
The answer here will be z=4
8 0
2 years ago
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1.Which statement about two negative numbers is true? .
Ugo [173]

Answer:

To negative numbers makes a positive number.

(only in dividing and multiplying)

Step-by-step explanation:

Please mark me brainliest

3 0
3 years ago
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Help please!!!llalslsnsnsnsn
salantis [7]

The answer you selected is correct

3x - 4y = 16

3x = 4y + 16

x = 4y/3 + 16/3

5 0
3 years ago
Write the equation of the line that passes through (5.-9) and (2.5).
Kipish [7]

Answer:

y = -14/3x + 43/3

Step-by-step explanation:

y2 - y1 / x2 - x1

5 - (-9) / 2 - 5

14 / -3

- 14/3

y = -14/3x + b

5 = -14/3(2) + b

5 = -28/3 + b

43/3 = b

y = -14/3x + 43/3

4 0
3 years ago
Solve step by step solution then only i can do it plxx ​
Anuta_ua [19.1K]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So \triangle ABC and \triangle ABD Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>
  • First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

  • We need to find base=b

According to Pythagoras thereon

{\boxed{\sf b^2=h^2-p^2}}

  • Substitutethe values

\longrightarrow\sf b^2=10^2-p^2

\longrightarrow\sf b={\sqrt {10^2-8^2}}

\longrightarrow\sf b={\sqrt{100-64}}

\longrightarrow\bf b={\sqrt {36}}

\longrightarrow\sf b=6

\therefore\overline{BC}=6cm

  • BD=BC+CD

\longrightarrowBD=9+6

\longrightarrowBD=15cm

  • Now in \triangle ABD

Perpendicular=p=8cm

Base =b=15cm

  • We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

{\boxed {\sf h^2=p^2+b^2}}

  • Substitute the values

\longrightarrow\sf h^2=8^2+15^2

\longrightarrow\sf h={\sqrt {8^2+15^2}}

\longrightarrow\sf h={\sqrt {64+225}}

\longrightarrow\sf h={\sqrt {289}}

\longrightarrow\sf h=17cm

\therefore{\underline{\boxed{\bf x=17cm}}}

3 0
3 years ago
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