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3 years ago

Heyy! i’ll give brainliest please help thanks

Mathematics

2 answers:

ankoles [38]3 years ago

8 0

Answer:

I think its A

hope this helps

have a good day :)

Step-by-step explanation:

BabaBlast [244]3 years ago

6 0

The answer to this is A

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Find the sum. Express your answer in simplest form.

Marina86 [1]

Answer:

$\frac{10x^2+3}{y^2+4}$

Step-by-step explanation:

When adding rational numbers, if the denominator is the same you simply keep the denominator (bottom of fraction) as it is and apply the operation given to the numerator ( top of fraction )

So we have $\frac{6x^2+5}{y^2+4} +\frac{4x^2-2}{y^2+4} =\frac{(6x^2+5)+(4x^2-2)}{y^2+4}$

==> remove parenthesis and apply signs

$\frac{x^2+5+4x^2-2}{y^2+4}$

==> simplify numerator by combining like terms

$\frac{10x^2+3}{y^2+4}$

and we are done!

Note:

like terms are terms with the same variable and exponent

An example of like terms are 6x^7 and 3x^7 as they have x as a variable and a power of 7

The like terms being combined here were (6x² and 4x²) and (5 and -2)

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2 years ago

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Can someone teach me how to solve this please

Alexeev081 [22]

Multiply it all together

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3 years ago

Please help me find the value of x. asap •

Juliette [100K]

How much after subtraction

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3 years ago

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A 20 kg bag of bananas for $10<br><br><br> ______ per kg

NeTakaya

Answer:

each bannana i think

Step-by-step explanation:

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3 years ago

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A particle moves along a line with acceleration a (t) = -1/(t+2)2 ft/sec2. Find the distance traveled by the particle during the

amm1812

Answer:

$s(t)=(ln|7|+ln|2|)\,ft$

Step-by-step explanation:

Acceleration is second derivative of distance and are related as:

$a(t)=\frac{d^2s}{dt^2}\\\\\frac{d^2s}{dt^2}=\frac{-1}{(t+2)^2}\\$

Integrating both sides w.r.to t

$v(t)=\frac{ds}{dt}=\frac{1}{t+2} +C\\$

Using initial value

$v(0)=\frac{1}{2}\\\\\frac{1}{2}=\frac{1}{0+2} +C\\\\C=0\\\\\frac{ds}{dt}=\frac{1}{t+2}$

We have to calculate the distance covered in time interval [0,5], so:

$\int\limits^5_0 \frac{ds}{dt}=\int\limits^5_0 {\frac{1}{t+2}} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft$

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3 years ago