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2 years ago

Plz help me, I'm not trying to fail math

Mathematics

1 answer:

spayn [35]2 years ago

8 0

The answer is a i believe

you can solve it

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You recently deposited $2000 in a bank account during your first year at work.

lara [203]

Answer:

A. $12,000

B. $147,000

Step-by-step explanation:

8 0

2 years ago

Please help me and pls do both

Nana76 [90]

Answer:

7/36

Step-by-step explanation:

find the LCD write all the numerators above the least common denominator

LCD is 36

4 +3 /36 =7/36

3 0

2 years ago

What additional information would allow you to prove the quadrilateral is a parallelogram?

slavikrds [6]

The additional information that would allow us to prove that the image is a parallelogram is that; Line EJ ≅ Line GJ

<h3>How to prove a Parallelogram?</h3>

The six basic properties of parallelograms are primarily;

Both pairs of opposite sides are parallel

Both pairs of opposite sides are congruent

Both pairs of opposite angles are congruent

Diagonals bisect each other

One angle is supplementary to both consecutive angles (same-side interior)

One pair of opposite sides are congruent AND parallel.

Now, looking at the parallelogram properties above and comparing with the given image of the quadrilateral attached, we can say that the additional information that would allow us to prove that the image is a parallelogram is that; Line EJ ≅ Line GJ

Read more about Parallelogram Proof at; brainly.com/question/24056495

#SPJ1

7 0

2 years ago

HELPP FOR BRAAINLISTT

Phoenix [80]

Answer:

The slope is approximately 2 and the y-intercept is approximately 3.

Step-by-step explanation:

I don't know if this is right but I hope it is have an amazing day.

5 0

1 year ago

Read 2 more answers

Factor completely <br> 4(x+1)^2/3 + 12(x+1)^-1/3

wolverine [178]

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
\left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}
\\\\
-------------------------------\\\\
%4(x+1)^2/3 + 12(x+1)^-1/3
4(x+1)^{\frac{2}{3}}+12(x+1)^{-\frac{1}{3}}\implies 4(x+1)^{\frac{2}{3}}+12\cfrac{1}{(x+1)^{\frac{1}{3}}}
\\\\\\$

$\bf 4(x+1)^{\frac{2}{3}}+\cfrac{12}{(x+1)^{\frac{1}{3}}}\impliedby \textit{so, our LCD is }(x+1)^{\frac{1}{3}}
\\\\\\
\cfrac{4(x+1)^{\frac{2}{3}}\cdot (x+1)^{\frac{1}{3}}+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+1)^{\frac{2}{3}+\frac{1}{3}}+12}{(x+1)^{\frac{1}{3}}}
\\\\\\
\cfrac{4(x+1)^{\frac{3}{3}}+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+1)+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4x+4+12}{(x+1)^{\frac{1}{3}}}
\\\\\\
\cfrac{4x+16}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+4)}{\sqrt[3]{x+1}}$

3 0

3 years ago

Read 2 more answers

Plz help me, I'm not trying to fail math ​

Plz help me, I'm not trying to fail math