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vladimir1956 [14]
4 years ago
8

To increase an amount by 3% what single multiplier would you use?

Mathematics
1 answer:
loris [4]4 years ago
3 0

3% is the decimal  0.03 .

To make a number 3% bigger, multiply it by  1.03 .

The ' 1 ' part is the amount that's already there, and
the ' .03 ' part is the amount to be added onto it.

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-7x^3-9x^2+8x this is factoring completely
loris [4]
Answer:
<span> (x + 6) • (x + 5) • (x - 2)

Is this what u were looking for? Sorry if not XDD

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3 0
3 years ago
Read 2 more answers
I need help on this
nasty-shy [4]

Answer:

the answer would be -3/4

Step-by-step explanation:

(-4,3)(0,0)

y2-y1/x2-x1

0-3/0+4

-3/4


8 0
3 years ago
Plz help me I have no idea how to do this
Leviafan [203]

Answer:

0.8

Step-by-step explanation:

cos z = 8/10 = 0.8

8 0
3 years ago
Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

8 0
3 years ago
Solve 3.4 [ 6 ( 4) +6^2]<br><br> Please show your work! &lt;3
VARVARA [1.3K]

Answer:

204

Step-by-step explanation:

3.4[6(4)+6^2]=

3.4[24+36]=

3.4[60]=

204

Hope this helps!

7 0
3 years ago
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