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4 years ago

Write 337,060 in expanded form using exponents

2 answers:

Crank4 years ago

8 0

337060 is the given number.
Expanded form is an expression that expands a given standard form equation.
In this given number we need to convert it into expanded form using anexponent.
=> 337 060
First, let’s simplify each values
=> 300 000 = 3 x 105
=> 30 000 = 3 x 104
=> 7 000 = 7 x 103
=> 60 = 6 x 10
=>3 x 105 + 3 x 104+ = 7 x 103 + 6 x 10 – expanded form for the given numbers withexponents.

Lerok [7]4 years ago

4 0

3 x 10 to the sixth power
3 x 10 to the fifth power
7 x 10 to the fourth power
6 x 10 to the first power

I might be wrong though. If so, sorry!

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Answer/Step-by-step explanation:

All of the solutions to the equation 3x^2 - 12 = 0 are x = 12 and x = -2

Answer: False

Explanation:

$3x^2-12+12=0+12$

$3x^2=12$

$\frac{3x^2}{3}=\frac{12}{3}$

$x^2=4$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{4},\:x=-\sqrt{4}$

$x=2,\:x=-2$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

There are two unique solutions to the equations (x-3)^2 = 16

Note: Each variable in the matrix can have only one possible value, and this is how you know that this matrix has one unique solution.

Answer: True

Explanation:

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=7,\:x=-1$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Ths solutions for the equation 2(x-3)^3 - 18 = 0 are x = 6 and x = 0

Answer: False

Explanation:

$2\left(x-3\right)^3-18+18=0+18$

$2\left(x-3\right)^3=18$

$\frac{2\left(x-3\right)^3}{2}=\frac{18}{2}$

$\left(x-3\right)^3=9$

$\mathrm{For\:}g^3\left(x\right)=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}$

$=\sqrt[3]{9}+3,\:x=\frac{6-\sqrt[3]{9}}{2}+i\frac{\sqrt[3]{9}\sqrt{3}}{2},\:x=\frac{6-\sqrt[3]{9}}{2}-i\frac{\sqrt[3]{9}\sqrt{3}}{2}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~The solutions for the equation 2(x-5)^2-8=0 are x = 7 and x = -7

Answer: False

Explanation:

$2\left(x-5\right)^2-8+8=0+8$

$2\left(x-5\right)^2=8$

$\frac{2\left(x-5\right)^2}{2}=\frac{8}{2}$

$\left(x-5\right)^2=4$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=7,\:x=3$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation (x + 3)^2-25 = -8 are x = 2 and x = -8

Answer: False

Explanation:

$\left(x+3\right)^2-25+25=-8+25$

$\left(x+3\right)^2=17$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{17}-3,\:x=-\sqrt{17}-3$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation 2(2x-1)^2=18 are x = 5 and x = -4

Answer: False

Explanation:

$\frac{2\left(2x-1\right)^2}{2}=\frac{18}{2}$

$\left(2x-1\right)^2=9$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=2,x=-1$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The only solution for equation (2x-1)^2-49=0 is x = 4

Answer: False
Explanation:

$\left(2x-1\right)^2-49+49=0+49$

$\left(2x-1\right)^2=49$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=4,\:x=-3$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation 3(x+2)^2 - 3 = 0 are x = -3 and x = -1

Answer: True
Explanation:

$3\left(x+2\right)^2-3+3=0+3$

$3\left(x+2\right)^2=3$

$\frac{3\left(x+2\right)^2}{3}=\frac{3}{3}$

$\left(x+2\right)^2=1$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=-1,\:x=-3$