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3 years ago

I need help with 1-43

Mathematics

1 answer:

Liono4ka [1.6K]3 years ago

8 0

1-43 =

-42

You can always subtract the opposite and add the negative

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Which value is NOT a solution of 8x3 – 1 = 0?

Tpy6a [65]

Note: As you may have unintentionally missed to add the value choices. But, I would make sure to explain the concept so that you may improve your understanding in terms of solving these type of questions.

Answer:

Any value other than the values $x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$ will not be a solution of $8x^3\:-\:1\:=\:0$ .

Step-by-step explanation:

Considering the equation

$8x^3\:-\:1\:=\:0$

Steps to solve the equation

$8x^3-1=0$

$\mathrm{Add\:}1\mathrm{\:to\:both\:sides}$

$8x^3-1+1=0+1$

$\mathrm{Simplify}$

$x^3=\frac{1}{8}$

$\mathrm{Divide\:both\:sides\:by\:}8$

$\frac{8x^3}{8}=\frac{1}{8}$

$\mathrm{Simplify}$

$x^3=\frac{1}{8}$

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

$x=\sqrt[3]{\frac{1}{8}},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1+\sqrt{3}i}{2},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1-\sqrt{3}i}{2}$

So,

$x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$

Therefore,

Any value other than the values $x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$ will not be a solution of $8x^3\:-\:1\:=\:0$ .

Keywords: solution, value

Learn more about equation solution from brainly.com/question/1679491

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3 years ago

Which might be a better way to evaluate 1150 divided by 46: drawing base-ten diagrams or using the partial quotients method ? Ex

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Answer:

idk sounds like a good question

Step-by-step explanation:

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Write the expression as a single logarithm

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$3\log_7y + 4(\log_7x - 5\log_7z)\\\log_7y^{3} + 4(\log_7x - 5\log_7z)\\\log_7y^{3} + 4(\log_7x - \log_7z^{5})\\\log_7y^{3} + 4(\log_7(\frac{x}{z^{5}}))\\\log_7y^{3} + \log_7(\frac{x}{z^{5}})^{4}\\\log_7(y^{3} \times (\frac{x}{z^5{5}})^{4})$

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3 years ago

Find the zeros of polynomial function and solve polynomials equations. f(x)=24x^3-64x^2-21x+56

8090 [49]

Since this polynomial has 4 terms, factoring by grouping should be the first thing we try here.

So, we have:

$8x^2(3x-8)-7(3x-8)=0 \implies\\ (8x^2-7)(3x-8)=0$

So, we can use ZPP to find out roots:

$3x-8 =0 \implies\\ x=\frac{8}{3}\\ 8x^2-7=0 \implies \\ 8x^2=7 \implies \\ x^2=\frac{7}{8} \implies \\ x=\pm\frac{\sqrt{7}}{\sqrt{{8}}}\\ \text{Rationalizing denominator:}\\ x=\pm\frac{\sqrt{56}}{8}$

So our three roots are:

$x \in \{\frac{8}{3},\frac{\sqrt{56}}{8},-\frac{\sqrt{56}}{8}\}$

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3 years ago