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2 years ago

Show how to solve the equation in the picture

Mathematics

1 answer:

MArishka [77]2 years ago

3 0

$\left(3\cdot8\cdot x\right)^7\\\\\left(3\cdot2\cdot4\cdot x\right)^7\\\\\left((3\cdot2)\cdot4\cdot x\right)^7\\\\\left(6\cdot4\cdot x\right)^7\\\\6^7\cdot4^7\cdot x^7\\\\$

In the second step, I rewrote 8 as 2*4. Then after that I regrouped terms so that the (3*2) can pair up together. This of course turns into 6. After that point, we have 6, 4 and x multiplied together as the base.

The last step uses the rule that $(a*b*c)^d = a^d*b^d*c^d$ , in other words, we apply the exponent d to each term inside the parenthesis. Each term inside gets its own exponent.

An example: $(2*4*6)^3 = 2^3*4^3*6^3$

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Three students were given the expression shown and were asked to take a common factor out of two of the terms. Use the drop-down

Tomtit [17]

Answer:

Calvin: incorrect, sign error

Greg: incorrect, has no relationship to the given expression

Ruben: incorrect, sign error

Equivalent expression: 7 92(4x -11)

Step-by-step explanation:

I hope this helps you :)

-KeairaDickson

3 0

3 years ago

Draw as many different triangles as possible that have two sides of length 4 cm and a 45∘ angle. How many can you draw?

zmey [24]

The answer is 3 sorry if it’s wrong

7 0

2 years ago

Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficie

Dvinal [7]

For part (a), you have

$\dfrac x{x^2+x-6}=\dfrac x{(x+3)(x-2)}=\dfrac a{x+3}+\dfrac b{x-2}$
$x=a(x-2)+b(x+3)$

If $x=2$ , then $2=b(2-3)\implies b=-2$ .

If $x=-3$ , then $-3=a(-3-2)\implies a=\dfrac35$ .

So,

$\dfrac x{x^2+x-6}=\dfrac 3{5(x+3)}-\dfrac 2{x-2}$

For part (b), since the degrees of the numerator and denominator are the same, you first need to find the quotient and remainder upon division.

$\dfrac{x^2}{x^2+x+2}=\dfrac{x^2+x+2-x-2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}$

In the remainder term, the denominator $x^2+x+2$ can't be factorized into linear components with real coefficients, since the discriminant is negative $(1-4\times1\times2=-7)$ . However, you can still factorized over the complex numbers, so a partial fraction decomposition in terms of complexes does exist.

$x^2+x+2=0\implies x=-\dfrac12\pm\dfrac{\sqrt7}2i$
$\implies x^2+x+2=\left(x-\left(-\dfrac12+\dfrac{\sqrt7}2i\right)\right)\left(x-\left(-\dfrac12-\dfrac{\sqrt7}2i\right)\right)$
$\implies x^2+x+2=\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)$

Then you have

$\dfrac{x+2}{x^2+x+2}=\dfrac a{x+\dfrac12-\dfrac{\sqrt7}2i}+\dfrac b{x+\dfrac12+\dfrac{\sqrt7}2i}$
$x+2=a\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)+b\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)$

When $x=-\dfrac12-\dfrac{\sqrt7}2i$ , you have

$-\dfrac12-\dfrac{\sqrt7}2i+2=b\left(-\dfrac12-\dfrac{\sqrt7}2i+\dfrac12-\dfrac{\sqrt7}2i\right)$
$\dfrac32-\dfrac{\sqrt7}2i=-\sqrt7ib$
$b=\dfrac12+\dfrac3{2\sqrt7}i=\dfrac1{14}(7+3\sqrt7i)$

When $x=-\dfrac12+\dfrac{\sqrt7}2i$ , you have

$-\dfrac12+\dfrac{\sqrt7}2i+2=a\left(-\dfrac12+\dfrac{\sqrt7}2i+\dfrac12+\dfrac{\sqrt7}2i\right)$
$\dfrac32+\dfrac{\sqrt7}2i=\sqrt7ia$
$a=\dfrac12-\dfrac3{2\sqrt7}i=\dfrac1{14}(7-3\sqrt7i)$

So, you could write

$\dfrac{x^2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}=1-\dfrac {7-3\sqrt7i}{14\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)}-\dfrac {7+3\sqrt7i}{14\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)}$

but that may or may not be considered acceptable by that webpage.

5 0

2 years ago

Read 2 more answers

Which equation could represent "the area of a square as a function of a side"?

Katen [24]

Let

s-------> the side length of a square

we know that

The area of a square is equal to

$A=s^{2}$