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Tcecarenko [31]
3 years ago
8

Does anyone know how to solve this ?

Mathematics
1 answer:
crimeas [40]3 years ago
3 0

9514 1404 393

Answer:

  see below

Step-by-step explanation:

Make use of the quadratic formula.

For equation ax² +bx +c = 0, the solutions are ...

  x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

Here, you have a=1, b=-1, c=1, so the solutions are ...

  x=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(1)}}{2(1)}=\dfrac{1\pm\sqrt{1-4}}{2}=\dfrac{1\pm i\sqrt{3}}{2}\\\\\boxed{x=\dfrac{1+i\sqrt{3}}{2}\text{ and }x=\dfrac{1-i\sqrt{3}}{2}}

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Can someone answer this please
fgiga [73]
<h3>Answer:  24 square meters</h3>

The notation "m^2" is a shorthand way of saying "square meters".

Chances are you'll simply type in the number and leave out the units, as that's already taken care of.

==========================================================

Work Shown:

area = base*height

area = 8*3

area = 24 square meters

---------

Explanation:

We simply multiply the base 8 and height 3 and that's our answer. It's not a trick question even if it might seem like it.

The side that's 7 meters long is never used. The base and height are always perpendicular to one another, which means that they form a right angle (aka 90 degree angle).

The area formula of a parallelogram and rectangle are the same. We can think of a non-rectangular parallelogram as a slanted rectangle in a sense. Imagine you had stacks of blocks to form a rectangle. Now imagine nudging that stack slightly to one side. The blocks themselves don't change, so the area of one wall doesn't change. All that's changed is the shape.

5 0
3 years ago
Which expression is equivalent to log w (x^2 -6)^4/ 3 sqrt x^2+8?
evablogger [386]

Answer:

C 4\log_w(x^2-6)-\dfrac{1}{3}\log_w(x^2+8)

Step-by-step explanation:

First use the property of logarithms

\log _ab-\log_ac=\log_a\dfrac{b}{c}.

For the given expression you get

\log_w\dfrac{(x^2-6)^4}{\sqrt[3]{x^2+8} }=\log_w(x^2-6)^4-\log_w\sqrt[3]{x^2+8}=\log_w(x^2-6)^4-\log_w(x^2+8)^{\frac{1}{3}}

Now use property of logarithms

\log_ab^k=k\log_ab.

For your simplified expression, you get

\log_w(x^2-6)^4-\log_w(x^2+8)^{\frac{1}{3}}=4\log_w(x^2-6)-\dfrac{1}{3}\log_w(x^2+8).

3 0
3 years ago
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