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3 years ago

6 = -5 -x solve for x

Mathematics

1 answer:

makkiz [27]3 years ago

7 0

Answer:

Step-by-step explanation:

In algebra, the goal is always to isolate the variable, so that its value can be determined.

6 + x = -5

<h3>Step 2: Subtract 6</h3>

x = -11

<h3>Step 3: Check</h3>

6 = -5 - -11

6 = -5 + 11

6 = 6 ✔

<h3>Step 4: Answer</h3>

x = -11

I'm always happy to help :)

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Tia performed an experiment where she flipped a coin 200 times. The coin landed heads up 92 times and tails up 108 times. Which

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Total number of times she flipped a coin =200.

Total number of heads in the experiment=92

Total number of tails in the experiment=108

Therefore probability of the coun landing heads up in this experiment is 0.5

In the above experiment, the probability of the coin landing heads up is

P(H)= 92/200 = 0.46

In the above experiment, the probability of the coin landing tails up is P(T) = 108/200 = 0.54

The ratio obj represent the experimental probability of the coin landing heads up in this experiment

Therefore the correct option is 1.

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(t-2)(t+1)(t+1)
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Scorpion4ik [409]

Answer: 91/167

Explanation:

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3 years ago

Help.........................<br><br>

garri49 [273]

<h3>Answer: Choice A</h3>

$x^2\left(\sqrt[4]{x^2}\right)$

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

Explanation:

The fourth root of x is the same as x^(1/4)

I.e,

$\sqrt[4]{x} = x^{1/4}$

The same applies to x^10 as well

$\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4}$

Multiply the exponents 10 and 1/4 to get 10/4

$\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4} = x^{10*1/4} = x^{10/4}$

$\sqrt[4]{x^{10}} = x^{10/4}$

-----------------------

If we have an expression in the form x^(m/n), with m > n, then we can simplify it into an equivalent form as shown below

$x^{m/n} = x^a\sqrt[n]{x^b}$

The 'a' and 'b' are found through dividing m/n

m/n = a remainder b

'a' is the quotient, b is the remainder

-----------------------

The general formula can easily be confusing, so let's replace m and n with the proper numbers. In this case, m = 10 and n = 4

m/n = 10/4 = 2 remainder 2

We have a = 2 and b = 2

$x^{m/n} = x^a\sqrt[n]{x^b}$

turns into

$x^{10/4} = x^2\sqrt[4]{x^2}$

which means

$\sqrt[4]{x^{10}} = {x^2} \sqrt[4]{x^2}$

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valkas [14]

Hope this helps lol :)

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