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3 years ago

Solve for X, will mark as brainliest! 6(1+x)+x=-14

Mathematics

2 answers:

Nutka1998 [239]3 years ago

8 0

Answer:

X is equal to -10

Step-by-step explanation:

#1. Multiply 6 to (1+x). Doing this makes this equation: 6+6x+x=-14

#2. Then add 6x and x to get: 7x+6=-14

#3. Then move 6 to other side of equation, while at the same time, changing it's absolute value: 7x=-14-6

#4. now you have: 7x=-20, now just divide 7 by -20 WooHoo! You finished it! you have: -2.85714285714 or 2 and 43/50.

VARVARA [1.3K]3 years ago

8 0

Answer:

x= $-2\frac{6}{7}$

Step-by-step explanation:

6(1+x)+x=-14

6+6x+x=-14

7x+6=-14

7x=-20

x=-2 $\frac{6}{7}$

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Ok so can y'all do this? 8n = 240

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Answer:

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Step-by-step explanation:

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Which number line represents the solutions to |x – 5| = 1? A number line from negative 7 to 7 in increments of 1. Two points, on

pochemuha

Answer:

Two points, one at 4 and one at 6.

Step-by-step explanation:

<u>Equations with absolute value </u>

The absolute value of a number x denoted as |x| is a number with the same magnitude with a positive sign. When we know the result of an absolute value, we cannot know if the original number is positive or negative.

If |x| = a, a>=0, then x=a or x=-a

The equation to solve is |x – 5| = 1. If we apply the above criteria, then

x-5=1 or x-5=-1

Which gives us two possible solutions for x

x=6

x=4

The correct answer is

Two points, one at 4 and one at 6.

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Answer:

1. her weekly allowance is 16 dollars

Step-by-step explanation:

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Please help if you can. I will give Brainiest to best answer. This question is a Calculus/Trigonometry question. I ask that you

Ierofanga [76]

7 is incorrect. The answer should be -4. Here's how I'd derive it:

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

With $\pi$ , we should expect $\cos x$ . If $\tan x=\dfrac8{15}$ , then

$\sec x=-\sqrt{1+\tan^2x}=-\dfrac{17}{15}\implies\cos x=-\dfrac{15}{17}$

Also,

$\pi$

so we should expect $\tan\dfrac x2$ and

$\tan\dfrac x2=-\sqrt{\dfrac{1-\cos x}{1+\cos x}}=-4$

You seem to be taking

$\tan\dfrac x2=\dfrac{1+\cos x}{-\sin x}$

but this is not an identity.

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8 is incorrect. Just a silly mistake, you swapped the order of the terms in the numerator. It should be

$\dfrac{\sqrt6-\sqrt2}4$

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9. Use the identity from (7). $\dfrac{7\pi}{12}$ lies in the second quadrant, so

$\tan\dfrac{7\pi}{12}=-\sqrt{\dfrac{1-\cos\frac{7\pi}6}{1+\cos\frac{7\pi}6}}=-\sqrt{7+4\sqrt3}=-2-\sqrt3$

###

12.

$\dfrac{\cot x-1}{1-\tan x}=\dfrac{\frac{\cos x}{\sin x}-1}{1-\frac{\sin x}{\cos x}}$

$=\dfrac{\cos x(\cos x-\sin x)}{\sin x(\cos x-\sin x)}$

$=\dfrac{\cos x}{\sin x}$

$=\dfrac{\csc x}{\sec x}$

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13.

$\dfrac{1+\tan x}{\sin x+\cos x}=\dfrac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}$

$=\dfrac{\cos x+\sin x}{\cos x(\sin x+\cos x)}$

$=\dfrac1{\cos x}$

$=\sec x$

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14.

$\sin2x(\cot x+\tan x)=2\sin x\cos x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right)$

$=2\sin x\cos x\dfrac{\cos^2x+\sin^2x}{\sin x\cos x}$

$=2$

###

15.

$\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}$

$=\dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}$

$=\cos2\theta$

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sammy [17]

Radically symmetrical

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