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2 years ago

6(x-2)=5x+3x+8 solve for x.

Mathematics

2 answers:

Aneli [31]2 years ago

6 0

Answer: x=-10

Step-by-step explanation:

6(x-2)=5x+3x+8

6(x−2)=8x+8

6x−12=8x+8

−12=8x+8−6x

−12=2x+8

−12−8=2x

−20=2x

-20/2=x

-10=x

x=-10

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY!

- Cutiepatutie ☺❀❤

Mnenie [13.5K]2 years ago

4 0

Answer:

x = -10

Step-by-step explanation:

6 • (x - 2) - (8x + 8) = 0

Pulling out like terms :

3.1 Pull out like factors :

-2x - 20 = -2 • (x + 10)

Equation at the end of step 3 :

-2 • (x + 10) = 0

Equations which are never true :

4.1 Solve : -2 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation :

4.2 Solve : x+10 = 0

Subtract 10 from both sides of the equation :

x = -10

One solution was found :

x = -10

Processing ends successfully

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If m< 1 = 10x + 4 and m <2 = 5x-4, find m< 2

BartSMP [9]

Answer:

m<2 = 56°

Step-by-step explanation:

m<1 and m<2 are supplementary angles and sum = 180

m<1 + m< 2 = 180

If m< 1 = 10x + 4 and m <2 = 5x-4, find m< 2

10x + 4 + 5x - 4 = 180

15x = 180

x = 12

m<2 = 5(12) -4 = 56

Answer

m<2 = 56°

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2 years ago

Use the given information to find the exact value of the trigonometric function

eimsori [14]

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Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

Also,

$\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}$

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Sedbober [7]

A = $ 861.69

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A = P(1 + rt)
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