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4 years ago

2/3x - 1 = 1 Solve for X.

Mathematics

1 answer:

Novosadov [1.4K]4 years ago

6 0

2/3x-1=1
+1 +1
2/3x=2
•3 •3
2x=6
/2 /2
x=3

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3 years ago

What is the arc length of the arc sub tended in a circle with radius 6 and an angle of 7 pi/ 8

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Answer:

Formula for the Arc length is given by:

$\text{Arc length} = 2 \pi r \cdot \frac{\theta}{360^{\circ}}$

As per the statement:

radius of circle(r) = 6 units

Angle ( $\theta$ ) = $\frac{7 \pi}{8}$ radian

Use conversion:

$1 rad = \frac{180}{\pi}$

$\frac{7 \pi}{8}$ = $\frac{180}{\pi} \cdot \frac{7 \pi}{8} = \frac{1260}{8} = 157.5^{\circ}$

then;

substitute these given values we have;

Use value of $\pi = 3.14$

$\text{Arc length} = 2\cdot 3.14 \cdot (6) \cdot \frac{157.5^{\circ}}{360^{\circ}}$

$\text{Arc length} = 2\cdot 3.14 \cdot (6) \cdot 0.4375$

Simplify:

$\text{Arc length}=16.485$

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3 years ago

Which of the following best describes the algebraic expression 5(x+2)-3

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D. Three less than five times the quantity of a number plus two.

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Answer:

Our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

Step-by-step explanation:

We want to find where the two graphs given by the equations:

$\displaystyle (x+1)^2+(y+2)^2 = 16\text{ and } 3x+4y=1$

Intersect.

When they intersect, their x- and y-values are equivalent. So, we can solve one equation for y and substitute it into the other and solve for x.

Since the linear equation is easier to solve, solve it for y:

$\displaystyle y = -\frac{3}{4} x + \frac{1}{4}$

Substitute this into the first equation:

$\displaystyle (x+1)^2 + \left(\left(-\frac{3}{4}x + \frac{1}{4}\right) +2\right)^2 = 16$

Simplify:

$\displaystyle (x+1)^2 + \left(-\frac{3}{4} x + \frac{9}{4}\right)^2 = 16$

Square. We can use the perfect square trinomial pattern:

$\displaystyle \underbrace{(x^2 + 2x+1)}_{(a+b)^2=a^2+2ab+b^2} + \underbrace{\left(\frac{9}{16}x^2-\frac{27}{8}x+\frac{81}{16}\right)}_{(a+b)^2=a^2+2ab+b^2} = 16$

Multiply both sides by 16:

$(16x^2+32x+16)+(9x^2-54x+81) = 256$

Combine like terms:

$25x^2+-22x+97=256$

Isolate the equation:

$\displaystyle 25x^2 - 22x -159=0$

We can use the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = 25, b = -22, and c = -159. Substitute:

$\displaystyle x = \frac{-(-22)\pm\sqrt{(-22)^2-4(25)(-159)}}{2(25)}$

Evaluate:

$\displaystyle \begin{aligned} x &= \frac{22\pm\sqrt{16384}}{50} \\ \\ &= \frac{22\pm 128}{50}\\ \\ &=\frac{11\pm 64}{25}\end{aligned}$

Hence, our two solutions are:

$\displaystyle x_1 = \frac{11+64}{25} = 3\text{ and } x_2 = \frac{11-64}{25} =-\frac{53}{25}$

We have our two x-coordinates.

To find the y-coordinates, we can simply substitute it into the linear equation and evaluate. Thus:

$\displaystyle y_1 = -\frac{3}{4}(3)+\frac{1}{4} = -2$

And:

$\displaystyle y _2 = -\frac{3}{4}\left(-\frac{53}{25}\right) +\frac{1}{4} = \frac{46}{25}$

Thus, our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

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3 years ago