QUESTION IMAGE
Question
- what value of x satisfies the equation -3(4x - 5)=2(1 - 5x)?
f. -\frac{17}{2}
g. -\frac{17}{22}
h. -1
j. \frac{3}{17}
k. \frac{13}{2}
- in right triangle \triangle abc shown below, the given lengths are in millimeters. what is sin a?
a. \frac{4\sqrt{2}}{9}
b. \frac{4\sqrt{2}}{7}
c. \frac{7\sqrt{2}}{8}
d. \frac{7}{9}
e. \frac{9}{7}
- (\frac{27}{64})^{-\frac{2}{3}}=?
f. -\frac{9}{16}
g. -\frac{9}{32}
h. \frac{9}{32}
j. \frac{16}{9}
k. \frac{32}{9}
- loto begins at his back door and walks 8 yards east, 6 yards north, 12 yards east, and 5 yards north to the barn door. about how many yards less would he walk if he could walk directly from the back door to the barn door?
a. 8
b. 19
c. 23
d. 26
e. 31
- for a given set of data, the standard score, z, corresponding to the raw score, x, is given by z = \frac{x - \mu}{\sigma}, where \mu is the mean of the set and \sigma is the standard deviation. if, for a set of scores, \mu = 78 and \sigma = 6, which of the following is the raw score, x, corresponding to z = 2?
f. 90
g. 84
h. 80
j. 76
k. 66
- in the figure below, a, b, c, and d lie on the circle centered at o.
which of the following does not appear in the figure?
a. acute triangle
b. equilateral triangle
c. isosceles triangle
d. right triangle
e. scalene triangle
- what is the slope of a line, in the standard (x,y) coordinate plane, that is parallel to x + 5y = 9?
f. -5
g. -\frac{1}{5}
h. \frac{1}{5}
j. \frac{9}{5}
k. 9
- given y = \frac{x}{x - 1} and x>1, which of the following is a possible value of y?
a. -1.9
b. -0.9
c. 0.0
d. 0.9
e. 1.9
16.
Step1: Expand both sides
Expand \(-3(4x - 5)\) to \(-12x+15\) and \(2(1 - 5x)\) to \(2 - 10x\). So the equation becomes \(-12x + 15=2-10x\).
Step2: Move \(x\) - terms to one side
Add \(12x\) to both sides: \(15=2 - 10x+12x\), which simplifies to \(15=2 + 2x\).
Step3: Solve for \(x\)
Subtract 2 from both sides: \(15 - 2=2x\), so \(13 = 2x\). Then \(x=\frac{13}{2}\).
In a right - triangle, the sine of an angle is defined as the ratio of the length of the opposite side to the length of the hypotenuse. For \(\angle A\) in right - triangle \(ABC\), the opposite side to \(\angle A\) is \(BC = 7\) and the hypotenuse is \(AC = 9\). So \(\sin A=\frac{BC}{AC}=\frac{7}{9}\).
Step1: Rewrite the negative exponent
\((\frac{27}{64})^{-\frac{2}{3}}=(\frac{64}{27})^{\frac{2}{3}}\).
Step2: Use the power - of - a - power rule
\((\frac{64}{27})^{\frac{2}{3}}=((\frac{64}{27})^{\frac{1}{3}})^2\).
Step3: Find the cube - root
\((\frac{64}{27})^{\frac{1}{3}}=\frac{\sqrt[3]{64}}{\sqrt[3]{27}}=\frac{4}{3}\).
Step4: Square the result
\((\frac{4}{3})^2=\frac{16}{9}\).
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K. \(\frac{13}{2}\)