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ad-work [718]
2 years ago
13

2 An equation is shown. 2/3x - 7=11 What value of x makes the equation true? MARK BRAINLY

Mathematics
1 answer:
shusha [124]2 years ago
3 0

\huge\text{Hey there!}

\mathsf{\dfrac{2}{5}x-7=11}

\large\textsf{ADD 7 to BOTH SIDES}

\mathsf{\dfrac{2}{5}x -7+7=11+7}

\large\textsf{CANCEL out: -7 + 7 because that gives you 0}

\large\textsf{KEEP: 11 + 7 because it helps you solve for your x-value}

\mathsf{11 + 7 = 18}

\large\text{New EQUATION: }\mathsf{\dfrac{2}{5}x = 18}

\large\textsf{MULTIPLY }\mathsf{\dfrac{5}{2}}\large\textsf{ to BOTH SIDES }

\mathsf{\dfrac{2}{5}x \times \dfrac{5}{2}=\dfrac{5}{2}\times18}

\large\textsf{CANCEL out: }\mathsf{\dfrac{5}{2}\times\dfrac{2}{5}}\large\textsf{ because it gives you 1}

\large\textsf{KEEP: }\mathsf{\dfrac{5}{2}\times 18}\large\textsf{ because it solves for x}

\large\textsf{New EQUATION: }\mathsf{x =18\times\dfrac{5}{2}}

\mathsf{18\times\dfrac{5}{2}=x}

\mathsf{18\times\dfrac{5}{2}=\bf 45}

\boxed{\boxed{\large\textsf{Answer: \huge \bf Option J. 45}}}\huge\checkmark

\text{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

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Answer:

Third table

Step-by-step explanation:

For the relationship to be proportional

y/x = constant

First table y/x = 8/4 = 2    then 7/11 doesn't =2  so it is not

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Third table y/x = 3/6=1/2    then 5/10 = 1/2   then 7/14 = 1/2  it is proportional

Last table y/x = 6/3 = 2    then 11/8 doesn't =2  so it is not

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Which of the following has a graph that is a straight line? (4 points) Equation 4: y = 3x3 Equation 3: y = 2x2 + 4 Equation 1: y
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Find the numbers b such that the average value of f(x) = 7 + 10x − 9x2 on the interval [0, b] is equal to 8.
barxatty [35]

Answer:

The numbers b such that the average value of f(x) = 7 +10\cdot x - 9\cdot x^{2} on the interval [0, b] is equal to 8 are b_{1} \approx 1.434 and b_{2} \approx 0.232.

Step-by-step explanation:

The mean value of function within a given interval is given by the following integral:

\bar f = \frac{1}{b-a}\cdot \int\limits^b_a {f(x)} \, dx

If f(x) = 7 +10\cdot x - 9\cdot x^{2}, a = 0, b = b and \bar f = 8, then:

\frac{1}{b}\cdot \int\limits^b_0 {7+10\cdot x -9\cdot x^{2}} \, dx = 8

\frac{7}{b}\int\limits^b_0 \, dx  + \frac{10}{b}  \int\limits^b_0 {x}\, dx - \frac{9}{b}  \int\limits^b_0 {x^{2}}\, dx = 8

\left(\frac{7}{b} \right)\cdot b + \left(\frac{10}{b} \right)\cdot \left(\frac{b^{2}}{2} \right)-\left(\frac{9}{b} \right)\cdot \left(\frac{b^{3}}{3} \right) = 8

7 + 5\cdot b - 3\cdot b^{2} = 8

3\cdot b^{2}-5\cdot b +1 = 0

The roots of this polynomial are determined by the Quadratic Formula:

b_{1} \approx 1.434 and b_{2} \approx 0.232.

The numbers b such that the average value of f(x) = 7 +10\cdot x - 9\cdot x^{2} on the interval [0, b] is equal to 8 are b_{1} \approx 1.434 and b_{2} \approx 0.232.

7 0
3 years ago
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