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astra-53 [7]
2 years ago
10

The equation 6y = 30 - 8x has solutions of (b, 1) and (a, b). What are the values of a and b?

Mathematics
1 answer:
victus00 [196]2 years ago
6 0

The values of a and b are a = \frac{3}{2} and b = 3, respectively.

In this question we need to solve the linear equation for each point, first for (x,y) = (b, 1) and later for (x,y) = (a, b).

If we know that (x,y) = (b, 1), then the value of b is:

6 = 30-8\cdot b

8\cdot b = 24

b = 3

If we know that (x,y) = (a, b) and b = 3, then the value of a is:

6\cdot b = 30 - 8\cdot a

18 = 30-8\cdot a

12 = 8\cdot a

a = \frac{3}{2}

The values of a and b are a = \frac{3}{2} and b = 3, respectively.

To learn more on linear functions, we kindly invite to check this question: brainly.com/question/5101300

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Answer:

(1, 4) and (1,3), because they have the same x-value

Step-by-step explanation:

For a relation to be regarded as a function, there should be no two y-values assigned to an x-value. However, two different x-values can have the same y-values.

In the relation given in the equation, the ordered pairs (1,4) and (1,3), prevent the relation from being a function because, two y-values were assigned to the same x-value. x = 1, is having y = 4, and 3 respectively.

Therefore, the relation is not a function anymore if both ordered pairs are included.

<em>The ordered pairs which make the relation not to be a function are: "(1, 4) and (1,3), because they have the same x-value".</em>

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3 years ago
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Read 2 more answers
Please help 20 points
12345 [234]

Answer:

x =± (√ 317/ 2 )+( 15 /2)

6 0
3 years ago
John Doe estimates that the price elasticity of demand for new SUVs to be 0.75. If the price of SUVs rose by 15%; would the quan
Charra [1.4K]
If the coefficient of demand for the SUV is 0.75 this means that it has a relatively inelastic demand since it is less than. In other words, there is only a little alteration in demand when prices change.
So when the price of SUV rise by 15%, and it has a coefficient of 0.75, we can anticipate only 11.25% decrease in its demand. This is for the reason that the SUVs do not have many substitutes for it.
So to solve:
(x/15%) = 0.75
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4 0
3 years ago
Can someone check whether its correct or no? this is supposed to be the steps in integration by parts​
Gwar [14]

Answer:

\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}

Step-by-step explanation:

\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}

Given integral:

\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x

\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:

\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x

Using <u>integration by parts</u>:

\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)

\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}

Therefore:

\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}

\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:

\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)

\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}

\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}

Therefore:

\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x

\textsf{Subtract }\: \displaystyle \int e^{-2x}\sin(2x)\:\text{d}x \quad \textsf{from both sides and add the constant C}:

\implies \displaystyle -2\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+\text{C}

Divide both sides by 2:

\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{4}e^{-2x}\sin (2x) +\dfrac{1}{4}e^{-2x}\cos(2x)+\text{C}

Rewrite in the same format as the given integral:

\displaystyle \implies - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}

5 0
2 years ago
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