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3 years ago

Slope -3/8y+-3/4x=12

Mathematics

1 answer:

grandymaker [24]3 years ago

6 0

Answer:

10 − 2 7 - 3

Step-by-step explanation:

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The one wirh my mouse over js 55

musickatia [10]

Answer:

55 is the answer

Step-by-step explanation:

180-76-49= 55 !!!!!

3 0

3 years ago

Read 2 more answers

How can 6(5 + 9) be expressed using the distributive property?

Vilka [71]

D. (6 + 5) x (5 + 9)

6 0

3 years ago

Find the equation of the line between the points (10, – 5) and ( - 2,0).

Natali5045456 [20]

Answer <u>(assuming it can be in slope-intercept form)</u>:

$y = -\frac{5}{12} x-\frac{5}{6}$

Step-by-step explanation:

1) First, find the slope of the line between the two points by using the slope formula, $m = \frac{y_2-y_1}{x_2-x_1}$ . Substitute the x and y values of the given points into the formula and solve:

$m = \frac{(0)-(-5)}{(-2)-(10)} \\m = \frac{0+5}{-2-10} \\m=\frac{5}{-12}$

Thus, the slope of the line is $-\frac{5}{12}$ .

2) Next, use the point-slope formula $y-y_1 = m (x-x_1)$ to write the equation of the line in point-slope form. Substitute values for $m$ , $x_1$ , and $y_1$ in the formula.

Since $m$ represents the slope, substitute $-\frac{5}{12}$ in its place. Since $x_1$ and $y_1$ represent the x and y values of one point the line intersects, choose any of the given points (it doesn't matter which one, it will equal the same thing) and substitute its x and y values into the formula as well. (I chose (-2,0), as seen below.) Then, isolate y and expand the right side in the resulting equation to find the equation of the line in slope-intercept form:

$y-(0)=-\frac{5}{12} (x-(-2))\\y-0 = -\frac{5}{12} (x+2)\\y = -\frac{5}{12} x-\frac{5}{6}$

3 0

3 years ago

What is the value of 6(2b-4) when b =5

Sindrei [870]

Answer: The value of $6(2b-4)$ is 36.

Step-by-step explanation:

Given expression: $6(2b-4)$

To find the value of $6(2b-4)$ at b= 5, we need to substitute the b=5 in the expression, we get

$6(2(5)-4)\\=6(10-4).......[\text{solve parentheses}]\\=6(6)\\=6\times6=36$

$\\\Rightarrow6(2(5)-4)=36$

Therefore, the value of $6(2b-4)$ is 36, when b=5.

5 0

4 years ago

Read 2 more answers

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