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3 years ago

Can some find the the answer than explain it Thank you

Mathematics

1 answer:

STALIN [3.7K]3 years ago

7 0

Answer:

2 week ago so you dont need so free point

Step-by-step explanation:

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Write an equation (a) in slope-intercept form and (b) in standard form for the line passing through (-3,5) and parallel to x + 4

Dmitry [639]

Answer:

$\boxed{\sf Standard-form :x + 4y -17=0 }\\$

$\boxed{\sf Slope-intercept\ form :y =\dfrac{-1}{4}x +\dfrac{17}{4}}$

Step-by-step explanation:

Here a equation of the line is given to us and we need to find out the equation of line which passes through the given point and parallel to the given line , the given equation is ,

$\longrightarrow x + 4y = 7\\$

Firstly convert it into slope intercept form of the line which is y = mx + x , as ;

$\longrightarrow 4y = -x + 7 \\$

$\longrightarrow y =\dfrac{-x}{4}+\dfrac{7}{4}\\$

On comparing it to y = mx + c , we have ,

$\longrightarrow m =\dfrac{-1}{4}\\$

$\longrightarrow c =\dfrac{7}{4}\\$

Now as we know that the slope of two parallel lines is same . Therefore the slope of the parallel line will be ,

$\longrightarrow m_{||)}=\dfrac{-1}{4}\\$

Now we may use point slope form of the line as ,

$\longrightarrow y - y_1 = m(x-x_1) \\$

On substituting the respective values ,

$\longrightarrow y - 5 =\dfrac{-1}{4}\{ x -(-3)\}\\$

$\longrightarrow y -5=\dfrac{-1}{4}(x+3)\\$

$\longrightarrow 4(y -5 ) =-1(x +3) \\$

$\longrightarrow 4y -20 = - x -3 \\$

$\longrightarrow x + 4y -20+3=0\\$

$\longrightarrow \underset{Standard \ Form }{\underbrace{\underline{\underline{ x + 4y -17=0}}}} \\$

Again the equation can be rewritten as ,

$\longrightarrow y - 5 = \dfrac{-1}{4}(x +3) \\$

$\longrightarrow y = \dfrac{-1}{4}x -\dfrac{3}{4}+5 \\$

$\longrightarrow y = \dfrac{-1}{4}x -\dfrac{20-3}{4} \\$

$\longrightarrow \underset{Slope-Intercept\ form }{\underbrace{\underline{\underline{ y =\dfrac{-1}{4}x +\dfrac{17}{4}}}}}\\$

6 0

2 years ago

Evaluate the iterated integral 2 0 2 x sin(y2) dy dx. SOLUTION If we try to evaluate the integral as it stands, we are faced wit

nignag [31]

Answer:

Step-by-step explanation:

Given that:

$\int^2_0 \int^2_x \ sin (y^2) \ dy dx \\ \\ \text{Using backward equation; we have:} \\ \\ \int^2_0\int^2_0 sin(y^2) \ dy \ dx = \int \int_o \ sin(y^2) \ dA \\ \\ where; \\ \\ D= \Big\{ (x,y) | }0 \le x \le 2, x \le y \le 2 \Big\}$

$\text{Sketching this region; the alternative description of D is:} \\ D= \Big\{ (x,y) | }0 \le y \le 2, 0 \le x \le y \Big\}$

$\text{Now, above equation gives room for double integral in reverse order;}$

$\int^2_0 \int^2_0 \ sin (y^2) dy dx = \int \int _o \ sin (y^2) \ dA \\ \\ = \int^2_o \int^y_o \ sin (y^2) \ dx \ dy \\ \\ = \int^2_o \Big [x sin (y^2) \Big] ^{x=y}_{x=o} \ dy \\ \\= \int^2_0 ( y -0) \ sin (y^2) \ dy \\ \\ = \int^2_0 y \ sin (y^2) \ dy \\ \\ y^2 = U \\ \\ 2y \ dy = du \\ \\ = \dfrac{1}{2} \int ^2 _ 0 \ sin (U) \ du \\ \\ = - \dfrac{1}{2} \Big [cos \ U \Big]^2_o \\ \\ = - \dfrac{1}{2} \Big [cos \ (y^2) \Big]^2_o \\ \\ = - \dfrac{1}{2} cos (4) + \dfrac{1}{2} cos (0) \\ \\$

$= - \dfrac{1}{2} cos (4) + \dfrac{1}{2} (1) \\ \\ = \dfrac{1}{2}\Big [1- cos (4) \Big] \\ \\ = \mathbf{0.82682}$

5 0

3 years ago

Evaluate g (x) = 4- 3x when x = -3, 0, and 5

Nataliya [291]

Answer:

13, 4, -11

Step-by-step explanation:

g(-3) = 4-3(-3) = 4+9 = 13

g(0) = 4-3(0) =4-0 = 4

g(5) = 4-3(5) = 4-15 = -11

7 0

3 years ago

No links please chdudnvd

Alex_Xolod [135]

Its the third option y-9=-1/4(x+2) because the equation for point slope form is y-y1= m(x-x1) so because there is a negative 2 it is a positive 2. Remember a negative minus a negative is a positive. Hope this helped :)!

7 0

2 years ago

Read 2 more answers

Can someone please answer this for me and with solution and "Checking"( to see if your answer is correct) please.

mamaluj [8]

Answer:

√9/√13=5

√10-3r=r

Step-by-step explanation:

this must be the answer

4 0

3 years ago