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2 years ago

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Mathematics

1 answer:

Lady bird [3.3K]2 years ago

8 0

10 because is the total hours was spent walking

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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

2 years ago

Which is the equation of a line that is parallel to 2x + 5y = 4 and goes through the point (5, –4) ?

barxatty [35]

we know that

if two lines are parallel

then

their slopes are equal

Step 1

Find the slope of the line 2x + 5y = 4

2x + 5y = 4------> 5y=4-2x-----> y=(4/5)-(2/5)x------> slope m=(-2/5)

step 2

with m=-2/5 and the point (5, –4) find the equation of the line

y-y1=m*(x-x1)------> y+4=(-2/5)*(x-5)----> y=(-2/5)x+2-4

y=(-2/5)x-2

therefore

the answer is the option D

4 0

3 years ago

Read 2 more answers

-8z+(4.5)+3.5z+7y-1.5

bogdanovich [222]

Answer:

$\boxed{\sf{7y+3-4.5z}}$

Step-by-step explanation:

In order to combine like terms, you have to isolate x and y from one side of the equation.

$\sf{-8z+\left(4.5\right)+3.5z+7y-1.5}$

First, thing you do is remove parentheses.

$\Longrightarrow: \sf{-8z+4.5+3.5z+7y-1.5}$

Solve.

Then, you combine like terms.

$\Longrightarrow:\sf{-8z+3.5z+7y+4.5-1.5}$

Add/subtract the numbers from left to right.

-8z+3.5z=-4.5z

Rewrite the problem down.

$\Longrightarrow: \sf{-4.5z+7y+4.5-1.5}$

Solve.

4.5-1.5=3

$\Longrightarrow: \boxed{\sf{7y+3-4.5z}}$

Therefore, the correct answer is 7y+3-4.5z.

I hope this helps, let me know if you have any questions.

6 0

1 year ago

Am I correct or wrong on my answer

astraxan [27]

Answer:

wrong

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Find the x and y intercepts for 3x-3y=12

Viktor [21]

$\huge \boxed{\sf C}\\\\\\\displaystyle \sf Substituting\ x=0 \\\\3(0)-3y=12\\\\-3y=12\\\\Dividing\ each\ side\ by\ -3 \\\\\frac{-3y}{-3} =\frac{12}{-3} \\\\y=-4 \\\\\\Substituting\ y=0 \\\\3x-3(0)=12\\\\3x=12\\\\Dividing\ each\ side\ by\ 3 \\\\\frac{3x}{3} =\frac{12}{3} \\\\x=4$

7 0

1 year ago