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2 years ago

How to solve this problem?

Mathematics

1 answer:

arsen [322]2 years ago

7 0

The exact answer is 336

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What is the answer to this equation? 4x+39

RSB [31]

Answer:

U CAN'T SOLVED IT BECAUSE IT HAS A VARIABLE AND A NUMBER U CANNOT COMBINED THEM

8 0

3 years ago

Read 2 more answers

Megan's family had cakes on two days. The first day, they had 2 3/4 cakes left. The second day, they had 1 1/4 cakes left. How m

vlabodo [156]

Answer: 4 cakes left in all.

Step-by-step explanation:

Given: Cakes left on first day= $2\frac{3}{4}=\frac{11}{4}$

Cakes left on second day = $1\frac{1}{4}=\frac{5}{4}$

The number of cakes left in all= $\frac{11}{4}+\frac{5}{4}$

⇒The number of cakes left in all= $\frac{11+5}{4}=\frac{16}{4}$

Therefore, The number of cakes left in all=4

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2 years ago

Write an equation of the line that passes through the given point and is (a) parallel and (b) perpendicular to the given line.

inessss [21]

Answer:

(a). y = 3x - 11 ; (b). y = $-\frac{1}{3}$ x - 1

Step-by-step explanation:

Parallel lines have the same slopes

Slopes of perpendicular lines are opposite reciprocals.

(3, - 2)

The slope of given line is 3

(a). Equation of ║ line is

y + 2 = 3(x - 3)

y = 3x - 11

(b). Slope of perpendicular line is $-\frac{1}{3}$

y + 2 = $-\frac{1}{3}$ (x - 3)

y =  $-\frac{1}{3}$  x - 1

5 0

3 years ago

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Can someone please help me quick I have no Idea what this means and it is due tomorrow

Tems11 [23]

12563 is the correct answer

6 0

3 years ago

Let C = C1 + C2 where C1 is the quarter circle x^2+y^2=4, z=0,from (0,2,0) to (2,0,0), and where C2 is the line segment from (2,

trapecia [35]

Not much can be done without knowing what $\mathbf F(x,y,z)$ is, but at the least we can set up the integral.

First parameterize the pieces of the contour:

$C_1:\mathbf r_1(t_1)=(2\sin t_1,2\cos t_1,0)$
$C_2:\mathbf r_2(t_2)=(1-t_2)(2,0,0)+t_2(3,3,3)=(2+t_2, 3t_2, 3t_2)$

where $0\le t_1\le\dfrac\pi2$ and $0\le t_2\le1$ . You have

$\mathrm d\mathbf r_1=(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1$
$\mathrm d\mathbf r_2=(1,3,3)\,\mathrm dt_2$

and so the work is given by the integral

$\displaystyle\int_C\mathbf F(x,y,z)\cdot\mathrm d\mathbf r$
$=\displaystyle\int_0^{\pi/2}\mathbf F(2\sin t_1,2\cos t_1,0)\cdot(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1$
${}\displaystyle\,\,\,\,\,\,\,\,+\int_0^1\mathbf F(2+t_2,3t_2,3t_2)\cdot(1,3,3)\,\mathrm dt_2$

5 0

3 years ago