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4 years ago

The distance between 15 and 6

2 answers:

5 0

9 because if you do 15, any number and 6, any number same as the first number ,the distance would be 9 because they are not doing distance between negative numbers and positive, they are doing positive and postitive. :)

Keith_Richards [23]4 years ago

3 0

Its 9 if u are asking 15 minus 6

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Help me with school (math)!!!

Oliga [24]

Answer: y=-4x+7

Step-by-step explanation: Move 8x to the right side to isolate y. 2y=14-8x. Divide the whole equation by 2 to get y by itself. y=7-4x

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2 years ago

Help me find the equation and tell me if 22 or 12 is wrong

shtirl [24]

Answer:

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5 0

3 years ago

What is the quotient of 1 1/3 and 5/6?

Musya8 [376]

7 0

3 years ago

Read 2 more answers

A triangle has a base of 18.1 kilometres and a height of 17.4 kilometres. What is the area?

vampirchik [111]

Answer:

157.47km²

Step-by-step explanation:

Given the following

base of 18.1 kilometres

height of 17.4 kilometres

Area of the triangle = 1/2 * base * height

Area of the triangle = 1/2 * 18.1 * 17.4

Area of the triangle = 314.94/2

Area of the triangle = 157.47km²

Hence the area of the triangle is 157.47km²

7 0

3 years ago

Evaluate the surface integral. s (x + y + z) ds, s is the parallelogram with parametric equations x = u + v, y = u − v, z = 1 +

Stolb23 [73]

$S$ is given to be parameterized by

$\mathbf r(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle=\langle u+v,u-v,1+2u+v\rangle$

with $0\le u\le3$ and $0\le v\le2$ . We have

$\mathbf r_u=\langle1,1,2\rangle$
$\mathbf r_v=\langle1,-1,1\rangle$
$\mathbf r_u\times\mathbf r_v=\langle3,1,-2\rangle$
$\left\|\mathbf r_u\times\mathbf r_v\right\|=\sqrt{14}$

The surface integral is then

$\displaystyle\iint_S(x+y+z)\,\mathrm dS=\iint_S(x(u,v)+y(u,v)+z(u,v))\left\|\mathbf r_u\times\mathbf r_v\right\|\,\mathrm du\,\mathrm dv$
$=\displaystyle\sqrt{14}\int_{u=0}^{u=3}\int_{v=0}^{v=2}((u+v)+(u-v)+(1+2u+v))\,\mathrm dv\,\mathrm du$
$=\displaystyle\sqrt{14}\int_{u=0}^{u=3}\int_{v=0}^{v=2}(4u+v+1)\,\mathrm dv\,\mathrm du$
$=\displaystyle\sqrt{14}\left(8\int_{u=0}^{u=3}u\,\mathrm du+3\int_{v=0}^{v=2}v\,\mathrm dv+6\right)$
$=\displaystyle\sqrt{14}\left(8\int_{u=0}^{u=3}u\,\mathrm du+3\int_{v=0}^{v=2}v\,\mathrm dv+6\right)$
$=48\sqrt{14}$

4 0

4 years ago