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3 years ago

Please help me on this question ASAP. i need it dont by midnight and its 10:40.

Mathematics

2 answers:

faltersainse [42]3 years ago

6 0

X= 6 y=9 and the last one is -40 i believe. the middle ones i'm not sure of

IrinaVladis [17]3 years ago

4 0

Answer:

i) x = 6 and y = 9

ii) x = 10 and y = 5

iii) x = 9 and y = 10

not entirely sure about the last one

Step-by-step explanation:

not sure how to explain but trust

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Find the value of the following expression below 8(15-4)/4^2-5

Sonja [21]

Answer:

The correct answer is 1/2 or 0.5

Step-by-step explanation:

8(15-4)/4^2-5

8(11)/4^2-5

Now i'm going to multiply 8 by 11

88/4^2-5

Now solve 4^2

Which is 16.

Now we have 88/16-5

Divide 88 and 16 which is 5.5.

Now minus 5.5 - 5 which gets us our answer: 0.5

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3 years ago

Bennett has 107 photographs to place in the school yearbook. He will put the same number of photos on each of the 13 pages. if h

HACTEHA [7]

Answer:

Step-by-step explanation: if he has 2 rows of four on each page and 13 pages 8 * 13 is 104 and 107 - 104 is 3

7 0

3 years ago

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Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an

Leya [2.2K]

Parameterize the lateral face $T_1$ of the cylinder by

$\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v$

where $0\le u\le2\pi$ and $0\le v\le3$ , and parameterize the disks $T_2,T_3$ as

$\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)$
$\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)$

where $0\le r\le2$ and $0\le\theta\le2\pi$ .

The integral along the surface of the cylinder (with outward/positive orientation) is then

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$

7 0

3 years ago

What’s the answer ?

xxMikexx [17]

B 5x+20 hope this helps!!

8 0

3 years ago

Write the circumference of a circle with a diameter of 49 inches in terms of pi

AleksAgata [21]

Answer:

153.94

Step-by-step explanation:

c=2r*r

The radius is half of the diameter.

To find the radius you will divide 49 inches by 2.

You should get a total of 24.5

Now that you have your radius you are going to use the circumference formula to solve.

C= 2r*r

C=2(24.5)*(24.5)

type that into the calculator (without the parenthesis) and boom your done!

You answer should be 153.938

then you round to get a total of 153.94

I hope i was able to help!

5 0

3 years ago

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