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3 years ago

How many numbers between 1 and 100 are multiples of both 2 and 7

1 answer:

6 0

7. The formula is : $98-14=84$ and then $\frac{84}{14}$ and then $6+1=7$ .

The reason behind it is 14 is the least common multiple of 7 and 2 and 98 is the largest 2 digit multiple of 14. Subtract 98 and 14 and then divide by 14 to get 6 and then add 1. You will get 7 which is our answer!

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A burro is standing near a cactus. The burro is 58 in. tall. His shadow is 4 ft long. The shadow of the cactus is 8 ft long. Wha

Sever21 [200]

To find the height of the cactus, make a ratio.

58 inches divided by 4 feet equals x inches divided by 8 feet.

58/4 = x/8

14.5 = x/8

116 = x

The height of the cactus is 116 inches.

8 0

3 years ago

What is the value of n? enter your answer in the box.

ankoles [38]

Answer:

n = 8

Step-by-step explanation:

The product of the parts of one chord is equal to the product of the parts of the other chord, that is

3n = 4 × 6 = 24 ( divide both sides by 3 )

n = 8

4 0

2 years ago

Evaluate the expression when a = -3 and b=5. a-4b x I ?

Juliette [100K]

-3-4(5)
-4(5)=-20
-3-20
The answers is -23

8 0

3 years ago

Wyatt has 32 marbles in his collection. Twenty of his marbles are red, four are blue, and the

dmitriy555 [2]

Answer:

6 and 2/5

Step-by-step explanation:

20% of 32 is 6.4

4 0

3 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

Tomtit [17]

Apparently my answer was unclear the first time?

The flux of F across S is given by the surface integral,

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S$

Parameterize S by the vector-valued function r(u, v) defined by

$\mathbf r(u,v)=7\cos u\sin v\,\mathbf i+7\sin u\sin v\,\mathbf j+7\cos v\,\mathbf k$

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2. Then the surface element is

dS = n • dS

where n is the normal vector to the surface. Take it to be

$\mathbf n=\dfrac{\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}}{\left\|\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}\right\|}$

The surface element reduces to

$\mathrm d\mathbf S=\mathbf n\,\mathrm dS=\mathbf n\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\implies\mathbf n\,\mathrm dS=-49(\cos u\sin^2v\,\mathbf i+\sin u\sin^2v\,\mathbf j+\cos v\sin v\,\mathbf k)\,\mathrm du\,\mathrm dv$

so that it points toward the origin at any point on S.

Then the integral with respect to u and v is

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S=\int_0^{\pi/2}\int_0^{\pi/2}\mathbf F(x(u,v),y(u,v),z(u,v))\cdot\mathbf n\,\mathrm dS$

$=\displaystyle-49\int_0^{\pi/2}\int_0^{\pi/2}(7\cos u\sin v\,\mathbf i-7\cos v\,\mathbf j+7\sin u\sin v\,\mathbf )\cdot\mathbf n\,\mathrm dS$

$=-343\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}\cos^2u\sin^3v\,\mathrm du\,\mathrm dv=\boxed{-\frac{343\pi}6}$

4 0

3 years ago