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3 years ago

Simplify the expression: (1 + 5h)(4)

Mathematics

1 answer:

Anon25 [30]3 years ago

4 0

Answer:

20h + 4

Step-by-step explanation:

It's easy. Just disperse the multiplying by 4 to the others in the first group.

Multiply 1 by 4, and you get 4.

Multiply 5h by 4, and you get 20h.

The answer is 20h + 4.

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Please help will mark Brainly

Troyanec [42]

Answer:

not

Step-by-step explanation:

6 0

3 years ago

Are 25 milligrams lighter or heavier than 25 centigrams

masya89 [10]

Answer:

25 mg. is lighter than 25 cg.

Step-by-step explanation:

Well,

The prefix "milli" means thousandth and the prefix "centi" means hundredth. So a milligram will be one-tenth of a centigram.

10 mg. = 1 cg.

25 mg. = 2.5 cg.

2.5 cg. < 25 cg.

25 mg. is lighter than 25 cg.

8 0

3 years ago

Read 2 more answers

What is the scale factor from AABC to ADEF?

8_murik_8 [283]

Answer: 8

Step-by-step explanation:

We are going from a smaller triangle to a larger triangle, so the scale factor is greater than 1.

Eliminate A and D.

We know scale factor = (image)/(preimage), so the scale factor is 32/4 = 8

7 0

2 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

What is the point-slope form of a line with slope -5 that contains the point a (2,-1)?

gulaghasi [49]

The point-slope form the line which has the value of slope -5 and contains a point as A(2,-1) is (y+1)=-5(x-2).

<h3>What is point slope form?</h3>

The point slope form of a line is the expression of line which has a specified slope and passes through a point.

The point slope form is givne as,

(y-y₁)=m(x-x₁)

Here, m is the slope of the line, x₁ is the x coordinate of the point by which line passes and y₁ is the y coordinate of the same point.

The slope of a line is -5. This line contains the point A (2,-1). Thus, the point slope form is,

(y-y₁)=m(x-x₁)

(y-(-1))=-5(x-2)

(y+1)=-5(x-2)

Thus, the point-slope form the line which has the value of slope -5 and contains a point as A(2,-1) is (y+1)=-5(x-2).

Learn more about the point slope form here;

brainly.com/question/6497976

#SPJ1

8 0

1 year ago