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3 years ago

Solve the compound inequality 6b < 42 or 4b + 12 > 8. (1 point)

Mathematics

1 answer:

Nitella [24]3 years ago

8 0

6b<42
b<7
4b+12>8
4b>-4
b>-1
b>-1 or b<7

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Solve the equation -7=y/7

Mars2501 [29]

Answer:

y=-49

Step-by-step explanation:

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2 years ago

martin is ordering dvds online. each dvd is priced at $8.50. the wedsite also charges a flat fee of $5.00 for shipping and handl

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Y=8.50x+5.00
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3 years ago

14. 3/64 + 2/32 + 5/16 = what is the answer

Anit [1.1K]

Answer:

27/64

Step-by-step explanation:

(7×16)+(5×64)64×16

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2 years ago

Read 2 more answers

Graph y=8.5xy=8.5xy, equals, 8, point, 5, x.

Morgarella [4.7K]

Answer:

A, B, E

Step-by-step explanation:

Given equation

y = 8.5x

Answer choices

A. The equation represents a proportional relationship.

TRUE, it is in the form of y = kx

B. The unit rate of change of y with respect to x is 8.5

TRUE, y = mx + b, the slope m = 8.5 is the rate of change

C. The slope of the line is 2/17

False, the slope is 8.5

D. A change of 17 units in x results in a change of 2 units in y.

False, a change of x = 17 results in 17*8.5 = 144.5 units in y

E. A change of 4 units in x results in a change of 34 units in y.

TRUE, because 4*8.5 = 34

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2 years ago

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt$

Complete the square in the quadratic term of the integrand: $t^2-2t+2=(t-1)^2+1=(1-t)^2+1$ , then in the integral we substitute $u=1-t$ :

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt$

$\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du$

Make another substitution of $v=u^2$ :

$\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv$

Integrate by parts, taking

$r=v+1\implies\mathrm dr=\mathrm dv$

$\mathrm ds=e^v\,\mathrm dv\implies s=e^v$

$\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv$

$\displaystyle=-(2e-1)+(e-1)=-e$

So, we have by the fundamental theorem of calculus that

$\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)$

$\implies-e=f(1,1,1)-2$

$\implies f(1,1,1)=2-e$

3 0

3 years ago