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2 years ago

If two lines are perpendicular, their slopes are negative reciprocals. A. True B.False

Mathematics

2 answers:

Ainat [17]2 years ago

8 0

A. If the slopes of two lines are negative then the reciprocals, of the lines are perpendicular.

igor_vitrenko [27]2 years ago

6 0

THE ANSWER IS TRUE I’M 100%

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Describe the rays of an angle that measures 1/2 turn?

mel-nik [20]

180 degrese or some thing flat

8 0

2 years ago

If you have a 12 side die and a 6 side die what is the probrability it lands on 5

umka2103 [35]

Answer:

For the 12 sided one it would be 1/12

For the 6 sided one it would be 1/6

Combine those together and it would be 3/12 which is equal to 1/4

Step-by-step explanation:

5 0

1 year ago

Simplify . (1/c + 1/h)/(1/(c ^ 2) - 1/(r ^ 2))

Marrrta [24]

Answer:

$\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Step-by-step explanation:

$\frac{\frac{1}{c}+\frac{1}{h}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine $\frac{1}{c} + \frac{1}{h}$

$\frac{\frac{h+c}{ch}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine the bottom, too.

$=\frac{\frac{h+c}{ch}}{\frac{r^2-c^2}{c^2r^2}}$

Apply the fraction rule

$=\frac{\left(h+c\right)c^2r^2}{ch\left(r^2-c^2\right)}$

Cancel

$=\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Therefore, $\frac{\left(\frac{1}{c}+\frac{1}{h}\right)}{\left(\frac{1}{\left(c^2\right)}-\frac{1}{\left(r^2\right)}\right)}:\quad \frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

5 0

2 years ago

The line tangent to the graph of g(x) = x^{3}-4x+1 at the point (2, 1) is given by the formula

Usimov [2.4K]

well, about A and D, I just plugged the values on the slope formula of

$\bf \begin{array}{llll} g(x)=x^3-4x+1\\ L(x) = 8(x-2)+1 \end{array} \qquad \begin{cases} x_1=1.9\\ x_2=2.1 \end{cases}\implies \cfrac{f(b)-f(a)}{b-a}$

for A the values are 8.01 and 8.0, so indeed those "slopes" are close. $\textit{\huge \checkmark}$

for D the values are -2.25 and 8.0, so no dice on that one.

for B, let's check the y-intercept for g(x), by setting x = 0, we end up g(0) = 0³-4(0)+1, which gives us g(0) = 1.

checking L(x) y-intercept, well, L(x) is in slope-intercept form, thus the +1 sticking out on the far right is the y-intercept, so, dice. $\textit{\huge \checkmark}$

for C, well, the slope if L(x) is 8, since it's in slope-intercept form, the derivative of g(x) is g'(x) = 3x² - 4, and thus g'(0) = -4, so no dice.

for E, do they intercept at (2,1)? well, come on now, L(x) is a tangent line to g(x), so that's a must for a tangent. $\textit{\huge \checkmark}$

for F, we know the slope of the line L(x) is 8, is g'(2) = 8? let's check

recall that g'(x) = 3x² - 4, so g'(2) = 3(2)² - 4, meaning g'(2) = 8, so, dice. $\textit{\huge \checkmark}$

6 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