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3 years ago

What is the solution to this equation?

Mathematics

1 answer:

vampirchik [111]3 years ago

5 0

Steps to solve:

x - 5 = 2

~Add 5 to both sides

x = 7

Best of Luck!

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A hiker averages 1.4 mph downhill on a mountain trail and 0.8 mph uphill. Find the total time spent hiking in terms of the dista

nadya68 [22]

Answer:

V=d/t.

V=(1.4)(0.8)÷8m/s

=0.14

Step-by-step explanation:

bali yung divide ay guhit sa baba

3 0

3 years ago

How to make this y=21x-7 into standard form

Galina-37 [17]

The standard form is 21x-y-7=0

6 0

3 years ago

Avery has 125 animal stickers. She gave 5 animal stickers to each of her 10 friends. HoW many animal stickers she have left? Wha

Angelina_Jolie [31]

Answer:

Step-by-step explanation:

125-(5*10)

125-(50)

3 0

3 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

An arc of length 15ft subtends a central angle 0 in a circle of radios 9ft. Find measure of 0 in degrees

Kisachek [45]

Answer:

95.49°

Step-by-step explanation:

The arc length formula is s = rФ, where r is the radius and Ф is the central angle in radians. Here, r = 9 ft and s = 15 ft. Thus, the central angle Ф is

Ф = (15 ft) / (9 ft) = 5/3 radians, or

5 rad 180°

------- * ---------- = 95.49°

3 π rad

6 0

4 years ago