Subtract: 8 3/8 - 3 1/4 Answers: A: 5 1/8 B: 5 5/8 C: 5 1/4 D: 5 1/2

boyakko [2]

Use distributive property
5(4x + 3) - 2x
20x + 15 - 2x
Simplify by liked terms
18x + 15
The solution is A

3 0

3 years ago

John had to pay a combined electric and gas bill $150. The electric portion of the bill was $90. What is the ratio of money spen

tatuchka [14]

Given:

John had to pay a combined electric and gas bill $150.

The electric portion of the bill was $90.

To find:

The ratio of money spent on electric to that on gas.

Solution:

Let x be the amount of gas bill.

The electric portion of the bill was $90 and John had to pay a combined electric and gas bill $150. So,

$90+x=150$

$x=150-90$

$x=60$

The gas portion of the bill was $60.

The ratio of money spent on electric to that on gas is:

$\text{Required Ratio}=\dfrac{\text{Electric bill}}{\text{Gas bill}}$

$\text{Required Ratio}=\dfrac{90}{60}$

$\text{Required Ratio}=\dfrac{3}{2}$

$\text{Required Ratio}=3:2$

Therefore, the ratio of money spent on electric to that on gas is 3:2.

7 0

3 years ago

Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li

Sophie [7]

Consider the closed region $V$ bounded simultaneously by the paraboloid and plane, jointly denoted $S$ . By the divergence theorem,

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$

And since we have

$\nabla\cdot\mathbf f(x,y,z)=1$

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

$\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr$
$=\dfrac\pi2$

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by $D$ , we have

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS$

Parameterize $D$ by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k$
$\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k$

which would give a unit normal vector of $\mathbf k$ . However, the divergence theorem requires that the closed surface $S$ be oriented with outward-pointing normal vectors, which means we should instead use $\mathbf s_v\times\mathbf s_u=-u\,\mathbf k$ .

Now,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
$=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du$
$=-2\pi$

So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

6 0

3 years ago

0.8) 498 what is the answer

Lina20 [59]

Step 1: We make the assumption that 498 is 100% since it is our output value.

Step 2: We next represent the value we seek with $x$x.

Step 3: From step 1, it follows that $100\%=498$100%=498.

Step 4: In the same vein, $x\%=4$x%=4.

Step 5: This gives us a pair of simple equations:

$100\%=498(1)$100%=498(1).

$x\%=4(2)$x%=4(2).

Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS

(left hand side) of both equations have the same unit (%); we have

$\frac{100\%}{x\%}=\frac{498}{4}$

100%

x%=

498

4

Step 7: Taking the inverse (or reciprocal) of both sides yields

$\frac{x\%}{100\%}=\frac{4}{498}$

x%

100%=

4

498

$\Rightarrow x=0.8\%$⇒x=0.8%

Therefore, $4$4 is $0.8\%$0.8% of $498$498.

7 0

2 years ago

Find the solution to the following system of linear equations.

bearhunter [10]

Infinitely many solutions

3 0

3 years ago

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