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2 years ago

If the standard volume of a

Mathematics

1 answer:

Alla [95]2 years ago

4 0

Answer:

6.7 %

which agrees with answer A in your list.

Step-by-step explanation:

If the standard volume is 1.5 cubic cm, and the measured volume is 1.6 cubic cm, then the difference between these two is :

1.6 - 1.5 = 0.1 cubic cm. and therefore the percentage error is given by:

0.1/1.5 = 0.066666...which corresponds to a rounding of 6.7%

Recall as well that percent error is always expressed as a positive number.

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Mrac [35]

Uh is there more? If not it’s 2

3 0

3 years ago

Read 2 more answers

Applying Inverse Properties In Exercise,apply the inverse properties of logarithmic and exponential functions to simplify the ex

Mazyrski [523]

Answer:

$x^2$

Step-by-step explanation:

apply the inverse properties of logarithmic and exponential functions to simplify

$ln(e^{x^2})$

all logs has base 'e'

Inverse property of log says that

$ln(e^x)=x$ , the value of ln e=1

we apply this property in our problem. ln has same base 'e' . ln and 'e' gets cancelled

$ln(e^{x^2})$

$x^2$

5 0

3 years ago

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

3 years ago

*80 POINTS!!!! PLEASE HELP ME I HAVE BEEN WAITING FOR MORE THAN AN HOUR!!!!!!!!!! MARKING AS BRAINLIEST!! PLEASE SHOW WORK AND E

ruslelena [56]

Answer:

D. tan(-π/6)

Step-by-step explanation:

The tangent function is periodic with a period of π, so ...

tan(5π/6) = tan(5π/6 ± kπ) . . . for any integer k

For k = -1, we have ...

tan(5π/6) = tan(5π/6 -π) = tan(-π/6)

7 0

3 years ago

In the diagram, point P is a point of tangency. Find the radius r of circle O.

vlabodo [156]

Answer:

r = 12

Step-by-step explanation:

From the figure attached,

QP is a tangent to the circle O at the point P.

Therefore, by the property of tangency,

OP ⊥ QP

By applying Pythagoras theorem In right triangle QPO,

(Hypotenuse)² = (Leg 1)² + (Leg 2)²

(OQ)² = (OP)² + (PQ)²

(25 + r)² = (35)² + r²

625 + r² + 50r = 1225 + r²

50r = 1225 - 625

50r = 600

r = 12

Therefore, r = 12 units is the answer.

4 0

2 years ago