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3 years ago

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What is the value of y in the product of powers below? 8 cubed times 8 Superscript negative 5 Baseline times 8 Superscript negat

ive 2 Baseline = StartFraction 1 Over 8 squared EndFraction

2 answers:

seropon [69]3 years ago

4 0

Answer: $y=0$

Step-by-step explanation:

<h3> The complete exercise is attached.</h3><h3> </h3>

For this exercise it is important to remember:

1. The Product of powers property. This property states that:

$(a^m)(a^n)=a^{(m+n)}$

2. The multiplication of signs:

$(+)(+)=+\\(-)(-)=+\\(-)(+)=-\\(+)(-)=-$

Then, given:

$8^3*8^{-5}*8^y=8^{-2}=\frac{1}{8^2}$

You can identify that $8^{-2}$ is obtained by applying the Product of powers property:

$8^{3+(-5)+y}=8^{-2}=\frac{1}{8^2}$

Based on the explained above, you can write the following equation:

$3+(-5)+y=-2$

Therefore you must solve for the variable "y" in order to find its value. You get that this is:

$3-5+y=-2\\\\-2+y=-2\\\\y=-2+2\\\\y=0$

Brilliant_brown [7]3 years ago

4 0

Answer:

You're trying to find the value of y that would result in 8^-2. When you multiply terms with the same base, you can add the exponents. 3+-5=-2 and -2-(-2) equals 0, so therefore y is equal to 0.

Step-by-step explanation:

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Meg rowed her boat upstream a distance of 12 mi and then rowed back to the starting point. The total time of the trip was 8 hour

nadya68 [22]

Answer:

3mph

Step-by-step explanation:

Boat speed x

current speed 2

against current x-2 mph

with current x+ 2 mph

Distance= 5 miles

Time against + time with = 6 hours

5 /( x + 2 ) + 5 /(x - 2 ) = 6

LCD = (x - 2 ) ( x + 2 )

5 *( x - 2 ) + 5 (x + 2 ) = 6

5 x - 10 + 5 x + 10 = 6 ( x ^2 - 4 )

10 x = 6 x ^2 - 24

6 x ^2 - -10 x - 24

Find the roots of the equation by quadratic formula

a= 6 , b= -10 , c= -24

b^2-4ac= 100 + 576

b^2-4ac=676

x1=(10+26)/12

x1=3

x2=(10-26)/12

x2= -1.33

Ignore negative value

Speed of boat is 3 mph

8 0

3 years ago

The formula for the volume of a sphere is V=4/3πr^3. What is the formula solved for r?

lorasvet [3.4K]

<span>The formula for the volume of a sphere is V=4/3πr^3. Find the radius of a sphere with each volume.
1. 10 in^3

Solve for r

r =~ 1.3365 inches</span>

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3 years ago

Read 2 more answers

Determine whether the integral converges.

Kryger [21]

You have one mistake which occurs when you integrate $\dfrac1{1-p^2}$ . The antiderivative of this is not in terms of $\tan^{-1}p$ . Instead, letting $p=\sin r$ (or $\cos r$ , if you want to bother with more signs) gives $\mathrm dp=\cos r\,\mathrm dr$ , making the indefinite integral equality

$\displaystyle-\frac12\int\frac{\mathrm dp}{1-p^2}=-\frac12\int\frac{\cos r}{1-\sin^2r}\,\mathrm dr=-\frac12\int\sec r\,\mathrm dr=\ln|\sec r+\tan r|+C$

and then compute the definite integral from there.

$-\dfrac12\ln|\sec r+\tan r|\stackrel{r=\sin^{-1}p}=-\dfrac12\ln\left|\dfrac{1+p}{\sqrt{1-p^2}}=\ln\left|\sqrt{\dfrac{1+p}{1-p}}\right|$
$\stackrel{p=u/2}=-\dfrac12\ln\left|\sqrt{\dfrac{1+\frac u2}{1-\frac u2}}\right|=-\dfrac12\ln\left|\sqrt{\dfrac{2+u}{2-u}}\right|$
$\stackrel{u=x+1}=-\dfrac12\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|$
$\implies-\dfrac12\displaystyle\lim_{t\to\infty}\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|\bigg|_{x=2}^{x=t}=-\frac12\left(\ln|-1|-\ln\left|\sqrt{\frac5{-1}}\right|\right)=\dfrac{\ln\sqrt5}2=\dfrac{\ln5}4$

Or, starting from the beginning, you could also have found it slightly more convenient to combine the substitutions in one fell swoop by letting $x+1=2\sec y$ . Then $\mathrm dx=2\sec y\tan y\,\mathrm dy$ , and the integral becomes

$\displaystyle\int_2^\infty\frac{\mathrm dx}{(x+1)^2-4}=\int_{\sec^{-1}(3/2)}^{\pi/2}\frac{2\sec y\tan y}{4\sec^2y-4}\,\mathrm dy$
$\displaystyle=\frac12\int_{\sec^{-1}(3/2)}^{\pi/2}\csc y\,\mathrm dy$
$\displaystyle=-\frac12\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2}}^{y=\pi/2}$
$\displaystyle=-\frac12\lim_{t\to\pi/2^-}\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2)}^{y=t}$
$\displaystyle=-\frac12\left(\lim_{t\to\pi/2^-}\ln|\csc t+\cot t|-\ln\frac5{\sqrt5}\right)$
$=\dfrac{\ln\sqrt5}2-\dfrac{\ln|1|}2$
$=\dfrac{\ln5}4$

Another way to do this is to notice that the integrand's denominator can be factorized.

$x^2+2x-3=(x+3)(x-1)$

So,

$\dfrac1{x^2+2x-3}=\dfrac1{(x+3)(x-1)}=\dfrac14\left(\dfrac1{x-1}-\dfrac1{x+3}\right)$

There are no discontinuities to worry about since you're integrate over $[2,\infty)$ , so you can proceed with integrating straightaway.

$\displaystyle\int_2^\infty\frac{\mathrm dx}{x^2+2x-3}=\frac14\lim_{t\to\infty}\int_2^t\left(\frac1{x-1}-\frac1{x+3}\right)\,\mathrm dx$
$=\displaystyle\frac14\lim_{t\to\infty}(\ln|x-1|-\ln|x+3|)\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\lim_{t\to\infty}\ln\left|\frac{x-1}{x+3}\right|\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\left(\lim_{t\to\infty}\ln\left|\frac{t-1}{t+3}\right|-\ln\frac15\right)$
$=\displaystyle\frac14\left(\ln1-\ln\frac15\right)$
$=-\dfrac14\ln\dfrac15=\dfrac{\ln5}4$

Just goes to show there's often more than one way to skin a cat...

7 0

3 years ago

✨BRAINLIEST IF YOUR RIGHT✨

KengaRu [80]

Answer:

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Step-by-step explanation:

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zaharov [31]

Answer:

d² ≥ 17

Step-by-step explanation:

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