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3 years ago

I need to create my own area word problem for extra credit (algebra 1)

1 answer:

3 0

Bob has a garden
the area is 75 square feet
the legnth is 3 times the width
solve
75=lw
l=3w
subsitute
75=3w times w=3w^2
75=3w^2
sdivide by 3
25=w^2
square root
5=width
subsitute
3w=l
3 times 5=l
l=15

width=5
legnth=15

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The black graph is the graph of y = f(x).<br> Choose the equation for the red graph.<br> (Picture)

Vesnalui [34]

What picture? Nothing is attached.

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2 years ago

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FromTheMoon [43]

Answer:

$f^{-1}(x)=\dfrac{1}{3}(\ln x +1)$

Step-by-step explanation:

Given function:

$f(x)=e^{3x-1}$

Replace f(x) with y:

$\implies y=e^{3x-1}$

Take natural logs of both sides:

$\implies \ln y = \ln e^{3x-1}$

Apply the Power Log Law $\ln x^n=n\ln x$ :

$\implies \ln y = (3x-1)\ln e$

As $\ln e=1$ then:

$\implies \ln y = 3x-1$

Rearrange to make x the subject:

$\implies 3x=\ln y +1$

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Swap x for $f^{-1}(x)$ and y for x:

$\implies f^{-1}(x)=\dfrac{1}{3}(\ln x +1)$

8 0

2 years ago

A company that produces fine crystal knows from experience that 17% of its goblets have cosmetic flaws and must be classified as

Alex73 [517]

Answer:

0.3891 = 38.91% probability that only one is a second

Step-by-step explanation:

For each globet, there are only two possible outcoes. Either they have cosmetic flaws, or they do not. The probability of a goblet having a cosmetic flaw is independent of other globets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

17% of its goblets have cosmetic flaws and must be classified as "seconds."

This means that $p = 0.17$

Among seven randomly selected goblets, how likely is it that only one is a second

This is P(X = 1) when n = 7. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{7,1}.(0.17)^{1}.(0.83)^{6} = 0.3891$

0.3891 = 38.91% probability that only one is a second

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3 years ago

Each day Alyssa jogs for 25 Minutes on the treadmill she jogs 135 Meters each day .How many kilometers will Alyssa in 25 minutes

velikii [3]

0.135
1 meter = 0.001 kilometers

4 0

3 years ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is

$\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6$

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2 years ago