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3 years ago

Solve the Inverse Function

Mathematics

1 answer:

QveST [7]3 years ago

8 0

Answer:

Step-by-step explanation:

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David is selling floral arrangements. Each arrangement uses 1 vase and 12 roses. Each vase costs David $2.00. Let C be the total

Oxana [17]

The cost of 1 vase is $2.0,
the cost of 12 roses is 12 * r $ per rose = 12r dollars.

so the cost of 1 vase with 12 roses is (2.0+12r) $

Each arrangement uses 1 vase and 12 roses, so the cost C of one arrangement is :

C=(2.0+12r) $

5 0

3 years ago

I need to know what kind of triangle this is

gayaneshka [121]

Answer: The triangle is a right triangle.

Step-by-step explanation: A right triangle is a triangle with a 90 degrees angle, and it usually has a square showing it.

7 0

2 years ago

Please help asap, brainliest,thanks, and 50 points. Thank you soooo much! <3

Anuta_ua [19.1K]

Answers:

1) $x^{8} y^{8}$

2) $y^{3} \sqrt{y}$

3) $5x^{4} \sqrt{6}$

4) $\sqrt{7}$

5) $\frac{\sqrt{z}}{z}$

Step-by-step explanation:

1) $\sqrt{x^{16} y^{36}}$

Rewriting the expression:

$(x^{16} y^{36})^{\frac{1}{2}}$

Multiplying the exponents:

$x^{\frac{16}{2}} y^{\frac{36}{2}}$

Simplifying:

$x^{8} y^{8}$

2) $\sqrt{y^{7}}$

Rewriting the expression:

$\sqrt{y^{6} y}=(y^{6} y)^{\frac{1}{2}}$

Multiplying the exponents:

$y^{\frac{6}{2}} y^{\frac{1}{2}}$

Simplifying:

$y^{3} y^{\frac{1}{2}}=y^{3} \sqrt{y}$

3) $\sqrt{150 x^{8}}$

Rewriting the expression:

$\sqrt{(6)(25) x^{8}}$

Since $\sqrt{25}=5$ :

$5x^{4}\sqrt{6}$

4) $\frac{7}{\sqrt{7}}$

Multiplying numerator and denominator by $\sqrt{7}$ :

$\frac{7}{\sqrt{7}} (\frac{\sqrt{7}}{\sqrt{7}})=\frac{7}{7\sqrt{7}}$

Simplifying:

$\sqrt{7}$

5) $\frac{5z}{\sqrt{25 z^{3}}}$

Rewriting the expression:

$\frac{5z}{5z \sqrt{z}}$

Simplifying:

$\frac{1}{\sqrt{z}}$

Since we do not want the square root in the denominator, we can multiply numerator and denominator by $\sqrt{z}$ :

$\frac{1}{\sqrt{z}}(\frac{\sqrt{z}}{\sqrt{z}})$

Finally:

$\frac{\sqrt{z}}{z}$

3 0

4 years ago

Find the two intersection points

bogdanovich [222]

Answer:

Our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

Step-by-step explanation:

We want to find where the two graphs given by the equations:

$\displaystyle (x+1)^2+(y+2)^2 = 16\text{ and } 3x+4y=1$

Intersect.

When they intersect, their x- and y-values are equivalent. So, we can solve one equation for y and substitute it into the other and solve for x.

Since the linear equation is easier to solve, solve it for y:

$\displaystyle y = -\frac{3}{4} x + \frac{1}{4}$

Substitute this into the first equation:

$\displaystyle (x+1)^2 + \left(\left(-\frac{3}{4}x + \frac{1}{4}\right) +2\right)^2 = 16$

Simplify:

$\displaystyle (x+1)^2 + \left(-\frac{3}{4} x + \frac{9}{4}\right)^2 = 16$

Square. We can use the perfect square trinomial pattern:

$\displaystyle \underbrace{(x^2 + 2x+1)}_{(a+b)^2=a^2+2ab+b^2} + \underbrace{\left(\frac{9}{16}x^2-\frac{27}{8}x+\frac{81}{16}\right)}_{(a+b)^2=a^2+2ab+b^2} = 16$

Multiply both sides by 16:

$(16x^2+32x+16)+(9x^2-54x+81) = 256$

Combine like terms:

$25x^2+-22x+97=256$

Isolate the equation:

$\displaystyle 25x^2 - 22x -159=0$

We can use the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = 25, b = -22, and c = -159. Substitute:

$\displaystyle x = \frac{-(-22)\pm\sqrt{(-22)^2-4(25)(-159)}}{2(25)}$

Evaluate:

$\displaystyle \begin{aligned} x &= \frac{22\pm\sqrt{16384}}{50} \\ \\ &= \frac{22\pm 128}{50}\\ \\ &=\frac{11\pm 64}{25}\end{aligned}$

Hence, our two solutions are:

$\displaystyle x_1 = \frac{11+64}{25} = 3\text{ and } x_2 = \frac{11-64}{25} =-\frac{53}{25}$

We have our two x-coordinates.

To find the y-coordinates, we can simply substitute it into the linear equation and evaluate. Thus:

$\displaystyle y_1 = -\frac{3}{4}(3)+\frac{1}{4} = -2$

And:

$\displaystyle y _2 = -\frac{3}{4}\left(-\frac{53}{25}\right) +\frac{1}{4} = \frac{46}{25}$

Thus, our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

6 0

3 years ago

The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approxi

VikaD [51]

Answer:

$512.90 should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Normally distributed with mean $480 and standard deviation $20.

This means that $\mu = 480, \sigma = 20$

How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05?

This is the 100 - 5 = 95th percentile, which is X when Z has a pvalue of 0.95, so X when Z = 1.645.

$Z = \frac{X - \mu}{\sigma}$

$1.645 = \frac{X - 480}{20}$

$X - 480 = 1.645*20$

$X = 512.9$

$512.90 should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05

5 0

3 years ago