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3 years ago

PLZ HELPI DON'T UNDERSTAND THIS ONE.

Mathematics

2 answers:

seraphim [82]3 years ago

7 0

Answer:

$50 to produce each and $150 would be her income

Step-by-step explanation:

Nastasia [14]3 years ago

5 0

Answer:

250 dollars

Step-by-step explanation:

5x50

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The tangents to the curve with equation y = x^3-3x at the points A and B with x coordinates −1 and 4 respectively meet at the po

sladkih [1.3K]

Answer:

$\displaystyle C=\left(\frac{26}{9}, 2\right)$

Step-by-step explanation:

We need to find the equation of the tangent lines of Points A and B.

Differentiate the equation:

$\displaystyle \frac{dy}{dx}=3x^2-3$

Point A has an x-coordinate of -1. Hence, the slope of its tangent line is:

$\displaystyle \frac{dy}{dx}\Big|_{x=-1}=3(-1)^2-3=0$

Find the y-coordinate of Point A using the original equation:

$y(-1)=(-1)^3-3(-1)=2$

Hence, Point A is (-1, 2).

Thus, the tangent line at Point A is:

$y-2=0(x-(-1))$

Simplify:

$y=2$

Point B has an x-coordinate of 4. Hence, the slope of its tangent line is:

$\displaystyle \frac{dy}{dx}\Big|_{x=4}=3(4)^2-3=45$

Find the y-coordinate of Point B:

$y(4)=(4)^3-3(4)=52$

Thus, Point B is at (4, 52).

So, the tangent line at Point B is:

$y-52=45(x-4)$

Simplify:

$y=45x-128$

Point C occurs at the intersections of the tangent lines of Points A and B. Set the two equations equal to each other and solve for x:

$2=45x-128$

We acquire:

$\displaystyle x=\frac{26}{9}$

Since one of our equations is y = 2, the y-coordinate is 2.

Hence, Point C is:

$\displaystyle C=\left(\frac{26}{9}, 2\right)$

5 0

3 years ago

Read 2 more answers

1)How many ways can the letters in the word BOOKKEEPER be arranged? (Must show set-up )

sergeinik [125]

Question 1

There are 5 letters (B, O, K, E, R) and there is a total of 10 letters to make up the word.
There are $\frac{10!}{(10-6)!6!}$ ways of arranging the letters, which equal to 210 ways

Question 2

There are seven swimmers in total.
There are $\frac{7!}{(7-1)!1!}$ ways of choosing the first winner, which is 7 ways
There are $\frac{6!}{(6-1)!1!}$ ways of choosing the second winner, which is 6 ways
There are $\frac{5!}{(5-1)!1!}$ ways of choosing the third winner, which is 5 ways
There are 7×6×5=210 ways of choosing first, second, and third winner

Question 3

The probability of eating an orange and a red candy is $\frac{15}{31}$ × $\frac{9}{30}$ , which equals to $\frac{9}{62}$

The probability of eating two green candies is $\frac{7}{31}$ × $\frac{6}{30}$ which equals to $\frac{7}{155}$