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3 years ago

11

Consider the parametric curve:

1 answer:

11Alexandr11 [23.1K]3 years ago

7 0

By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}=\dfrac{4\sec t\tan t}{3\sec^2t}=\dfrac43\sin t$

When $t=\pi$ , we have

$\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{t=\pi}=\dfrac43\sin\pi=0$

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jane lives 1 3/8 miles from the airport. She drove from her house to the airport and back 2 times. How far did Jane drive? Move

Ket [755]

Answer:

$5\frac{1}{2}\ mi$

Step-by-step explanation:

we know that

$1\frac{3}{8}\ mi=\frac{1*8+3}{8}=\frac{11}{8}\ mi$

She drove from her house to the airport and back 2 times

so

Multiply the distance from her house to the airport by 4

$(4)*\frac{11}{8}=\frac{11}{2}=5\frac{1}{2}\ mi$

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3 years ago

A line is parallel to y = x - 8 and

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Answer:

y=x-13

Step-by-step explanation:

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3 years ago

The output is the cube of the input.

chubhunter [2.5K]

Yes it is right because you are measuring by either the base to height or length times width

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3 years ago

When the members of a family discussed where their annual reunion should take place, they found that out of all the family mem

Doss [256]

Answer:

The total number of family members is 21.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of these sets.

I am going to say that:

-The set A represents those that would not go to a park.

-The set B represents those who would not go to a beach.

-The set C represents those who would not go to the family cottage.

The value d represents those who would go to all three places.

We have that:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

In which a are those that would only not go to a park, $A \cap B$ are those who would not got to a park or to the beach, $A \cap C$ are those who would not go to a park or to the famili cottage. And $A \cap B \cap C$ are those that would not go to any of these places.

By the same logic, we have:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

This diagram has the following values:

$a,b,c,d,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)$

The total number of family members is the sum of all these values:

$T = a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

We start finding the values from the intersection of the three sets

5 would not go to a park or a beach or to the family cottage.

This means that $A \cap B \cap C = 5$

1 would go to all three places. This means that $d = 1$ .

8 would go to neither a park nor the family cottage

This means that:

$A \cap C + (A \cap B \cap C) = 8$

$A \cap C = 3$

8 would go to neither a beach nor the family cottage

$B \cap C + (A \cap B \cap C) = 8$

$B \cap C = 3$

7 would go to neither a park nor a beach

$A \cap B + (A \cap B \cap C) = 7$

$A \cap B = 2$

15 would not go to the family cottage

$C = 15$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$15 = c + 3 + 3 + 5$

$c = 4$

12 would not go to a beach

$B = 12$

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$12 = b + 3 + 2 + 5$

$b = 2$

11 would not go to a park

$A = 11$

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

$11 = a + 2 + 3 + 5$

$a = 1$

Now, we can find the total number of family members.

$T = a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$T = 1 + 2 + 4 + 1 + 2 + 3 + 3 + 5$

$T = 21$

The total number of family members is 21.

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2 years ago

If you double president Reagan’s age at the time of his first inauguration(69) and subtract his age at the time he died the resu

e-lub [12.9K]

It should be:

69(2)=138
138-48=90

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3 years ago