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3 years ago

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Joshua has 7 different varieties of tulips in his garden. If he has 385 tulip plants in total and equal numbers of each variety,

how many of each tulip variety does he have?

1 answer:

Harrizon [31]3 years ago

3 0

Answer:

55

Step-by-step explanation:

385 ÷ 7 = 55

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Which explanation can be used to derive the formula for the circumference of a circle?

Gelneren [198K]

Answer: The answer is the first explanation.

Step-by-step explanation: We are given five different options and we are to select which explanation is correct to derive the formula for a circumference of a circle.

Let 'C' be the circumference and 'd' be the diameter of a circle. Now, we will write the ratio of the circumference to the diameter as

$\dfrac{\textup{C}}{\textup{d}}.$

Also, we know that

$\dfrac{\textup{C}}{\textup{d}}=\pi.$

And diameter of a circle is twice the radius, so

$d=2r.$

Therefore,

$\dfrac{\textup{C}}{2\textup{r}}=\pi\\\\\Rightarrow \textup{C}=2\pi \textup{r}.$

This is the formula for the circumference of a circle. Since this explanation matches exactly with the first option, so the correct option is

(a). Find the relationship between the circumference and the diameter by dividing the length of the circumference and length of the diameter. Use this quotient to set up an equation to showing the ratio of the circumference over the diameter equals to π . Then rearrange the equation to solve for the circumference. Substitute 2 times the radius for the diameter.

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What is the base of a rectangle with a height of 12.6 meters and an area of 28.98 square meters? Enter your answer in the box.

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Mr Torres has seven dollars in $1 bill and fifteen dollars in $5 bills. If he gives his daughter three bills that are not all th

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3 years ago

For the equation given below, evaluate y′ at the point (−2,2)<br> xe^y−4y=2x−4−2e^2

jeka94

Implicit differentiation
chain rule is important here

I'll show the steps partially
$e^y+xe^yy'-4y'=2$
$xe^yy'-4y'=2-e^y$
$y'(xe^y-4)=2-e^y$
$y'=\dfrac{2-e^y}{xe^y-4}$
now evaluate for (-2,2)
x=-2 and y=2
$y'=\dfrac{2-e^2}{-2e^2-4}$
$y'=\dfrac{2-e^2}{-2e^2-4}$
$y'=\dfrac{e^2-2}{2e^2+4}$
that's it, simplest form

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3 years ago