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3 years ago

find the height and of a parallelogram if its base is 35 centimeters and its area is 700 square centimeters

1 answer:

5 0

Hello!

We know what the base and area of the parallelogram are, but the height is still unknown. We also know that the formula for the area is:

A = bh

Meaning we multiply the base by the height to find the area. The opposite of multiplication is division, so we must divide the area by the base to find the height. (We work in the opposite way).

B = a / h

B = 700 cm² / 35 cm

B = 20

We can double check by multiplying 20 by 35, and the product should result in 700.

20 * 35 = 700

Therefore, the height of the parallelogram is 20 centimetres.

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Given the relationship x2 + 3y2 =12, with y > 0 and dx, dt = 2 units/min., find the value of dy, dt at the instant y = 1 unit

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6y\frac{dy}{dt}}
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$\bf \textit{now, when y = 1, what is \underline{x}?}\qquad x^2+3y^2=12\implies x^2+3(1)^2=12
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\begin{cases}
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3 years ago

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earnstyle [38]

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3 years ago

NEED HELP ASAP!! can someone please explain this to me!

34kurt

Answer:

BG = 3; GE = 6

Step-by-step explanation:

The centroid of a triangle divides the median into two parts in the ratio 1 : 2. That is, the short segment is 1/3 the length of the median, and the long segment is 2/3 the length of the median.

BG = 1/3·BE = 9/3 = 3

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3 years ago

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Step-by-step explanation:

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3 years ago

Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the co

jeka57 [31]

Answer:

<h2>A. The series CONVERGES</h2>

Step-by-step explanation:

If $\sum a_n$ is a series, for the series to converge/diverge according to ratio test, the following conditions must be met.

$\lim_{n \to \infty} |\frac{a_n_+_1}{a_n}| = \rho$

If $\rho$ < 1, the series converges absolutely

If $\rho > 1$ , the series diverges

If $\rho = 1$ , the test fails.

Given the series $\sum\left\ {\infty} \atop {1} \right \frac{n^2}{5^n}$

To test for convergence or divergence using ratio test, we will use the condition above.

$a_n = \frac{n^2}{5^n} \\a_n_+_1 = \frac{(n+1)^2}{5^{n+1}}$

$\frac{a_n_+_1}{a_n} = \frac{{\frac{(n+1)^2}{5^{n+1}}}}{\frac{n^2}{5^n} }\\\\ \frac{a_n_+_1}{a_n} = {{\frac{(n+1)^2}{5^{n+1}} * \frac{5^n}{n^2}\$

$\frac{a_n_+_1}{a_n} = {{\frac{(n^2+2n+1)}{5^n*5^1}} * \frac{5^n}{n^2}\\$

aₙ₊₁/aₙ =

$\lim_{n \to \infty} |\frac{ n^2+2n+1}{5n^2}| \\\\Dividing\ through\ by \ n^2\\\\\lim_{n \to \infty} |\frac{ n^2/n^2+2n/n^2+1/n^2}{5n^2/n^2}|\\\\\lim_{n \to \infty} |\frac{1+2/n+1/n^2}{5}|\\\\$

note that any constant dividing infinity is equal to zero

$|\frac{1+2/\infty+1/\infty^2}{5}|\\\\$

$\frac{1+0+0}{5}\\ = 1/5$

$\rho = 1/5$

Since The limit of the sequence given is less than 1, hence the series converges.

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3 years ago