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3 years ago

Which ratio is proportional to 18/28 ? Help me plzzzz i will give brainliest

Mathematics

1 answer:

maks197457 [2]3 years ago

3 0

Answer:

18/21 = 30/35

Step-by-step explanation:

so the problem says 18/21 so if that's the correct ratio to proportion to the answer would be. 30/35. the reason is that you can divide 18 and 21 by 3 to get 6/7 then if you multiply by 5 you get 30/35. Dividing and multiplying both the denominator and numerator keeps the ratio proportional.

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The pattern 96, 92, 88, 84, ________ follows the "subtract 4" rule. Study the pattern to find traits that are not obvious in the

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Step-by-step explanation:

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3 years ago

90 3/7 was a demical

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90.42857143 or
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3 years ago

Megan Hikes 15.12 mi in 6.3 hours. If Megan hike the same number of miles each hour, how many miles did she hike each hour.

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Write the following as a root in radical form 2^-4/5

erastovalidia [21]

Answer:

Given that:

=2^(-4/5)

This means that 2 is the base and -4/5 is the exponent.

To change it in radical form:

firstly the denominator would be written with radical sign
Then the negative sign with 4 will be removed by inverting the base.
Then the fraction will be simplified according to the exponent (power)
All the steps are performed below:

$=\sqrt[5]{2^{-4} } \\=\sqrt[5]{1/2^4}\\ =\sqrt[5]{1/16}$

i hope it will help you!

6 0

3 years ago

A one-parameter family of solutions of the DE P' = P( 1 - P) is given below. P = c1et/1 + c1et Does any solution curve pass thro

topjm [15]

Answer:

a. The curve $P(t) = -\frac{6e^t}{5-6e^t}$ passes through the point (0, 6)

b. No solution of the curve $P(t)$ passes through the point (0, 1)

Step-by-step explanation:

Consider the family of the solution of DE $P' = P(1 - P)$ is $P = \frac{c_1e^t}{1 + c_1e^t}$

a. If any solution passes through the point (0, 6), then there is $c_1$ such that the point (0, 6) satisfies the solution $P = \frac{c_1e^t}{1 + c_1e^t}$

Substitute $t = 0, P = 6$ in $P = \frac{c_1e^t}{1 + c_1e^t}$ and then solve the equation to obtain $c_1$

$P(t) = \frac{c_1e^t}{1 + c_1e^t}\\P(0) = \frac{c_1e^0}{1+c_1e^0}\\ 6 = \frac{c_1}{1 + c_1}\\ c_1 = -\frac{6}{5}$

Therefore, the curve $P(t) = -\frac{6e^t}{5 - 6e^t}$ passes through the point (0, 6)

b. If any solution passes through the point(0, 1), then there is $c_1$ such that the point (0, 1) satisfies the solution $P = \frac{c_1e^t}{1+c_1e^t}$

$P(t) = \frac{c_1e^t}{1 + c_1e^t}\\ P(0) = \frac{c_1e^0}{1 + c_1e^0}\\ 1 = \frac{c_1}{1+c_1} \\1 + c_1 = c_1$

this is not possible

Hence, there is no curve $P(t)$ that exists which passes through the point (0, 1)

4 0

3 years ago