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3 years ago

If a scale factor is grater than 1, then your figure gets

Mathematics

2 answers:

blsea [12.9K]3 years ago

6 0

Answer:

multiplied

Step-by-step explanation:

jok3333 [9.3K]3 years ago

6 0

Answer:

If a scale factor is greater than 1, then your figure gets bigger.

Step-by-step explanation:

For scale factors, you are multiplying the scale factor to get the desired outcome. If the scale factor is greater than 1, you are multiplying by a whole number, which would be ≥ the original number. If the scale is 1, then it would be = too the original number. If the scale factor is a fraction, then the figure would be scaled down.

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Explain why 0.04/3.6 has the same answer as 4/360

lisov135 [29]

Because when you divide you get the same answer or something to do with percents

7 0

3 years ago

A bag with 8 marbles has 4 red marbles, 3 blue marbles, and 1 yellow marble. A marble is chosen from the bag at random. What is

Andrei [34K]

Answer:

P = 4 / ( 4 + 3 +1 )

= 4 / 8

= 4 / 8 = 1 / 2

Hope it helps

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8 0

3 years ago

Read 2 more answers

1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a

katovenus [111]

1. Let a and b be coefficients such that

$\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}$

Combining the fractions on the right gives

$\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}$

$\implies 1 = (2a+b)x + 3a$

$\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23$

so that

$\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}$

2. a. The given ODE is separable as

$x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}$

Using the result of part (1), integrating both sides gives

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C$

Given that y = 1 when x = 1, we find

$\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)$

so the particular solution to the ODE is

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)$

We can solve this explicitly for y :

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)$

$\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|$

$\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|$

$\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}$

2. b. When x = 9, we get

$y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}$

8 0

2 years ago

Whats the common ratio of the sequence 7.04 Q14

julsineya [31]

Answer:

$\frac{1}{4}$

Step-by-step explanation:

The given sequence is a geometric sequence.

We can find the common ratio using any two consecutive terms of the sequence.

$r=\frac{\frac{-1}{10} }{\frac{x-2}{5} }=\frac{-1}{10}\times \frac{-5}{2}=\frac{1}{4}$

The common ratio is $\frac{1}{4}$

4 0

3 years ago

Read 2 more answers

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belka [17]

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8 0

2 years ago