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Mariana [72]
2 years ago
12

PLEASE HELP, EASY QUESTION

Mathematics
2 answers:
alukav5142 [94]2 years ago
5 0
I’m pretty sure it’s a/c=b/d. Bc don’t a=c and b=d?
GaryK [48]2 years ago
4 0

Answer:

All of the following proportions are equivalent except a/d=c/b.

a/b=c/d

b/a=d/c => a/b = c/d

a/d=c/b not equal to a/b = c/d

a/c=b/d => a/b = c/d

Step-by-step explanation:

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Keevyn has a bag of Jolly Ranchers. Keevyn gave 3/8 of the bag to Zavion and 1/4 of the bag to Qeisha. Then keeyvn gave 1/2 of t
sergejj [24]

Answer:

1.5/8 of the bag

Step-by-step explanation:

1/4 * 2 = 2/8

----------------------

3/8 + 2/8 = 5/8

1 - 5/8 = 3/8

3/8 divided by 1/2 = <u>1.5/8</u>

3 0
3 years ago
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A plane left Atlanta at 4:50 and arrived at Boston in 3 hours and 15 minutes. When did it arrive in Boston?
sashaice [31]

Answer: 8:05 pm

Step-by-step explanation: 4 hours + 3 hours = 7 hours

50 min + 15 min = 1 hour, 5 min

add those together

watch out for time zones when crossing

4 0
3 years ago
How can i score best mark uee in 2019?
jolli1 [7]

Answer:

You need to study to do good on those exams bud.

8 0
3 years ago
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Select the equation that is the inverse of the function: f(x)= x-3/5
andrezito [222]
-1(x)= 5x+15inverse of the function
5 0
3 years ago
Find a power series representation for the function. (Give your power series representation centered at x = 0.) f(x) = x/6x^2 +
timama [110]

Looks like your function is

f(x)=\dfrac x{6x^2+1}

Rewrite it as

f(x)=\dfrac x{1-(-6x^2)}

Recall that for |x|, we have

\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n

If we replace x with -6x^2, we get

f(x)=\displaystyle x\sum_{n=0}^\infty\frac(-6x^2)^n=\sum_{n=0}^\infty (-6)^n x^{2n+1}

By the ratio test, the series converges if

\displaystyle\lim_{n\to\infty}\left|\frac{(-6)^{n+1} x^{2(n+1)+1}}{(-6)^n x^{2n+1}}\right|=6|x^2|\lim_{n\to\infty}1=6|x|^2

Solving for x gives the interval of convergence,

|x|^2

We can confirm that the interval is open by checking for convergence at the endpoints; we'd find that the resulting series diverge.

3 0
3 years ago
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