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3 years ago

We have 4 3/4 pounds of apples in each bag. If we have 4 bags, how many pounds of Apple do we have in all? Explain

1 answer:

7 0

So in each bag there is 4.75 pounds and you want to know how much is in 4 bags so you times 4.75 by 4 which is 19 pounds so the answer is d

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What is 1/5 expressed as a percent? A. 10% B. 20% C. 30% D. 50%

Serggg [28]

Answer:

I'm pretty sure it's B.20%

8 0

2 years ago

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Please help 10 Points

QveST [7]

Answer:

Step-by-step explanation:

8 0

2 years ago

When the positive integer x is divided by 11, the quotient is y and the remainder 3. When x is divided by 19, the remainder is a

Lyrx [107]

Answer:

A) 0

Step-by-step explanation:

When x is divided by 11, we have a quotient of y and a remainder of 3

x/11 = y + 3

x = 11y + 3 ........(1)

When x is divided by 19, we have a remainder of 3 also

x/19 = p + 3 (p = quotient)

x = 19p + 3 ..........(2)

Equate (1) and (2)

x = 11y + 3 = 19p + 3

11y + 3 = 19p + 3

11y = 19p + 3 -3

11y = 19p

Divide both sides by 11

11y/11 = 19p/11

y = 19p/11

y and p are integers. 19 is a prime number. P/11 is also an integer

y = 19(integer)

This implies that y is a multiple of 19. When divided by 19, there is no remainder. The remainder is 0

8 0

3 years ago

Find the value of x.

photoshop1234 [79]

Answer:

Step-by-step explanation:

When two chords intersect inside the circle, the product of their segments are equal.

BE * ED = AE * EC

x *3x = 4 *3

3x² = 12 {Divide both sides by 3}

x² = 12/3

x² = 4

x = √4

x = 2

5 0

2 years ago

Read 2 more answers

Test the series for convergence or divergence (using ratio test)

Triss [41]

Answer:

$\lim_{n \to \infty} U_n =0$

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

Step(i):-

Given series is an alternating series

∑ $(-1)^{n} \frac{n^{2} }{n^{3}+3 }$

Let $U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }$

By using Leibnitz's rule

$U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }$

$U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}$

Uₙ-Uₙ₋₁ <0

Step(ii):-

$\lim_{n \to \infty} U_n = \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }$

$= \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }$

= $= \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }$

$=\frac{1}{infinite }$

$\lim_{n \to \infty} U_n =0$

∴ Given series is converges

3 0

2 years ago