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skad [1K]
3 years ago
14

4 1/5 ÷ 2/3 this is a

Mathematics
1 answer:
wel3 years ago
7 0

Answer:

27/4 also written as 6.75 and 6 3/4


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In △JKL, solve for x. <br><br> (A)74.89<br> (B)30.29<br> (C)38.16<br> (D)66.73
Stels [109]

ANSWER

(C)38.16

EXPLANATION

The acute angle given in the right triangle is 27°.

The side length adjacent to the 27° angle is 34 units.

The side length we want to find is x units, which is the hypotenuse of the right triangle.

We use the cosine ratio to obtain:

\cos(27 \degree)  =  \frac{adjacent}{hypotenuse}

\cos(27 \degree)  =  \frac{34}{x}

Solve for x,

x  =  \frac{34}{\cos(27 \degree)}

x = 38.16

to the nearest hundredth.

4 0
4 years ago
Read 2 more answers
Againnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
ipn [44]

Answer:

B

Step-by-step explanation:

i hope this helps

7 0
3 years ago
During a sale, Cindy found boxes of dog biscuits on sale for $0.69 that had previously cost $2.23. What percentage is the discou
N76 [4]

Answer:

The dog biscuits are approximately 69% off.

Step-by-step explanation:

(1 - 0.69/2.23) * 100

(223/223 - 69/223) * 100

154/223 * 100 = 69%

7 0
3 years ago
A store sold a certain brand of jeans for $38 one day, the store sold 6 pair of jeans of that brand. How much did the 6 pairs of
baherus [9]
6 pairs of jeans would cost $228.
4 0
3 years ago
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Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges.
Arisa [49]

Answer: S_n=5(1-\dfrac{1}{n+1}) ; 5

Step-by-step explanation:

Given series : [\dfrac{5}{1\cdot2}]+[\dfrac{5}{2\cdot3}]+[\dfrac{5}{3\cdot4}]+....+[\dfrac{5}{n\cdot(n+1)}]

Sum of series = S_n=\sum^{\infty}_{1}\ [\dfrac{5}{n\cdot(n+1)}]=5[\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}]

Consider \dfrac{1}{n\cdot(n+1)}=\dfrac{n+1-n}{n(n+1)}

=\dfrac{1}{n}-\dfrac{1}{n+1}

⇒ S_n=5\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}=5\sum^{\infty}_{1}[\dfrac{1}{n}-\dfrac{1}{n+1}]

Put values of n= 1,2,3,4,5,.....n

⇒ S_n=5(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......-\dfrac{1}{n}+\dfrac{1}{n}-\dfrac{1}{n+1})

All terms get cancel but First and last terms left behind.

⇒ S_n=5(1-\dfrac{1}{n+1})

Formula for the nth partial sum of the series :

S_n=5(1-\dfrac{1}{n+1})

Also, \lim_{n \to \infty} S_n = 5(1-\dfrac{1}{n+1})

=5(1-\dfrac{1}{\infty})\\\\=5(1-0)=5

4 0
3 years ago
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