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Evgen [1.6K]
3 years ago
13

Which of the following expressions is equivalent to 1/3 ÷ 3/4

Mathematics
1 answer:
REY [17]3 years ago
7 0
The right answer is D 1/3x4/3
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Naddika [18.5K]
You need to add a picture in order for others to understand what you are referring to.
4 0
3 years ago
a wholesaler allows a discount of 25kobo for every 25 naira goods purchased . find the percentage discount allowed.
ZanzabumX [31]

The percentage discount allowed by the wholesaler is 1%

<h3>Percentage</h3>

  • 1 naira = 100 Kobo

25 naira = 25 × 100

= 2500 Kobo

Percentage discount allowed = 25 / 2500 × 100

= 0.01 × 100

= 1%

Therefore, the percentage discount allowed is 1%

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5 0
1 year ago
Help me on my question please <br><br> Giving Brainlist
astra-53 [7]

Answer:

0.8

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
AB has a vértices at (-5,1) and (-3,7). After a transformation, the image had endpoints at (3,-1) and (5,5). Describe the transf
Scrat [10]
That looks like a translation; let's check.  We have

A(-5,1), B(-3,7), A'(3,-1), B'(5,5)

If it's a translation by T(x,y) we'd have

A' = A + T

B' = B + T

so 

T = A' - A = (3,-1) - (-5,1) = (8,-2)

and also 

T = B' - B = (5, 5) - (-3, 7) = (8,-2)

They're the same so we've verified this transformation is a translation by (8,-2), eight right, two down.

 



5 0
3 years ago
Read 2 more answers
A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021
egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

  • A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

  • \displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

  • \displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}

Therefore, for the last term we have;

  • \displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

\displaystyle \frac{A}{2}  = \mathbf{ 1011 \cdot  \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460}  \right)}

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2} +\frac{1}{2} -  \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022}  \right)

Which gives;

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2022}  \right)

\displaystyle  A = 2 \times 1011 \cdot  \left(1 - \frac{1}{2022}  \right) = \frac{1032231}{511} \approx \mathbf{2020.022}

  • A ≈ <u>2020.022</u>

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8 0
2 years ago
Read 2 more answers
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