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3 years ago

Which addition expression

Mathematics

1 answer:

mina [271]3 years ago

6 0

Answer:(9+41)+-1-71)

Step-by-step explanation:

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Can someone pls help me with this question

jeyben [28]

Answer:

I'll find X for you; the next step is by you.

Step-by-step explanation:

we have two similar pentagons

I will do it in 2 ways.

=> 48/4x = (x+2)/x

=> 12/x = (x+2)/x

Because of pentagons so x > 0

=> 12 = x + 2 => x = 10

k is unknown number

=> 4x = k.x and 48 = k.(x + 2) (x > 0)

=> k = 4 and 48 = 4.(x + 2)

=> 12 = x + 2 => x = 10

7 0

2 years ago

What is 1 Tablespoon as a fraction?

OLEGan [10]

In a cup, one tablespoon is 1/16. There are 16 tablespoons in a cup. 1 tablespoon also equals 3 teaspoons.

8 0

3 years ago

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Which of the choices shown are the lengths of the sides of a right triangle

Triss [41]

D because , 10+ 24 greater than 26
You need take the first and the second number add together must be greater the last one.
10,24,26
10+24 = 34
34>26

5 0

3 years ago

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Determine which of the lines are parallel and which of the lines are perpendicular.

hichkok12 [17]

This would be easier for you if you put it in a graph

6 0

3 years ago

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>

Learn more about the sum of a series here:

brainly.com/question/190295

8 0

2 years ago

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