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2 years ago

Quadrilateral RSTU is a parallelogram. What must be the value of x?

1 answer:

5 0

Well, that's pretty easy. The main thing you need to know to solve this task is that since the opposite sides of a parallelogram are congruent,sides RS = TU.
Then we can find the value of x :
 $2x+5 = x + 16. 

x = 11$
I am sure that helps!

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I need help with 15,16,17 please and thank you

Morgarella [4.7K]

Answer:

1= 23

2= 104

3= 90

Step-by-step explanation:

6 0

2 years ago

An archeologist discovered two artifacts buried beneath the ground. She found the first artifact at an elevation of -8:2/7 meter

8090 [49]

Answer:

Option D is the correct choice.

Step-by-step explanation:

We are provided angle of elevations of two artifacts buried beneath the ground and we are asked to find distance between these two elevations.

We will find distance between these elevations by taking the difference of two.

$35:\frac{4}{7}--8:\frac{2}{7}$

$35:\frac{4}{7}+8:\frac{2}{7}=43:\frac{6}{7}$

Since we know that distance is always positive so option A is incorrect.

Therefore, option D is the correct choice and distance between these two elevations is $43:\frac{6}{7}$ .

5 0

2 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 2020.022

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

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

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