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3 years ago

Use the given information to determine which segments must be parallel. If there are no segments, write none.

Mathematics

1 answer:

densk [106]3 years ago

6 0

Answer:

11. AW║BY

12. DX║YZ

13. AW║YZ

14. AW║DZ

Step-by-step explanation:

11. If alternate interior angles are congruent, the lines the transversal crosses are parallel.

12. The sum ∠5+∠6 is an alternate interior angle with ∠10. The reasoning of problem 11 applies.

13. If same-side interior angles are suppplementary, the lines the transversal crosses are parallel.

14. Lines perpendicular to the same line are parallel.

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Leto [7]

Answer:

p=p+25%p

Step-by-step explanation:

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3 years ago

What is the surface area of a box with the dimensions of length 4cm, width 3cm, and height 5cm? Only type the number - no units.

stich3 [128]

Answer:

heyy

Step-by-step explanation:

5 0

3 years ago

13. For the arithmetic sequence: 6, 12, 18, 24,...

serg [7]

a_15 = 90

a_n = 6n

Step-by-step explanation:

Given sequence is:

6, 12, 18, 24,...

First of all we have to find the common difference. The common difference is the difference between consecutive terms of an arithmetic sequence.

Here,

$d=a_2-a_1 = 12-6 = 6\\d=a_3-a_2= 18-12 = 6$

The general form for arithmetic sequence is:

$a_n=a_1+(n-1)d$

Putting the values for a_1 and d

$a_n=6+(n-1)(6)\\a_n=6+6n-6\\a_n=6n$

$a_{15} = 6(15)\\=90$

<u>b) Write an equation for the nth term.</u>

The equation is: a_n=6n

Keywords: Arithmetic sequence, Common Difference

Learn more about arithmetic sequence at:

#LearnwithBrainly

4 0

3 years ago

Need help on question 5 please.

Irina18 [472]

Answer:

the answer is a

Step-by-step explanation:

6 0

3 years ago

Please someone what is the total surface area of this pryamid

Dimas [21]

Answer:

$\large\boxed{C.\ (25+25\sqrt3)\ in^2}$

Step-by-step explanation:

We have the square and four equilateral triangles.

The formula of an area of a squre:

$A_S=a^2$

a - length of side

The formula of an area of an equilateral triangle:

$A_T=\dfrac{a^2\sqrt3}{4}$

a - length of side

Clculate the areas:

SQURE:

$A_S=x^2\ in^2$

TRIANGLE:

$A_T=\dfrac{x^2\sqrt3}{4}\ in^2$

The SURFACE AREA of a square pyramid:

$S.A.=A_S+4A_T\\\\S.A.=x^2+4\cdot\dfrac{x^2\sqrt3}{4}=x^2+x^2\sqrt3$

Put x = 5:

$S.A.=5^2+5^2\sqrt3=(25+25\sqrt3)\ in^2$

7 0

3 years ago