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3 years ago

LOOK AT PICTURE!! PLEASSEEEE HELP! (the one about the quadrilateral)

Mathematics

2 answers:

liraira [26]3 years ago

6 0

The answer too this question is D

pashok25 [27]3 years ago

5 0

Answer:

Step-by-step explanation:

They say it's a Parallelogram, meaning that you would make a square to make it fit in.

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Someeee one?????????????????

Ad libitum [116K]

Answer:

Option B) $a_{n} = 2\cdot 4^{n-1}$

Step-by-step explanation:

The given geometric sequence is

2, 8, 32, 128,....

The general form of a geometric sequence is given by

$a_{n} = a_{1}\cdot r^{n-1}$

Where n is the nth term that we want to find out.

a₁ is the first term in the geometric sequence that is 2

r is the common ratio and can found by simply dividing any two consecutive numbers in the sequence,

$r=\frac{8}{2} = 4$

You can try other consecutive numbers too, you will get the same common ratio

$r=\frac{32}{8} = 4$

$r=\frac{128}{32} = 4$

So the common ratio is 4 in this case.

Substitute the value of a₁ and r into the above general equation

$a_{n} = 2\cdot 4^{n-1}$

This is the general form of the given geometric sequence.

Therefore, the correct option is B

Note: Don't multiply the first term and common ratio otherwise you wont get correct results.

Verification:

$a_{n} = 2\cdot 4^{n-1}$

Lets find out the 2nd term

Substitute n = 2

$a_{2} = 2\cdot 4^{2-1} = 2\cdot 4^{1} = 2\cdot 4 = 8$

Lets find out the 3rd term

Substitute n = 3

$a_{3} = 2\cdot 4^{3-1} = 2\cdot 4^{2} = 2\cdot 16 = 32$

Lets find out the 4th term

Substitute n = 4

$a_{4} = 2\cdot 4^{4-1} = 2\cdot 4^{3} = 2\cdot 64 = 128$

Lets find out the 5th term

Substitute n = 5

$a_{5} = 2\cdot 4^{5-1} = 2\cdot 4^{4} = 2\cdot 256 = 512$

Hence, we are getting correct results!

6 0

3 years ago

I am having trouble with this problem. I need to find the lengths of this right triangle. How would you do this?

podryga [215]

You labeled the triangle wrong sides 'a' and 'b' are supposed to be the sides that make the right angle. the other side is called the hypotenuse which is the longest side which you should have labeled 'c'

so Pythagorean theorem says
a^2+b^2=c^2
so
(2x+1)^2+(11x+5)^2=(12x+1)^2
distribute
(4x^2+4x+1)+(121x^2+110x+25)=(144x^2+24x+1)
add like terms
125x^2+114x+26=144x^2+24x+1
subtract 125x^2 from both sides
114x+26=19x^2+24x+1
subtract 114x from both sides
26=19x^2-90x+1
subtract 26 from both sides
0=19x^2-90-25
factor
(x-5)(19x+5)=0
therefor x-5=0 and/or 19x+5=0
so
x-5=0 add 5 to both sides
x=5
19x+5=0
subtract 5 from both sides
19x=-5
divide both sides by 19
x=-5/19
since side legnths can't be negative, we can cross this solution out

so x=5
subtitute
1+2x
1+2(5)
1+10=11
side a=11

11x+5
11(5)+5
55+5=60
side b=60

12x+1
12(5)+1
60+1=60
side c=61
add them all up
side a+b+c=11+60+61=132=total legnth

5 0

3 years ago

A teacher gave her class two exams; 60% of the class passed the second exam, but only 48% of the class passed both exams. What p

Dahasolnce [82]

This is an example of conditional probability because we are trying to find the probability of an event occurring GIVEN the occurrence of some other event. There is a formula for this (see image attached).
If we follow this formula, the numerator would be the probability of (A AND B) which in this case is "48% of the class passed BOTH exams." The denominator in the formula would be that "60% of the class passed ONLY THE SECOND exam."
Therefore, P(A and B) = 0.48, which is 48% expressed as a decimal and P(B)= 0.60, which is 60% expressed as a decimal. Then, you can figure out the answer by dividing.