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2 years ago

PLEASE HELP I WILL MARK BRAINLIEST!!!!

Mathematics

1 answer:

Neporo4naja [7]2 years ago

8 0

Answer:

Step-by-step explanation:

I hope it helps

carryonlearning

have a great, good, nice, happy day

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How many terms dose the polynomial have 2x^9 + 3x^6 - x - 20?

riadik2000 [5.3K]

Answer:

Step-by-step explanation:

2x^9 = 1

3x^6 = 2

- x - 20 = 3

6 0

3 years ago

If arc AD = 62° and arc BC = 28°, then what is the measure of angle AED?

nata0808 [166]

Well according to the Inscribed Angle Theorem, the angle inscribed in a circle is equal to one-half of its intercepted arc. We can apply that theorem to this problem as well. So the measure of AED = 1/2( arc AD + arc BC). Plug in the numbers and simplify.
AED = 1/2( 62 + 28)
AED = 1/2( 90 )
AED = 45
So the measure of angle AED is B. 45 degrees.
Hope I helped!

6 0

3 years ago

Can you please show work if you can. Thank you

klemol [59]

17-34(304)$-5 this is what I put

7 0

3 years ago

A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)

Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

$36^6$

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

$5\cdot 36^5$

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

$\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}$

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

$P(A)=P(B)=\dfrac{5}{36}$

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

$axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8$

$exxxx0,\ exxxx2,\ exxxx4,\ exxxx6,\ exxxx8$

$ixxxx0,\ ixxxx2,\ ixxxx4,\ ixxxx6,\ ixxxx8$

$oaxxxx0,\ oxxxx2,\ oxxxx4,\ oxxxx6,\ oxxxx8$

$uxxxx0,\ uxxxx2,\ uxxxx4,\ uxxxx6,\ uxxxx8$

Where x can be any of the 36 characters.

So, we have 25 cases with 4 vacant slots, leading to a probability of

$P(A\cap B)=\dfrac{25\cdot 36^4}{36^6}=\dfrac{25}{1296}$

Finally, you can compute the probability of the union using the formula

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Since we already computed all these quantities.

7 0

2 years ago

What is the common difference between the elements of the arithmetic sequence below? –18, –22.5, –27, –31.5, –36

klasskru [66]

Answer 
Common difference = -4.5

Explanation. 
An arithmetic sequence is a sequence in which the next term is found by adding a constant to the current term. This constant is called the common difference.
If a, b, c, d, e, ….is an arithmetic sequence, then the common difference will be,
Common difference = b-a=c-b=d-c=e-d…..
Now the question above common difference is;
Common difference = (-22.5)-(-14)=-4.5

4 2

3 years ago