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3 years ago

WILL MARK BRAINLIEST!!! NEED QUICKLY Look at attachment

Mathematics

2 answers:

frosja888 [35]3 years ago

7 0

Answer:

the answer is V=20 .................

Step-by-step explanation:

Nastasia [14]3 years ago

5 0

Answer:

v = 25

Step-by-step explanation:

The crucial information you need to know to solve this is to realize that HI and GH are the same length. However, why they are equal is not immediately obvious.

Both sides of the middle line (HF) are symmetrical, since G and I are the same distance away from the line, and they both lie on a line perpendicular to the middle line.

Note: we know they're the same distance away due to the small red marks in the lines, indicating that they're the same length.

The angles at G and I in the triangles are also the same, as the lines from G and I both meet at H. If they were different angles, they would each hit a different point on the middle line.

Thus, we can conclude that GH and HI are the same length.

Since we know the following:

GH = 4v - 75

HI = v

We can set GH and HI equal to each other and solve the equation.

4v - 75 = v

Subtract v from both sides:

3v - 75 = 0

Add 75 to both sides:

3v = 75

Divide both sides by 3:

v = 25

Answer: v = 25

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3 years ago

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3 years ago

Describe the diffrence between algebraic expressions and numerical expressions

xxTIMURxx [149]

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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

$5\cdot 36^5$

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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:

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

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

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

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

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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

3 years ago

The population of algae in an

mezya [45]

The equation is y = 20(1.40)^x

where x is the number of days and y is the future amount after x days go by

The 1.40 is from the fact that 1+r = 1+0.40 = 1.40, where r is the decimal form of 40%, so r = 0.40

-----------------------------

Plug in x = 7 to get

y = 20(1.40)^x

y = 20(1.40)^7

y = 210.827008

y = 211

We round to the nearest whole number since it doesn't make much sense to have a fractional portion of an algae. After 7 days, there are about 211 algae.

5 0

3 years ago