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valentina_108 [34]
3 years ago
5

Determine the points for the graph of the relation y = -5x -3 using the x-values given in the table

Mathematics
1 answer:
Marina CMI [18]3 years ago
5 0

Answer:

A. Yes. Each x value has only one y value. No x value has 2 different y values.

B. For the table, y is 8 when x is 11.

Plug in the value to determine the y value for the equation.

f(x)= 5(11)-21

f(x)= 55-21

f(x)= 34

Since 34>8, the value of the equation f(x) is greater when x=11.

C. 99=5x-21

120=5x

x=24

Final answer: 24

Step-by-step explanation:

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Monica [59]

Answer:

i think its 10cm

Step-by-step explanation: brainlyest plz :)

90+80=170

180-170=10

7 0
3 years ago
Some people think it is unlucky if the 13th day of month falls on a Friday. show that in that there year (non-leap or leap) ther
Vlad1618 [11]
<span>There are several ways to do this problem. One of them is to realize that there's only 14 possible calendars for any year (a year may start on any of 7 days, and a year may be either a leap year, or a non-leap year. So 7*2 = 14 possible calendars for any year). And since there's only 14 different possibilities, it's quite easy to perform an exhaustive search to prove that any year has between 1 and 3 Friday the 13ths. Let's first deal with non-leap years. Initially, I'll determine what day of the week the 13th falls for each month for a year that starts on Sunday. Jan - Friday Feb - Monday Mar - Monday Apr - Thursday May - Saturday Jun - Tuesday Jul - Thursday Aug - Sunday Sep - Wednesday Oct - Friday Nov - Monday Dec - Wednesday Now let's count how many times for each weekday, the 13th falls there. Sunday - 1 Monday - 3 Tuesday - 1 Wednesday - 2 Thursday - 2 Friday - 2 Saturday - 1 The key thing to notice is that there is that the number of times the 13th falls upon a weekday is always in the range of 1 to 3 days. And if the non-leap year were to start on any other day of the week, the numbers would simply rotate to the next days. The above list is generated for a year where January 1st falls on a Sunday. If instead it were to fall on a Monday, then the value above for Sunday would be the value for Monday. The value above for Monday would be the value for Tuesday, etc. So we've handled all possible non-leap years. Let's do that again for a leap year starting on a Sunday. We get: Jan - Friday Feb - Monday Mar - Tuesday Apr - Friday May - Sunday Jun - Wednesday Jul - Friday Aug - Monday Sep - Thursday Oct - Saturday Nov - Tuesday Dec - Thursday And the weekday totals are: Sunday - 1 Monday - 2 Tuesday - 2 Wednesday - 1 Thursday - 2 Friday - 3 Saturday - 1 And once again, for every weekday, the total is between 1 and 3. And the same argument applies for every leap year. And since we've covered both leap and non-leap years. Then we've demonstrated that for every possible year, Friday the 13th will happen at least once, and no more than 3 times.</span>
5 0
2 years ago
Help pleaseeeeeeeeee
sergij07 [2.7K]

9514 1404 393

Answer:

  47 -6√10

Step-by-step explanation:

As you know, the area of a square is the square of the side length. It can be helpful here to make use of the form for the square of a binomial.

  (a -b)² = a² - 2ab + b²

  (√2 -3√5)² = (√2)² - 2(√2)(3√5) + (3√5)²

  = 2 - 6√10 + 3²(5)

  = 47 -6√10

__

<em>Check</em>

  √2-3√5 ≈ -5.29399 . . . . . . . . note that a negative value for side length makes no sense, so this isn't about geometry, it's about binomials and radicals

  (√2-3√5)² ≈ 28.02633

  47 -6√10 ≈ 28.02633

5 0
3 years ago
A ball is thrown in the air from a height of 3 m with an initial velocity of 12 m/s. To the nearest tenth of a second, how long
stepladder [879]

Answer:  1.4 seconds

<u>Step-by-step explanation:</u>

The equation is: h(t) = at² + v₀t + h₀          where

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Since we are trying to find out when it lands on the ground, h(t) = 0

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4 0
3 years ago
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algol [13]

Answer:

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7 0
2 years ago
Read 2 more answers
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