All categories

3 years ago

Please help me please I need help!!!!!!!!!!!!!!!!!!!hshdjdndndbdbnfjfkfkfkdkdkfkfjdjfiid

Mathematics

1 answer:

Serggg [28]3 years ago

6 0

y = (x - 2)(x + 5) First I would multiply (x - 2) and (x + 5) together

y = x² + 5x - 2x - 10

y = x² + 3x - 10

Next I would plug in a number for x to find its y value. I will plug in 0

y = 0² + 3(0) - 10

y = -10

(0, -10) [this is the y-intercept (the y value when x = 0)]

Your answer is Graph A because the y-intercept is (0,-10)

You might be interested in

A thin metal plate, located in the xy-plane, has temperature T(x, y) at the point (x, y). Sketch some level curves (isothermals)

Sophie [7]

Answer:

Step-by-step explanation:

Given that:

$T(x,y) = \dfrac{100}{1+x^2+y^2}$

This implies that the level curves of a function(f) of two variables relates with the curves with equation f(x,y) = c

here c is the constant.

$c = \dfrac{100}{1+x^2+2y^2} \ \ \--- (1)$

By cross multiply

$c({1+x^2+2y^2}) = 100$

$1+x^2+2y^2 = \dfrac{100}{c}$

$x^2+2y^2 = \dfrac{100}{c} - 1 \ \ -- (2)$

From (2); let assume that the values of c > 0 likewise c < 100, then the interval can be expressed as 0 < c <100.

Now,

$\dfrac{(x)^2}{\dfrac{100}{c}-1 } + \dfrac{(y)^2}{\dfrac{50}{c}-\dfrac{1}{2} }=1$

This is the equation for the family of the eclipses centred at (0,0) is :

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

$a^2 = \dfrac{100}{c} -1 \ \ and \ \ b^2 = \dfrac{50}{c}- \dfrac{1}{2}$

Therefore; the level of the curves are all the eclipses with the major axis:

$a = \sqrt{\dfrac{100 }{c}-1}$ and a minor axis $b = \sqrt{\dfrac{50 }{c}-\dfrac{1}{2}}$ which satisfies the values for which 0< c < 100.

The sketch of the level curves can be see in the attached image below.

7 0

3 years ago

Robert earned 10$ which is 4 less than twice what Eric earned ,how much did Eric earn?

horrorfan [7]

Dont really understand but i think 6 idk im not really a good mathematical person but who cares im on this app for a reason.

5 0

3 years ago

Read 2 more answers

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

Part a: The probability of no arrivals in a one-minute period is 0.000045.

Part b: The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c: The probability of no arrivals in a 15-second is 0.0821.

Part d: The probability of at least one arrival in a 15-second period is 0.9179.

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

The probability of no arrivals in a one-minute period is 0.000045.

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

The probability of no arrivals in a 15-second is 0.0821.

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

The probability of at least one arrival in a 15-second period is 0.9179.

7 0

3 years ago

If x+9.8=14.7, what is the value of 8(x-3.7?

yulyashka [42]

First find X by subtracting 9.8 from 14.7
X should equal 4.9
than fill in the equation for x
8(4.9-3.7)
you then subtract in the parenthasis
getting the answer of 1.2
finally multiply 1.2x8
Your final answer is 9.6

3 0

3 years ago

GEOMETRY PROBLEM THAT I CAN NEVER GET

alina1380 [7]

Answer:

Therefore the friends will have to buy 511 feet of fence (rounding up) or 510 feet of fence ( to the nearest foot).

Step-by-step explanation:

i) perimeter of rectangular yard = 319 feet

ii) perimeter of yard of friends $= \frac{8}{5} \times 319 = 510.4$

iii) Therefore the friends will have to buy 511 feet of fence or 510 feet of fence ( to the nearest foot).

6 0

3 years ago