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3 years ago

How are the suns rays like mathematical rays

1 answer:

4 0

The mathematical rays have only one starting poin and go enendlessly in one direction,sun rays also have only one starting point and go endlessly in one direction ( although the sun rays meet the earth but the distance is so long that we don't consider it)

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Select all of the ordered pairs (x,y) that are solutions to the linear equation 2x+3y=6.

Romashka-Z-Leto [24]

Answer: (0,2), (3,0), (6, -2)

6 0

3 years ago

PLEASE HELP NEED ASAP. I BEG PLZZZZZZZ

seropon [69]

Answer:

Triangular prism

Hope this helps!

8 0

3 years ago

Read 2 more answers

7-18=______+______=______

ra1l [238]

7-18= a negative number. We know that it'll equal a negative number because the minued(7) is less than the subtrahend(18), which results in a negative number (if both are positive).

We can make an addition problem from this. We can have 18 turn negative and 7 stay positive, and we'll get the same answer.

7+-18=7-18

7-18=7+-18=11

8 0

3 years ago

Sam attempts a 42-yard field goal in a football game. For his attempt to be a success, the football needs to pass through the up

Zigmanuir [339]

Answer:

The correct option is;

The kick is good! The ball clears the crossbar by nearly 15 feet.

Step-by-step explanation:

Here we have;

Velocity of ball = 70 ft/s

Angle of motion = 50°

Therefore;

Vertical component of velocity, $v_y$ = v×sinθ = v×sin×50 = 53.62 ft/s

$v_y$ = u $_y$ - g·t

At maximum height $v_y$ = 0, therefore;

u $_y$ = g·t and

t = u $_y$ /g

Where:

u $_y$ = Initial vertical velocity = 53.62 ft/s

g = Acceleration due to gravity = 32.1740 ft/s²

∴ t = 53.62/32.1740 = 1.66666 s

Total time of flight = 2 × 1.66666 = 3.33332 s

Max height is given by the following relation;

v $_y$ ² = u $_y$ ² - 2·g·s

v $_y$ = 0 ft/s at maximum height, therefore;

u $_y$ ² = 2·g·s

53.62² = 2×32.1740×s

s = 53.62² ÷(2×32.1740) = 44.69 ft

Horizontal velocity = 70×cos(50) = 44.995 ft/s

Hence the time it takes the ball to reach 42 yards is given by the following relation;

42 yards = 126 ft

$Time, t = \frac{Distance}{Velocity} = \frac{126}{44.995} = 2.8 \, s$

Height of ball at 2.8 s is given by;

Direction of ball at 2.8 s = downwards

Hence time after peak = 2.8 - 1.666 = 1.1336429 s

Position at 2.8 s is given by

s = u $_y$ ·t + 1/2·g·t²

u $_y$ = 0 ft/s as ball is on second half of flight

∴ s = 1/2·g·t² = 1/2×32.1740×1.1336429²= 20.674 ft below maximum height

∴ Height above ground = Maximum height - 20.674 ft = 44.69 - 20.674

Height above ground = 24.01 ft

Hence the ball clears the crossbar by 24.01 - 10 = 14.01 ft

The best option is therefore;

The kick is good! The ball clears the crossbar by nearly 15 feet.

8 0

4 years ago

Let X be the future lifetime random variable in years for a newborn (age 0). The probability density function of X is as follows

meriva

Answer:

Step-by-step explanation:

First we need to find the value of constant k in equation. To find k we use the formula:

$\int\limits^{100}_0 {ke^{-0.01x}} \, dx =1\\Integrating:\\k\int\limits^{100}_0 {e^{-0.01x}} \, dx =1\\\frac{k}{-0.01}[e^{-0.01x}]_0^{100}=1\\-100k [e^{-0.01x}]_0^{100}=1\\-100k[e^{-0.01*100}-e^{-100*0}]=1\\-100k[e^{-1}-e^0]=1\\-100k(-0.632)=1\\63.2k = 1\\k=0.0158$

the probability that a person age 20 will live for at least 20 more years = P(40 ≤ x <∞).

The person would live for at least 40 years

Therefore:

P(40 ≤ x <∞) =

$\int\limits^{\infty}_{40} 0.0158e^{-0.01x}} \, dx \\Integrating:\\0.0158\int\limits^{\infty}_{40} {e^{-0.01x}} \, dx =\frac{0.0158}{-0.01}[e^{-0.01x}]_{40}^{\infty}\\=-1.58 [e^{-0.01x}]_{40}^{\infty}=-1.58[e^{-\infty}-e^{-100*40}]\\=-1.58[e^{-\infty}-e^{-4000}]=-1.58(0-0)=0$

6 0

3 years ago