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All numbers bigger than five, or at most negative one

Kisachek [45]

Answer:

Hi there!

Your answer is:

-1 >=x>5

This can be broken down to:

x <= -1

x>5

Which fits your requirements!

6 0

3 years ago

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station. Find the ra

Anika [276]

Answer:

The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

Step-by-step explanation:

Given information:

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station.

$z=1$

$\frac{dx}{dt}=430$

We need to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

$y=2$

According to Pythagoras

$hypotenuse^2=base^2+perpendicular^2$

$y^2=x^2+1^2$

$y^2=x^2+1$ .... (1)

Put z=1 and y=2, to find the value of x.

$2^2=x^2+1^2$

$4=x^2+1$

$4-1=x^2$

$3=x^2$

Taking square root both sides.

$\sqrt{3}=x$

Differentiate equation (1) with respect to t.

$2y\frac{dy}{dt}=2x\frac{dx}{dt}+0$

Divide both sides by 2.

$y\frac{dy}{dt}=x\frac{dx}{dt}$

Put $x=\sqrt{3}$ , y=2, $\frac{dx}{dt}=430$ in the above equation.

$2\frac{dy}{dt}=\sqrt{3}(430)$

Divide both sides by 2.

$\frac{dy}{dt}=\frac{\sqrt{3}(430)}{2}$

$\frac{dy}{dt}=372.390923627$

$\frac{dy}{dt}\approx 372$

Therefore the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

6 0

3 years ago

Convert the fraction 1/3 into a repeating decimal

den301095 [7]

Answer:

0.33 with the bar on top

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the distance of this line? Which is the right answer?

Shtirlitz [24]

Answer:

Step-by-step explanation:

We can use the distance formula derived from the Pythagorean theorem

D = $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

the two points given are

(0, 3) and (-2, -3)

$(x_2-x_1) = (0-(-2)) = 2\\(y_2-y_1) = (3-(-3)) = 6\\D = \sqrt{(2)^2 + (6)^2} \\D = \sqrt{4 + 36} \\D = \sqrt{40} = 6.324$

6 0

3 years ago

Given that 2y^3+by-cy+d,where b,c and d are constants,leaves a remainder R when divided by (y+1) , (y-2) and (2y-1). Find the va

Eddi Din [679]

The polynomial remainder theorem says that a polynomial p(x) leaves a remainder of p(k) when it's divided by x - k.

We're given that dividing p(y) = 2y³ + by² - cy + d leaves the same remainder R after dividing it by y + 1, y - 2, and 2y - 1. So we have

p(-1) = 2(-1)³ + b(-1)² - c(-1) + d = R

==> R = -2 + b + c + d

p(2) = 2(2)³ + b(2)² - c(2) + d = R

==> R = 16 + 4b - 2c + d

p(1/2) = 2(1/2)³ + b(1/2)² - c(1/2) + d = R

==> R = 1/4 + b/4 - c/2 + d

We're also given that y + 2 is a factor, which means dividing p(y) by it leaves no remainder, and so

p(-2) = 2(-2)³ + b(-2)² - c(-2) + d = 0

==> 0 = -16 + 4b + 2c + d

Solve the system of equations in boldface. You can eliminate d from the first 3 to first solve for b and c, then solve for d :

(-2 + b + c + d) - (16 + 4b - 2c + d) = R - R

-18 - 3b + 3c = 0

b - c = -6

(-2 + b + c + d) - (1/4 + b/4 - c/2 + d) = R - R

-9/4 + 3b/4 + 3c/2 = 0

b + 2c = 3

(b - c) - (b + 2c) = -6 - 3

-3c = -9

c = 3

b - 3 = -6

b = -3

-16 + 4(-3) + 2(3) + d = 0

d = 22

7 0

3 years ago