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2 years ago

8

A rectangular park is 110 meters long and 60 meters wide.

1 answer:

Harlamova29_29 [7]2 years ago

3 0

Answer:

100 meters long 70 meters wide I think

Step-by-step explanation:

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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

2 years ago

nadya68 [22]

Answer:

Scalene, Acute

Step-by-step explanation:

Scalene because each side has a different length

obtuse because one angle is more than 90°

6 0

2 years ago

If 2k, 5k-1 and 6k+2 are the first 3 terms of an arithmetic sequence, find k and the 8th term

Klio2033 [76]

Answer:

see explanation

Step-by-step explanation:

The common difference d of an arithmetic sequence is

d = $a_{2}$ - $a_{1}$ = $a_{3}$ - $a_{2}$

Substitute in values and solve for k, that is

5k - 1 - 2k = 6k + 2 - (5k - 1)

3k - 1 = 6k + 2 - 5k + 1

3k - 1 = k + 3 ( subtract k from both sides )

2k - 1 = 3 ( add 1 to both sides )

2k = 4 ⇒ k = 2

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

The n th term of an arithmetic sequence is

$a_{n}$ = $a_{1}$ + (n - 1)d

$a_{1}$ = 2k = 2 × 2 = 4 and

d = 5k - 1 - 2k = 3k - 1 = (3 × 2) - 1 = 5

Hence

$a_{8}$ = 4 + (7 × 5) = 4 + 35 = 39

4 0

2 years ago

If F(x) = 5x-7 and G(x) = -2x+9 , what is F(G(x))?

dybincka [34]

Answer:

Step-by-step explanation:

7 0

2 years ago

Find an equation of the line that has the given slope and contains the given point. m=7/3 (1,6)

iris [78.8K]

The point-slope form:

$y-y_1=m(x-x_1)$

We have the point (1, 6) and the slope m = 7/3. Substitute:

$y-6=\dfrac{7}{3}(x-1)$ <em>use distributive property</em>

$y-6=\dfrac{7}{3}x-\dfrac{7}{3}$ <em>add 6 to both sides</em>

$y=\dfrac{7}{3}x+\dfrac{11}{3}$ <em>multiply both sides by 3</em>

$3y=7x+11$ <em>subtract 7x from both sides</em>

$-7x+3y=11$ <em>change the signs</em>

$7x-3y=-11$

Answer:

point-slope form: $y-6=\dfrac{7}{3}(x-1)$

slope-intercept form: $y=\dfrac{7}{3}x+\dfrac{11}{3}$

standard form: $7x-3y=-11$

5 0

2 years ago