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3 years ago

11

Write an expression to describe the sequence below. Use n to represent the position of a term in the sequence, where n = 1 for t

he first term. -35, -45, -55, -65

1 answer:

nydimaria [60]3 years ago

5 0

Answer: The Given sequence is 9, 10, 11, 12

We need to find the expression to describe the sequence.

Let be the term of the sequence.

Let n represent the position of the term.

Let a be the first term in the sequence.

and d be the common difference between the sequence

Hence the expression to find the above sequence is given below;

when n=1 d= 1 a = 9

when n=2 d= 1 a = 9

when n=3 d= 1 a = 9

when n=3 d= 1 a = 9

Hence the expression is

Step-by-step explanation:

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Tom spent 153 minutes completing a race. He walked 63 minutes and jogged the rest. What is the ratio of time he jogged to time h

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Order the numbers from least to greatest 41/50, 0.83, 80%

Tresset [83]

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80%, 41/50, 0.83

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3 years ago

5g + 4(-5+3g) = 1 - g

Korvikt [17]

5g + 4(-5 + 3g) = 1 - g
5g - 20 + 12g = 1 - g
17g - 20 = 1 - g
$\frac{18g}{18} = \frac{21}{18}$
there can be to answer
g = $\frac{21}{18}$ or
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3 years ago

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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

4 years ago