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andreyandreev [35.5K]

2 years ago

8

The area of a triangle can be represented by the expression 14x^5+63x^2. If the base is 7x^2, write an expression to represent i

ts height.

1 answer:

horrorfan [7]2 years ago

4 0

Please see pic for work and answer.

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A young couple purchases their first new home in 2002 for $120,000. They sell it to move into bigger home in 2007 for $150,000.

Fed [463]

Answer:

The exponential model for the value of the home is $V(t)=120000e^{0.045t}$ .

Step-by-step explanation:

According to the give information 2002 is the initial year and the value of the hom in 2002 is $120,000.

The model will have the form

$V(t)=V_0e^{kt}$

Where V₀ is initial value of home, k is a constant and t is number f years after 2002.

$V(t)=120000e^{kt}$

The value of home in 2007 is $150,000. Difference between 2007 and 2002 is 5 years. Therefore the value of function is 150000 at t=5.

$150000=120000e^{k(5)}$

$\frac{150000}{120000}=e^{5k}$

$\frac{5}{4}=e^{5k}$

Take ln both sides.

$ln(\frac{5}{4})=lne^{5k}$

$ln(\frac{5}{4})=5k$ ( $lne^a=a$ )

$\frac{ln(\frac{5}{4})}{5}=k$

$k=0.04462871\approx 0.045$

Therefore exponential model for the value of the home is $V(t)=120000e^{0.045t}$ .

Where t is number of years after 2002.

7 0

2 years ago

Read 2 more answers

The number f of miles a helicopter is from its destination x minutes after takeoff is given by f(x)=75−1.5x. The number f of mil

soldier1979 [14.2K]

Answer:

9

Step-by-step explanation:

10-1=9

5 0

2 years ago

A swimmer can swim 50.2 meters in 1 minute. How far can he swim in half an hour? *

belka [17]

Answer:

1506meters

Step-by-step explanation:

1hour-60minutes

which means half an hour is 30minutes,so it's more like they are asking you to find how far the swimmer can swim in 30minutes

50.2-1min

x -30min

x=50.2×30

x=1506meters

6 0

2 years ago

Read 2 more answers

Half angle tan 5pi/8

Bond [772]

The half-angle formula for tangent is:

tan(a/2) = (sin a / (1 + cos a)) = ((1 - cos a) / sin a)

Now we can plug in values:

tan(5π/8) = (sin(5π/4) / (1 + cos(5π/4)) = ((1 - cos(5π/4)) / sin(5π/4)

tan(5π/8) = (-√2/2) / (1 + (-√2/2)) = (1 - (-√2/2)) / (-√2/2)

tan(5π/8) = ((-√2/2)) / ((2 - √2)/2) = ((2 + √2)/2) / (-√2/2)

Now we can solve the first half:

(-√2/2)(2 / (2 - √2))
(-√2/2)((4 + 2√2) / 2)
(-√2/2)(2 + √2)
(-2√2 - 2)/2
-√2 - 1

tan(5pi/8) = -√2 - 1

4 0

2 years ago

In the month of June, the temperature in Johannesburg, South Africa, varies over the day in a periodic way that can be modeled a

UNO [17]

Answer:

T = - 7.5 Cos π/12( t - 4 ) + 10.5

Step-by-step explanation:

Given that the

Maximum temp. = 18 degree Celsius

Minimum temp. = 3 degree Celsius

The half way between 10 am and 10 pm is 4 am

The sine and cosine functions can be used to model fluctuations in temperature data through out the year. An equation that can be used to model these data is of the form:

T = A cos B(t - C) + D, where A,B,C,D, are constants, T is the temperature in °C and t is the hour (1–24)

A = amplitude = (Tmax - Tmin)/2

A = (3 - 18)/2 = - 15/2 = -7.5 ( note : after midnight)

B = 2π/24 = π/12

C = units translated to the right

C = 4

D = ymin + amplitude = units translated up

D = 7.5 + 3 = 10.5

The formula of the trigonometric function that models the temperature T in Johannesburg t hours after midnight we be

T = - 7.5 Cos π/12( t - 4 ) + 10.5

3 0

2 years ago