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3 years ago

How are trigonometric functions graphed to show the period, midline, and amplitude?

1 answer:

4 0

Trigonometric functions are graphed like repeated waves to show that the period is the horizontal length from one peak to the next, that the midline is the line in the middle of the graph, and that the amplitude is the height of the wave.

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The right circular cone below has a slant height of 16.2 centimeters and a base circumference of 44 centimeters.

Goryan [66]

Step-by-step explanation:

slant height² = height² + base²

16.2²= h²+44²

16.2²-44²=h²

1936-262.44= h²

=√1673.56=√h²

h=40.91

5 0

4 years ago

The circumference of a circle is 13π ft. What is the area, in square feet? Express your answer in terms of π.

Ksenya-84 [330]

Answer:

$\frac{169}{4} \pi$ square feet

Step-by-step explanation:

Circumference = 2πr

13π = 2πr

$\frac{13}{2}$ =r

Area = πr²

π × $\frac{13}{2}$ ² =

$\frac{169}{4}$ π square feet

5 0

3 years ago

A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, t

ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

$T(t)=T_a+(T_0-T_a)e^{-kt}$

where $T_a$ is the ambient temperature, $T_0$ is the initial temperature, $t$ is the time and $k$ is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say $T_0$

We know that the ambient temperature is $T_a=65$ , so

$T(t)=65+(T_0-65)e^{-kt}$

We also know that when $t=5 \:min$ the temperature is $T(5)=35$ and when $t=10 \:min$ the temperature is $T(10)=50$ which gives:

$T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}$

$T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}$

Rearranging,

$35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}$

$50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}$

If we divide these two equations we get

$\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}$

$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

so the beam started out at 5 °F.

6 0

4 years ago

AB = 6x + 2 and<br> BC = 5x + 3<br> Find AC.

uranmaximum [27]

Answer:

AC = 3x + 3 (answers may vary)

Step-by-step explanation:

AB = 6x + 2

Answers may vary for this question,

Let's say that,

B = 4x + 1

So that means that,

A = 2x + 1

BC = 5x + 3

Since we chose

B = 4x + 1,

we must subtract B from BC to get C,

C = x + 2

Now we add A and C to get AC,

A = 2x + 1

C = x + 2

AC = 3x + 3

5 0

3 years ago

Can I get some help with this question? Solve 3(x + 2) > x.

Montano1993 [528]

Answer:

$\large\boxed{x>-3\to\{x\ |\ x>-3\}\to x\in(-3,\ \infty)}$

Step-by-step explanation:

$3(x+2)>x\qquad\text{use the distributive property:}\ a(b+c)=ab+ac\\\\(3)(x)+(3)(2)>x\\\\3x+6>x\qquad\text{subtract 6 from both sides}\\\\3x+6-6>x-6\\\\3x>x-6\qquad\text{subtract x from both sides}\\\\3x-x>x-x-6\\\\2x>-6\qquad\text{divide both sides by 2}\\\\\dfrac{2x}{2}>\dfrac{-6}{2}\\\\x>-3$

8 0

4 years ago

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