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2 years ago

Choose all of the zeroes of f(x).

Mathematics

2 answers:

Korolek [52]2 years ago

8 0

Answer:

yes

Step-by-step explanation:

Sergeu [11.5K]2 years ago

8 0

Answer:

A, B, D on edge :)

Step-by-step explanation:

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2/3 + (-1/3) = ?????

amm1812

Answer: (+) 1/3

Step-by-step explanation: 2/3 + -1/3 is basically subtracting 1/3 from 2/3 so, 2/3 - 1/3 = 1/3.

6 0

2 years ago

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Where are my taurus’s At??

lozanna [386]

Answer:

im a scorpio but my sister in law is a taurus... :)

Step-by-step explanation:

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2 years ago

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Car A, traveling at a constant 10 m/s, crosses the start line 3 seconds before Car B, who is traveling at a constant 15 m/s. At

Margaret [11]

The distance between Car A and car B,When Car A crosses the start line is:
distance =speed car B* time
distance=(15 m/s)(3 s)=45 m

Distance traveled by car A =x, (when the car B is at the same distance from the start line)
time of car A=t

x=10 m/st ⇒ x=10t (1)

Distance traveled by car B=x
time of car B=t-3

x=15(t-3) (2)

With the equations (1) and (2) we make a system of equations:

x=10t
x=15(t-3)

We solve this system of equations:

10t=15(t-3)
10t=15t-45
-5t=-45
t=-45 / -5
t=9

t-3=9-3=6

x=10 t=10 (9)=90

Answer: The time would be 9 seconds for Car A and 6 seconds for car B and the distance would be 90 meters.

8 0

3 years ago

Please help ASAP!!!!!

LenKa [72]

Answer: first answer choice (1/3) (p+45)

Step-by-step explanation:

to do this, factor out 1/3 from each element

so divide 1/3 p by 1/3 and 15 by 1/3

(1/3 p) / (1/3 ) = p

15 / (1/3) = 15 * 3 = 45

now put 1/3 outside of the parentheses

(1/3) (p+45)

8 0

2 years ago

Find the average value of f(x+y)=sin(x+y) over

Troyanec [42]

For each given region $D$ , the average value of $f$ over $D$ is the integral of $f$ over $D$ divided by the area of $D$ . In both cases, $D$ is a rectangle so the area is trivial.

a. The area of $D$ is $\dfrac\pi3\cdot\dfrac{5\pi}3=\dfrac{5\pi^2}9$ . The integral of $f$ over $D$ is

$\displaystyle\int_0^{5\pi/3}\int_0^{\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$\displaystyle=\int_0^{5\pi/3}\left(\cos y-\cos\left(y+\dfrac\pi3\right)\right)\,\mathrm dy$

$=\sin\dfrac{5\pi}3-\sin\left(\dfrac{5\pi}3+\dfrac\pi3\right) - \sin0+\sin\dfrac\pi3=0$

So the average value of $f$ over this region is 0.

b. The area of $D$ is $\dfrac{2\pi}3\cdot\dfrac{7\pi}6=\dfrac{7\pi^2}9$ . The integral of $f$ is

$\displaystyle\int_0^{7\pi/6}\int_0^{2\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_0^{7\pi/6}\left(\cos y-\cos\left(y+\frac{2\pi}3\right)\right)\,\mathrm dy$

$=\sin\dfrac{7\pi}6-\sin\left(\dfrac{7\pi}6+\dfrac{2\pi}3\right) - \sin0+\sin\dfrac{2\pi}3=\dfrac{\sqrt3}2$

The average vale of $f$ over this region is then $\dfrac{\frac{\sqrt3}2}{\frac{7\pi^2}9}=\dfrac{9\sqrt3}{14\pi^2}$ .

4 0

2 years ago