Given f(x) = sin x and g(x) = cos x, find f(g(pi/2)) = 0. show all steps.

What is 3.14 x 11 x2

butalik [34]

Answer:jurrhthrhth

Step-by-step explanation:hdhdbbdfh

4 0

3 years ago

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-3 = 7 - BLANK pls tell me what blank is

liberstina [14]

Answer:

10

Step-by-step explanation:

-3 = 7 - x

Add x to both sides

x -3 = 7 - x +x

x - 3 = 7

Now, add 3 to both sides

x - 3 + 3 = 7 + 3

x = 10

4 0

3 years ago

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Riley solved and equation as shown in the table

Mama L [17]

She made the mistake in step 4

She should have divided by -2 not just 2 alone

6 0

3 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

3 years ago

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A scientist has a sample of bacteria that initially contains 10 million microbes. They observe the sample and finds that the num

sergiy2304 [10]

Answer:

M(t) = 6.7 * 10⁷ (67 million)

Minutes (t) =55

Step-by-step explanation:

1. Write an exponential equation that represents M, the total number of bacterial microbes in millions, as a function of t, the number of minutes the sample has been observed.

For answering this question, we will use the following formula:

M(t) = B₀ * g ^(t/m), where:

•M(t) represents the total number of bacterial microbes in millions.

• B₀ represents the initial population of bacteria in millions.

• g represents the growth factor.

• t represents the total number of minutes we will observe the bacteria growing.

• m represents the time in minutes it takes to the growth factor g to occur.

2. Then, determine how much time, to the nearest minute, will pass until there are 67 million bacterial microbes.

M(t) = B₀ * g ^(t/m)

Replacing with the values we know:

6.7 * 10⁷ = 10⁷ * 2 ^(t/20)

6.7 = 2 ^(t/20) (Dividing by 10⁷ at both sides)

ln 6.7 = ln 2 ^(t/20)

ln 6.7 = t/20 ln 2

ln 6.7/ ln 2 = t/20

t = ln 6.7/ln 2 * 20

t = 2.74 * 20

t = 54.88

t ≅ 55 (rounding to the nearest minute)

3 0

2 years ago