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3 years ago

Factor the expression. 6x + 54

Mathematics

2 answers:

vichka [17]3 years ago

8 0

6(x+9) is the factored expression.

Anna71 [15]3 years ago

7 0

Answer:

6(x+9)

Step-by-step explanation:

6x+54

2(3x+27)

2x3(x+9)

6(x+9)

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For 0 ≤ ϴ < 2π, how many solutions are there to tan(StartFraction theta Over 2 EndFraction) = sin(ϴ)? Note: Do not include va

Black_prince [1.1K]

Answer:

3 solutions:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

Step-by-step explanation:

So, first of all, we need to figure the angles that cannot be included in our answers out. The only function in the equation that isn't defined for some angles is $tan(\frac{\theta}{2})$ so let's focus on that part of the equation first.

We know that:

$tan(\frac{\theta}{2})=\frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})}$

therefore:

$cos(\frac{\theta}{2})\neq0$

so we need to find the angles that will make the cos function equal to zero. So we get:

$cos(\frac{\theta}{2})=0$

$\frac{\theta}{2}=cos^{-1}(0)$

$\frac{\theta}{2}=\frac{\pi}{2}+\pi n$

$\theta=\pi+2\pi n$

we can now start plugging values in for n:

$\theta=\pi+2\pi (0)=\pi$

if we plugged any value greater than 0, we would end up with an angle that is greater than $2\pi$ so, that's the only angle we cannot include in our answer set, so:

$\theta\neq \pi$

having said this, we can now start solving the equation:

$tan(\frac{\theta}{2})=sin(\theta)$

we can start solving this equation by using the half angle formula, such a formula tells us the following:

$tan(\frac{\theta}{2})=\frac{1-cos(\theta)}{sin(\theta)}$

so we can substitute it into our equation:

$\frac{1-cos(\theta)}{sin(\theta)}=sin(\theta)$

we can now multiply both sides of the equation by $sin(\theta)$

so we get:

$1-cos(\theta)=sin^{2}(\theta)$

we can use the pythagorean identity to rewrite $sin^{2}(\theta)$ in terms of cos:

$sin^{2}(\theta)=1-cos^{2}(\theta)$

so we get:

$1-cos(\theta)=1-cos^{2}(\theta)$

we can subtract a 1 from both sides of the equation so we end up with:

$-cos(\theta)=-cos^{2}(\theta)$

and we can now add $cos^{2}(\theta)$

to both sides of the equation so we get:

$cos^{2}(\theta)-cos(\theta)=0$

and we can solve this equation by factoring. We can factor $cos(\theta)$ to get:

$cos(\theta)(cos(\theta)-1)=0$

and we can use the zero product property to solve this, so we get two equations:

Equation 1:

$cos(\theta)=0$

$\theta=cos^{-1}(0)$

$\theta={\frac{\pi}{2}, \frac{3\pi}{2}}$

Equation 2:

$cos(\theta)-1=0$

we add a 1 to both sides of the equation so we get:

$cos(\theta)=1$

$\theta=cos^{-1}(1)$

$\theta=0$

so we end up with three answers to this equation:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

7 0

3 years ago

Solve the problem c=3.14d for d

Maksim231197 [3]

Answer:

$c = 3.14d \\ d = \frac{c}{3.14}$

It's change of subject. :)

3 0

2 years ago

All the fourth-graders in a certain elementary school took a standardized test. A total of 81% of the students were found to be

Dvinal [7]

Answer:

The probability that a student is proficient in mathematics, but not in reading is, 0.10.

The probability that a student is proficient in reading, but not in mathematics is, 0.17

Step-by-step explanation:

Let's define the events:

L: The student is proficient in reading

M: The student is proficient in math

The probabilities are given by:

$P (L) = 0.81\\P (M) = 0.74\\P (L\bigcap M) = 0.64$

$P (M\bigcap L^c) = P (M) - P (M\bigcap L) = 0.74 - 0.64 = 0.1\\P (M^c\bigcap L) = P (L) - P (M\bigcap L) = 0.81 - 0.64 = 0.17$

The probability that a student is proficient in mathematics, but not in reading is, 0.10.

The probability that a student is proficient in reading, but not in mathematics is, 0.17

5 0

4 years ago

In a survey of students , every student studies at least one of the three subjects Maths English and Science . The number of stu

Mandarinka [93]

Answer:

80x2=160 20x3=60 160+60=220

Step-by-step explanation:

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3 years ago

Suppose that x and y vary inversely and that y = 5 when x = 1. What is y when x = 15?

LiRa [457]

Answer:

Y=1/3

Step-by-step explanation:

X=k/y

So let's make k the subject of formula

k=xy

x=1,y=5

So let's substitute

k=1×5

k=5

So we are to determine the value for y when x is 15

So let's solve

Y=k/x

Y=5/15

Y=1/3

Y=0.333333~0.3

7 0

3 years ago