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Juli2301 [7.4K]
3 years ago
13

Someone please help me

Mathematics
1 answer:
Firdavs [7]3 years ago
4 0

Answer:

  x = -3/2

Step-by-step explanation:

The removable discontinuity is at the value of x where a denominator factor cancels a numerator factor, and those factors are zero.

  p(x)=\dfrac{2x+3}{4x^2-9}=\dfrac{2x+3}{(2x+3)(2x-3)}=\dfrac{2x+3}{2x+3}\cdot\dfrac{1}{2x-3}\\\\p(x)=\dfrac{1}{2x-3}\qquad x\ne-3/2

That cancellation occurs where 2x+3 = 0, at x=-3/2.

The function p(x) has a removable discontinuity when x = -3/2.

You might be interested in
Solve by substitution or elimination<br> 4X – 3y = 7<br> 4x + y = 11
Simora [160]

Answer:

(5/2 , 1)

Step-by-step explanation:

4X – 3y = 7

4x + y = 11

———————

Multiply a -1 to the bottom equation to get the 4x as a negative so it cancels out.

4x - 3y = 7

-4x - y = -11

——————

-4y = -4

Divide by -4

y = 1

Substitute the value of y into one of the equations and solve

4x - 3(1) = 7

4x -3 = 7

4x = 10

Divide by 4

X = 10/4

Simplify by dividing by 2

x = 5/2

Therefore the answer is (5/2, 1 )

8 0
3 years ago
Sam was given a bag of 100 red and blue marbles. He believes
Rashid [163]

Answer:

(D) There are probably more blue marbles than red marbles in the bag.

Step-by-step explanation:

There are a total of 100 marbles in the bag.

In the experiment of 50 trials, Sam had the following outcome:

Blue=35

Red=15

Relative Frequency of Blue Marbles =35/50=0.7

Relative Frequency of Red Marbles =15/50=0.3

Since the relative frequency of blue marbles is greater than the relative frequency of red marbles, <u>there are probably more blue marbles than red marbles in the bag.</u>

The correct option is D.

6 0
3 years ago
First question, thanks. I believe there should be 3 answers
zysi [14]

Given: The following functions

A)cos^2\theta=sin^2\theta-1B)sin\theta=\frac{1}{csc\theta}\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}

B

\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}

C

\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}

D

\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}

E

\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}

Hence, the following are identities

\begin{gathered} B)sin\theta=\frac{1}{csc\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}

The marked are the trigonometric identities

3 0
1 year ago
What is the value of (-2.8)+0+(-2.6)+6.4
Sever21 [200]
(-2.8)+0+(-2.6)+6.4
-2.8-2.6+6.4
-5.4+6.4
1
5 0
2 years ago
The distance d of a particle moving in a straight line is given by d(t) = 2t3 + 5t – 2, where t is given in seconds and d is mea
Sergeeva-Olga [200]

Answer:

(C)6t^2+5

Step-by-step explanation:

Given the distance, d(t) of a particle moving in a straight line at any time t is:

d(t) = 2t^3 + 5t - 2, $ where t is given in seconds and d is measured in meters.

To find an expression for the instantaneous velocity v(t) of the particle at any given point in time, we take the derivative of d(t).

v(t)=\dfrac{d}{dt}\\\\v(t) =\dfrac{d}{dt}(2t^3 + 5t - 2) =3(2)t^{3-1}+5t^{1-1}\\\\v(t)=6t^2+5

The correct option is C.

6 0
3 years ago
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