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muminat
4 years ago
8

Can someone please solve this fast? thanks !

Mathematics
1 answer:
Tpy6a [65]4 years ago
4 0

Answer:

x=2

Step-by-step explanation:

-3x-3y=0

so 3y = -3x

y=-3/3x

y=-x

plug into next eqn

x+(-2x) = -2

-x = -2

x=2

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The first three terms of an arithmetic series are 6p+2, 4p²-10 and 4p+3 respectively. Find the possible values of p. Calculate t
Doss [256]

Answer:

First Case:

\displaystyle p=\frac{5}{2}\text{ and } d=-2

Second Case:

\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}

Step-by-step explanation:

We know that the first three terms of an arithmetic series are:

6p+2, 4p^2-10, \text{ and } 4p+3

Since this is an arithmetic sequence, each subsequent term is <em>d</em> more than the previous term, where <em>d</em> is our common difference.

Therefore, we can write the second term as;

4p^2-10=(6p+2)+d

And, likewise, for the third term:

4p+3=(6p+2)+2d

Let's solve for <em>d</em> for each of the equations.

Subtracting in the first equation yields:

d=4p^2-6p-12

And for the second equation:

2d=-2p+1

To avoid fractions, let's multiply the first equation by 2. Hence:

2d=8p^2-12p-24

Therefore:

8p^2-12p-24=-2p+1

Simplifying yields:

8p^2-10p-25=0

Solve for <em>p</em>. We can factor:

8p^2+10p-20p-25=0

Factor:

2p(4p+5)-5(4p+5)=0

Grouping:

(2p-5)(4p+5)=0

Zero Product Property:

\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}

Then, we can use the second equation to solve for <em>d</em>. So:

2d_1=-2p_1+1

Substituting:

\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}

So, for the first case, <em>p</em> is 5/2 and <em>d</em> is -2.

Likewise, for the second case:

\begin{aligned} 2d_2&=-2(-\frac{5}{4})+1 \\ 2d_2&=\frac{5}{2}+1 \\ 2d_2&=\frac{7}{2} \\ d_2&=\frac{7}{4}\end{aligned}

So, for the second case, <em>p </em>is -5/4, and <em>d</em> is 7/4.

By using the values, we can determine our series.

For Case 1, we will have:

17, 15, 13.

For Case 2, we will have:

-11/2, -15/4, -2.

8 0
3 years ago
I suck at math really vad
Diano4ka-milaya [45]
I would say to check the 1st and 3rd
3 0
3 years ago
Consider the function f(x) = the quantity x squared minus x minus 12 all over the quantity x plus 3. Describe the graph of this
FromTheMoon [43]

Answer:

The graph of f(x) is discontinuous at x= -3. The x and y-intercepts or he function f(x) are (4,0) and (0,-4) respectively. The graph of the function is a straight line.

Step-by-step explanation:

The given function is

f(x)=\frac{x^2-x-12}{x+3}

f(x)=\frac{x^2-4x+3x-12}{x+3}

f(x)=\frac{x(x-4)+3(x-4)}{x+3}

f(x)=\frac{(x-4)(x+3)}{x+3}

Cancel out the common factor (x+3).

f(x)=x-4

It is a linear equation and it will give a straight line.

Put x=0, to find the y-intercept.

f(x)=0-4

f(x)=-4

Therefore the y-intercept is (0,-4).

Put x=0, to find the x-intercept.

0=x-4

4=x

Therefore the x-intercept is (4,0).

At x= -3 the value of denominator is 0, therefore the function is undefined for x= -3.

As x approaches toward x=-3 from left and right, the value of function approaches towards y=-7.

The function is discontinuous at x= -3.

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Answer:

C

Step-by-step explanation:

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