Find the value of x for which p is parallel to q , if m<1=(3x) and m<3=105. The diagram is not scale.

Please help and please slove and explain im not getting this

OLga [1]

Answer:

x-intercept= (3⅓, 0)

y-intercept= (0, 4²⁄₇)

Please see the attached picture for the graph.

Step-by-step explanation:

-9x -7y= -30

Let's simplify the equation by dividing both sides by -1.

9x +7y= 30

x- intercept occurs at y= 0.

When y= 0,

9x +7(0)= 30

9x= 30

x= 30 ÷9

$x = 3 \frac{1}{3}$

Thus, x- intercept occurs at (3⅓, 0).

y-intercept occurs at x= 0.

When x= 0,

9(0) +7y= 30

7y= 30

y= 30 ÷7

$y = 4 \frac{2}{7}$

Thus, y- intercept occurs at (0, 4²⁄₇).

_____

To graph the equation, draw the x and y axis on a graph paper. Use an appropriate scale to divide the line into equal parts. Next, plot the points (3⅓, 0) and (0, 4²⁄₇). Then, join the two points with a straight line.

Notes:

x- intercept is the point at which the graph cuts through the x- axis. In this case, your x- axis is the horizontal line that runs from left to right of your graph paper. In order for a point to be on this horizontal line, look at the y- axis and notice that it sits at y= 0. The same reason applies for why the y- intercept occurs at x= 0. This has to do with the two axis cutting each other at the point (0,0), resulting in the x and y coordinates of 0 for the y and x intercepts respectively.

Simplifying the equation in the first step is not necessary, but it is a good practice and might reduce carelessness.

5 0

2 years ago

What are all the common factors of 24, 64 and 88

Hunter-Best [27]

1,2,4&8 hopefully this helps

8 0

3 years ago

Read 2 more answers

Find the nth term of this quadratic sequence 3 8 15 24

olga2289 [7]

Answer:

35

Step-by-step explanation:

A quadratic sequence is a form. of mathematical progression of numbers in which the subsequent differences between each consecutive term differ by the exact amount. This is known as the common second difference.

Hence in this case,

we have

S= 3+8+15+24+…..+Tn - 1.

S=. 3+8+15+……………..Tn-1 + Tn

Deducting 2nd series from the first

We have 0=3+5+7+…….Tn

Hence, we have Tn= n (n+2)

= 24 (9+2)

= 24 + 11

= 35

6 0

3 years ago

Use the laplace transform to solve the given initial-value problem. y' 5y = e4t, y(0) = 2

Basile [38]

The Laplace transform of the given initial-value problem

$y' 5y = e^{4t}, y(0) = 2$ is mathematically given as

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$

<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>

Generally, the equation for the problem is mathematically given as

$&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}$

$\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}$

$put $s=-s \Rightarrow a_{1}=\frac{17}{9}$$

$\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}$

In conclusion, Taking inverse Laplace tranoform

$L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\$

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$