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2 years ago

Find the slope of a line parallel to each given line. 5)2x+3y=-3

Mathematics

1 answer:

Volgvan2 years ago

7 0

Answer:

-2/3

Step-by-step explanation:

2x + 3y = - 3 re-arrange to y = mx+b form: slope of this line will be 'm'

y = -2/3 x -1 slope = -2/3 parallel slope is the same

You might be interested in

Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

7 0

2 years ago

What are the real solutions of the equation x^4= 81?

zmey [24]

Use photomath it’s free

5 0

3 years ago

Look closely at the model.

Maslowich

Answer:

Perimeter: 2(x+3) + 2(x+2)

Step-by-step explanation:

6 0

3 years ago

Given the equation y=2x+4, what is the value of y when x=5? *

Law Incorporation [45]

Y=2(5)+ 4
Y= 10+4
Y = 14

6 0

2 years ago

It’s a new semester! Students are grouped into three clubs, which each has 10, 4 and 5 students. In how many ways can teacher se

ozzi

Here we must see in how many different ways we can select 2 students from the 3 clubs, such that the students <em>do not belong to the same club. </em>We will see that there are 110 different ways in which 2 students from different clubs can be selected.

So there are 3 clubs:

Club A, with 10 students.
Club B, with 4 students.
Club C, with 5 students.

The possible combinations of 2 students from different clubs are

Club A with club B
Club A with club C
Club B with club C.

The number of combinations for each of these is given by the product between the number of students in the club, so we get:

Club A with club B: 10*4 = 40
Club A with club C: 10*5 = 50
Club B with club C. 4*5 = 20

For a total of 40 + 50 + 20 = 110 different combinations.

This means that there are 110 different ways in which 2 students from different clubs can be selected.

If you want to learn more about combination and selections, you can read:

brainly.com/question/251701

6 0

2 years ago