All categories

3 years ago

PLEASE HELP....explain your answer, thank you so much! I will mark brainlyest!

Mathematics

2 answers:

sineoko [7]3 years ago

7 0

Answer:

Slope is y2-y1/x2-x1

Let’s make x1 and y1 (6,8)

Let’s make x2 and y2 (2,1)

1-8/2-6

-7/-4

Both are negative numbers, which cancel each other out, making the fraction a positive number.

7/4

A. Would be your answer

Hope I helped!

Tasya [4]3 years ago

6 0

Answer:

A. 7/4

Step-by-step explanation:

To find the slope when we have 2 points, we use the formula

m = (y2-y1)/(x2-x1)

where (x1,y1) and (x2,y2) are the given points

m = (1-8)/(2-6)

= -7/-4

=7/4

You might be interested in

If f(x)= 3x2- 2x + 3 then find what f(3) is equal to

Anettt [7]

Answer:

Answer is 24

Step-by-step explanation:

$given \: f(x) = 3 {x}^{2} - 2x + 3 \\ to \: find \: f(3) \ : \: ? \\ f(3) = 3( {3}^{2} ) - 2(3) + 3 \\ = 3(9) - 6 +3 \\ = 27 - 3 \\ = 24$

HAVE A NICE DAY!

THANKS FOR GIVING ME THE OPPORTUNITY TO ANSWER YOUR QUESTION.

5 0

3 years ago

Read 2 more answers

Help asap!!!! Divide x3/4 by X1/6

HACTEHA [7]

It should be 9/2 or 4.5

5 0

3 years ago

At a high school the student government is made up of 5 seniors, 4 juniors, 3 sophomores and 2 freshmen. If there are 8 seniors,

hjlf

Answer:

32598720

Step-by-step explanation:

Combination for the school government requires:

5 out of 8 seniors = 8! / 3!5! = 6*7*8/2*3 = 7*8 = 56
4 out of 9 juniors = 9! / 4!5! = 6*7*8*9/2*3*4= 7*2*9 = 126
3 out of 12 sophomores = 12! / 3!9! = 10*11*12/2*3 = 10*11*2 = 220
2 out of 7 fishermen = 7! / 2!5! = 6*7/2 = 21

So number of total ways is:

56*126*220*21 = 32598720

5 0

2 years ago

Help please !!!!!!!! ASAP

fgiga [73]

Answer:

Step-by-step explanation:

3 1/2 x 2 = 7

5 0

3 years ago

In Which Quadrant is this true

34kurt

Given:

$\sin \theta$

$\tan \theta$

To find:

The quadrant in which $\theta$ lie.

Solution:

Quadrant concept:

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II, only $\sin\theta$ and $\csc\theta$ are positive.

In Quadrant III, only $\tan\theta$ and $\cot\theta$ are positive.

In Quadrant IV, only $\cos\theta$ and $\sec\theta$ are positive.

We have,

$\sin \theta$

$\tan \theta$

Here, $\sin\theta$ is negative and $\tan\theta$ is also negative. It is possible, if $\theta$ lies in the Quadrant IV.

Therefore, the correct option is D.

5 0

3 years ago