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3 years ago

Find the slope between the pair of points: (-4,3) and (6,-2)

Mathematics

2 answers:

ira [324]3 years ago

8 0

Slope is always (y2-y1)/(x2-x1) in this case:

(-2-3)/(6--4)

-5/10

-1/2

pickupchik [31]3 years ago

3 0

(3-(-2))/(-4-6)= -1/2 hope that help

You might be interested in

Let sin A= -3/5 with 270°< A< 360°. Find the following. sin A/2

GREYUIT [131]

Answer:

1/sqrt10

Step-by-step explanation:

1) Find out cosA using formula (cosA)^2+(sinA)^2=1

The module of cosA= sqrt (1- (-3/5)^2)= sqrt 16/25=4/5

So cosA=-4/5 or cosA=4/5.

Due to the condition 270degrees< A<360 degrees, 0<cosA<1 that's why cosA=4/5.

2) Find sinA/2 using a formula cosA= 1-2sinA/2*sinA/2 where cosA=4/5.

(sinA/2)^2= 0.1

sinA= sqrt 0.1= 1/ sqrt10 or sinA= - sqrt 0.1= -1/sqrt10

But 270°< A< 360°, then 270/2°<A/2<360/2°

135°<A/2<180°, so sinA/2 must be positive and the only correct answer is

sin A/2= 1/sqrt10

6 0

3 years ago

In its first year of operation, Big Lake Camp had 40 campers. This year’s attendance was 180% of the first year’s. How many camp

Murrr4er [49]

Answer:

72 campers participated this year

Step-by-step explanation:

Percentages

To determine x% of y, we must calculate x*y/100.

The number of campers that participated in Big Lake Camp during its first year was 40. We now want to know the number of campers that participated this year, if the attendance was 180% of the first year's.

The new attendance is calculated as follows:

40 * 180 / 100 = 72

72 campers participated this year

4 0

3 years ago

Evaluate M^2 /p^2 for M = 10, N =-5p and P = -2

kondaur [170]

Answer:

The solution of $\bold{\frac{M^{2}}{P^{2}}}$ for M = 10, N = -5P and P = -2 is 25

Solution:

From question, given that the value of M is 10 and N is -5p and P is -2

We have to evaluate the value of $\frac{M^{2}}{P^{2}}$ ,

By substituting the values of M and N, we get

$\frac{M^{2}}{P^{2}} = \frac{10^{2}}{(-2)^{2}}$

Expanding $\bold{10^{2}}$ :

Here 10 is the base value and 2 is the exponent value. So the base term 10 is multiplied by itself two times.

$10^{2} = 10 \times 10 = 100$

Similarly expanding $\bold{(-2)^{2}}$ :

Here -2 is the base term and 2 is the exponent value. So the base term -2 is multiplied by itself two times.

$(-2)^{2} = -2 \times -2 = 4$

So the equation $\frac{M^{2}}{P^{2}} = \frac{10^{2}}{(-2)^{2}}$ becomes,

$\frac{M^{2}}{P^{2}} = \frac{100}{4}$

By dividing 100 by 4 , we get the result as 25

Hence the solution of $\bold{\frac{M^{2}}{P^{2}}}$ when M = 10 and P = -2 is 25

6 0

4 years ago

Prove that x-1 is a factor of x^n-1 for any positive integer n.

PolarNik [594]

Answer:

$x-1$ is a factor of $x^n - 1$

Step-by-step explanation:

$x-1$ is a factor of $x^n - 1$

We will prove this with the help of principal of mathematical induction.

For n = 1, $x-1$ is a factor $x-1$ , which is true.

Let the given statement be true for n = k that is $x-1$ is a factor of $x^k - 1$ .

Thus, $x^k - 1$ can be written equal to $y(x-1)$ , where y is an integer.

Now, we will prove that the given statement is true for n = k+1

$x^{k+1} - 1\\=(x-1)x^k + x^k - 1\\=(x-1)x^k + y(x-1)\\(x-1)(x^k + y)$

Thus, $x^k - 1$ is divisible by $x-1$ .

Hence, by principle of mathematical induction, the given statement is true for all natural numbers,n.

3 0

3 years ago

7 POINTS help help help help help help PLZZZZ!!!

pishuonlain [190]

-2 is the answer it’s very easy

8 0

3 years ago