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3 years ago

What is the slope of this line?

Mathematics

1 answer:

algol [13]3 years ago

4 0

Answer:

the slope is -1/1 or just -1

Step-by-step explanation:

so u start at the origin or say 0 then u go right 1 and down 1

since the line is going down its gonna be a negative 1

hope this helps

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the coordinates of point t are (0,2). the midpoint of st is (1,-6) find the coordinates of point s.

sasho [114]

Question

the coordinates of point t are (0,2). the midpoint of st is (1,-6) find the coordinates of point s.

Answer:

(10,-14)

Step-by-step explanation:

0 + x

--------- = 5 Multiply both sides by 2 ----> 0 + x = 10 or x = 10

2 + y

-------- = -8 Multiply both sides by 2 ----> 2 + y = -16 or y = -14

Therefore, the coordinates of point S should be at (10,-14)

4 0

3 years ago

Mr Thompson please answer. This is hw that is for mathematics Algebra.

ddd [48]

Answer:

question 6:

$x^{2}$ + 9x + 14 and (x + 2) (x + 7)

question 7:

$x^{2}$ - 7x + 10 and (x - 2)(x - 5)

question 8:

$x^{2}$ - 9x + 20 and (x - 4)(x - 5)

question 9:

(x + 1) (x - 17)

$x^{2}$ - 17x + 1x - 17

$x^{2}$ - 16x - 17

question 10:

(x - 1) (x + 17)

$x^{2}$ + 17x - 1x - 17

$x^{2}$ + 16x - 17

6 0

3 years ago

What theorems can be used to prove the triangles are congruent?<br>SSS<br>ASA<br>AAS<br>HL

Arada [10]

Answer:

1.) SAS | 2.) HL | 3.) HL | 4.) SSS (Not too sure about this one)

Step-by-step explanation:

3 0

3 years ago

Verify by substitution whetherthe given functions are solutions of the given DE. Primes denote derivatives with respect to x.y!!

julia-pushkina [17]

Complete Question

The complete question is shown on the first uploaded

Answer:

$y_1$ is not a solution of the differential equation

$y_2$ is not a solution of the differential equation

$y_3$ is not a solution of the differential equation

Step-by-step explanation:

The differential equation given is $y'' + y' = cos2x$

Let consider the first equation to substitute

$y_1 = cosx +sinx$

$y_1' = -sinx +cosx$

$y_1'' = -cosx -sinx$

$y_1'' - y_1' = -cosx -sinx -sinx +cosx$

$y_1'' + y_1' = -2sinx$

$-2sinx \ne cos2x$

This means that $y_1$ is not a solution of the differential equation

Let consider the second equation to substitute

$y_2 = cos2x$

$y_2' = -2sin2x$

$y_2'' = -4cos2x$

$y_2'' + y_2' = -4cos2x-2sin2x$

$-4cos2x-2sin2x \ne cos2x$

This means that $y_2$ is not a solution of the differential equation

Let consider the third equation to substitute

$y_3 = sin 2x$

$y_3' = 2cos 2x$

$y_3'' = -4sin2x$

$y_3'' + y_3' = -4sin2x - 2cos2x$

$-4sin2x - 2cos2x \ne cos2x$

This means that $y_3$ is not a solution of the differential equation

6 0

3 years ago

32°

ira [324]

I think it’s C

I hope it’s right

4 0

3 years ago