All categories

3 years ago

Find the slope between points (4, 6) and (-2, 6).

Mathematics

1 answer:

mash [69]3 years ago

7 0

The slope is 0

y2-y1/x2-x1

6-6/-2-4 = 0/-6

You might be interested in

The diameter d in inches of a rope required to lift a weight of w tons is given by the function d = 10w/ π . How much weight can

Ilia_Sergeevich [38]

Answer:

Option D) 1.26 tons

Step-by-step explanation:

we have

$d=\frac{10w}{\pi}$

where

d is the diameter of a rope in inches

w is the weight in tons

we have

$d=2(2)=4\ in$ ----> the diameter is two times the radius

substitute the value of d and solve for w

$4=\frac{10w}{\pi}$

$w=\frac{4\pi}{10}=0.4\pi\ tons$

assume

$\pi =3.14$

$0.4(3.14)=1.26\ tons$

7 0

4 years ago

Read 2 more answers

Somebody please help me Do any that you want you don’t have to do all.

Dmitrij [34]

Answer:

I don't see anything, but I can help you what is it about? If its science. I'm bad at science. If its math I can try to help you.

Step-by-step explanation:

6 0

3 years ago

Ricky types at a rate of 68 words per minute. The equation w = 68m can be used to find w, the number of words he has typed after

algol [13]

Answer:

2380

Step-by-step explanation:

3 0

3 years ago

The line goes through the ordered pair (10, 6) It has a y-intercept of -2.

VLD [36.1K]

Answer: the x-intercept would be (2,-2)

Step-by-step explanation:

6 0

3 years ago

Please dont ignore, Need help!!! Use the law of sines/cosines to find..

Ket [755]

Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

$\sin{A} = \sin{103\textdegree{}}$ ,
The opposite side of angle A $a = BC = 26$ ,
The angle C is to be found, and
The length of the side opposite to angle C $c = AB = 6$ .

$\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}$ .

$\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}$ .

$\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}$ .

Note that the inverse sine function here $\sin^{-1}()$ is also known as arcsin.

By the law of cosine,

$c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C}$ ,

where

$a$ , $b$ , and $c$ are the lengths of sides of triangle ABC, and
$\cos{C}$ is the cosine of angle C.

For triangle ABC:

$b = 21$ ,
$c = 30$ ,
The length of $a$ (segment BC) is to be found, and
The cosine of angle A is $\cos{123\textdegree}$ .

Therefore, replace C in the equation with A, and the law of cosine will become:

$a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}$ .

$\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}$ .

For triangle ABC:

$a = 14$ ,
$b = 9$ ,
$c = 6$ , and
Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

$b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}$ .

$\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}$ .

$\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree$ .

For triangle DEF:

The length of segment DF is to be found,
The length of segment EF is 9,
The sine of angle E is $\sin{64\textdegree}}$ , and
The sine of angle D is $\sin{39\textdegree}$ .

Apply the law of sine:

$\displaystyle \frac{DF}{EF} = \frac{\sin{E}}{\sin{D}}$

$\displaystyle DF = \frac{\sin{E}}{\sin{D}}\cdot EF = \frac{\sin{64\textdegree}}{39\textdegree} \times 9 = 12.9$ .

7 0

3 years ago