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1 year ago

PLEASE HELP 100 POINTS write an equation of the horizontal line that passes through (-7,-10)

Mathematics

2 answers:

Vedmedyk [2.9K]1 year ago

5 0

Answer:

y + 10 = 0

Step-by-step explanation:

Slope of horizontal line is zero (0)

m = 0

By point slope form of a line.

$y-y_1 =m(x - x_1) \\ \\ y - ( - 10) = 0 \{x - ( - 7) \} \\ \\ y + 10 = 0 \{x + 7\} \\ \\y + 10 = 0 \\ \\$

kotegsom [21]1 year ago

3 0

Answer:

y = -10

Step-by-step explanation:

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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

2 years ago

Pls help DUE TODAY (13points)

Brut [27]

Answer:

42 ft²

Step-by-step explanation:

Rectangle Area: L * W

$\bf A=(8ft)(6ft)$

$\bf 48 \; ft^2$

$\bf A= (3ft)(2ft)$

$\bf 6\;ft^2$

$\bf 48ft^2-6ft^2$

$\boxed{\bf 42\; ft^2}$

8 0

1 year ago

Read 2 more answers

Find the difference <br><br><br> 15 - (-15)

Yuki888 [10]

Answer:

Step-by-step explanation:

Note that, when two negative signs are directly next to each other, they mean addition. Note the rule:

A positive sign & A positive sign = Positive number.

A positive sign & A negative sign = Negative number.

A negative sign & A negative sign = Positive number.

In this case:

- (-15) = + 15

Simplify. Combine like terms:

15 + 15 = 30

30 is your answer.

8 0

2 years ago

Find four numbers that form a geometric progression such that the third term is greater than the first by 12 and the fourth is g

olchik [2.2K]

If $x$ is the first number in the progression, and $r$ is the common ratio between consecutive terms, then the first four terms in the progression are