Which number is greater 0.9 or 95 %

Im trying to do IXL and I dont understand the way my teacher teaches (ironic right?)

dlinn [17]

Use the formula slope= y-y/x-x
-10-r/-2-8= 1/10
-10-r/-10= 1/10
Multiply the denominator (-10) on both sides
-10-r= -1
Add -10 to both sides
-r= 9
r= -9
Hope this helps!

4 0

1 year ago

Read 2 more answers

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

Bill and Amy want to ride their bikes from their neighborhood to school which is 14.4 kilometers away. It takes Amy 40 minutes t

ruslelena [56]

Bill and Amy travel the same distance from home to school of 14.4 kilometers. Amy takes 40 minutes of time to arrive to school which means she has a rate of 14.4 km /40 minutes or 6 meters/ second. Bill takes a total 60 minutes to arrive to school. Hence Bill's rate is equal to 14.4 km/60 minutes or 4 meters/ second.Amy's is 2 m/s faster than Bill's speed, then.

8 0

2 years ago

SOLVE EACH PROBLEM: 1. Edwin pays tax at the rate of 25% of his income. What fraction of Edwin’s income is this? 2. Vida correct

Elena L [17]

Step-by-step explanation:

1/4

90%

1.5

0.005

6/10

25%

2/10

49.5

3/4

0.4

4 0

1 year ago

45 mi/h = how many ft/s

Inga [223]

237,600 ft is 45 miles

4 0

2 years ago