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3 years ago

A train traveled 30 miles at an average speed of 60 mi/h ind the time taken for the trip

Mathematics

2 answers:

Ronch [10]3 years ago

4 0

1/2 hour. 30/60 = 1/2 3 divided by 6 is 1/2 since it's not possible for it to be 1

spayn [35]3 years ago

4 0

Speed= distance / time Time = distance / speed = 30/60 = 1/2h

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Help please, I'm stuck on this :/

nlexa [21]

Do you have a calculator? you can solve it by substituting x.

y=16x^2

0: y = 16(0)^2 = 16(0) = 0

(x = 0 , y = 0)

0.5: y = 16(0.5)^2 = 16(0.25) = 4

(x = 0.5 , y = 4)

1: y = 16(1)^2 = 16(1) = 16

(x = 1 , y = 16)

1.5: y = 16(1.5)^2 = 16(2.25) = 36

(x = 1.5 , y = 36)

2: y = 16(2)^2 = 16(4) = 64

(x = 2 , y = 64)

2.5: y = 16(2.5)^2 = 16(6.25) = 100

(x = 2.5 , y = 100)

3 : y = 16(3)^2 = 16(9) = 144

(x = 3 , y = 144)

4: y = 16(4)^2 = 16(16) = 256

(x = 4 , y = 256)

if you multiply a negative number by itself, it will become positive. So, -4, -3, -2.5, -2, -1.5, -1, -0.5 will be the same as the positive 4, 3, 2.5, 2, 1.5, 1, 0.5.

I'm not sure about the pattern, but if you graph it, it'll be symmetrical across the y-axis.

8 0

3 years ago

25% of all vehicles sold at the dealership were trucks, If 32 trucks were sold, how many vehicles were sold?

SpyIntel [72]

Answer:

128 vehicles

Step-by-step explanation:

32 trucks being 25 % or 1/4 of the vehicles, you simply mulitply 32 x 4=128

4 0

3 years ago

Read 2 more answers

Please !!!!!! Help me

PtichkaEL [24]

Answer:

a. 1 b. 3/5 or 60%

Step-by-step explanation:

number of favourable events/number of total events

3x/5x = 3/5

5 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.

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

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

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