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3 years ago

Question

1 answer:

8 0

45 miles per hour is equal to 45 miles per 60 minutes. Let’s make it into a ratio, 45:60, and simplify it to 3:4 by dividing each by 15. Now, we can multiply the simplified ratio by 10 to get 30:40, which translates to 30 miles in 40 minutes.

Answer: 30 miles

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I’ll give brainiest to someone who answers all of the given answers below

Papessa [141]

Answer:

Below in bold.

Step-by-step explanation:

a. Distance = speed * time

d = st

So substituting in the given values:

d = 9t where t is the time it takes to travel from the hive to the flowerbed.

The total time that the bee is inflight = 19 - 16 = 3 minutes

= 180 seconds.

So for the return flight the time taken is (180-t) seconds.

So d = 6(180 - t).

and we can find the distance from the hive to flowerbed by solving the the 2 equations highlighted.

b, d = 9t

d = 6(180 - t).

Therefore:

9t = 6(180 - t)

9t = 1080 - 6t

9t + 6t = 1080

15t = 1080

t = 72 seconds

So Distance d = 9*72

= 648 ft.

c. d = 9t.

7 0

2 years ago

A nursery owner buys 9 panes of glass to fix some damage to his greenhouse. The 9 panes cost $21..15 Unfortunately, he breaks

soldier1979 [14.2K]

$9.4

$21.15 / 9 = 2.35
$2.35 x 4 = 9.4

7 0

3 years ago

Read 2 more answers

Is y=17 a linear function

timofeeve [1]

Answer:

No.

Explanation:

The term linear function refers to two distinct but related notions. This is just one point.

8 0

4 years ago

The length of a rectangle is 6 ft longer than its width.

Elenna [48]

Answer:

112 ft square

Step-by-step explanation:

Perimeter=2l+2w

l=w+6

Perimeter=2(w+6)+2w

44=2w+12+2w

44=4w+12

32=4w

8=w

l=8+6

l=14

area=lw

area=14x8

area=112 ft square

5 0

3 years ago

a. If cosθ=−45 where θ is in quadrant 3, find sin2θ. b. If cosθ=2√2 where θ is in quadrant 1, find cos2θ. c. If sinθ=817 where θ

sesenic [268]

Answer:

Part A) $sin(2\theta)=\frac{24}{25}$

Part B) $cos(2\theta)=0$

Part C) $tan(2\theta)=-\frac{240}{161}$

Step-by-step explanation:

Part A) we have $cos(\theta)=-\frac{4}{5}$

θ is in quadrant 3 ----> the sine is negative

Find $sin(2\theta)$

we know that

$sin(2\theta)=2sin(\theta)cos(\theta)$

Remember that

$cos^{2} (\theta)+sin^{2} (\theta)=1$

substitute

$(-\frac{4}{5})^{2}+sin^{2} (\theta)=1$

$(\frac{16}{25})+sin^{2} (\theta)=1$

$sin^{2} (\theta)=1-\frac{16}{25}$

$sin^{2} (\theta)=\frac{9}{25}$

$sin(\theta)=-\frac{3}{5}$ ---> remember that the sine is negative (3 quadrant)

Find $sin(2\theta)$

we have

$cos(\theta)=-\frac{4}{5}$

$sin(\theta)=-\frac{3}{5}$

$sin(2\theta)=2sin(\theta)cos(\theta)$

substitute

$sin(2\theta)=2(-\frac{3}{5})(-\frac{4}{5})$

$sin(2\theta)=\frac{24}{25}$

Part B) we have $cos(\theta)=\frac{\sqrt{2}}{2}$

θ is in quadrant 1

Find $cos(2\theta)$

we know that

$cos(2\theta)=2cos^{2} (\theta)-1$

substitute

$cos(2\theta)=2(\frac{\sqrt{2}}{2} )^{2}-1$

$cos(2\theta)=0$

Part C) we have $sin(\theta)=\frac{8}{17}$

θ is in quadrant 2 ----> the cosine is negative

Find $tan(2\theta)$

we know that

$tan(2\theta)=\frac{2tan(\theta)}{1-tan^{2} (\theta)}$

Remember that

$cos^{2} (\theta)+sin^{2} (\theta)=1$

substitute

$cos^{2} (\theta)+(\frac{8}{17})^{2}=1$

$cos^{2} (\theta)=1-\frac{64}{289}$

$cos^{2} (\theta)=\frac{225}{289}$

$cos(\theta)=-\frac{15}{17}$

Find $tan(\theta)$

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

substitute

$tan(\theta)=\frac{(8/17)}{(-15/17)}$

$tan(\theta)=-\frac{8}{15}$

Find $tan(2\theta)$

$tan(2\theta)=\frac{2tan(\theta)}{1-tan^{2} (\theta)}$

substitute

$tan(2\theta)=\frac{2(-\frac{8}{15})}{1-(-\frac{8}{15})^{2}}$

$tan(2\theta)=\frac{(-\frac{16}{15})}{1-(\frac{64}{225})}$

$tan(2\theta)=\frac{(-\frac{16}{15})}{1-\frac{64}{225}}$

$tan(2\theta)=\frac{(-\frac{16}{15})}{\frac{161}{225}}$

$tan(2\theta)=-\frac{240}{161}$

3 0

3 years ago