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2 years ago

11

During a snowstorm, the temperature drops from 0°F to -10°F in 2 hours. On average,by how many degrees is the temperature changi

ng per hour?

1 answer:

masha68 [24]2 years ago

6 0

Answer:

-5°F

Step-by-step explanation:

As, in 2 hrs = - 10°F

For 1 hr = - 10°F/2 = - 5°F

So, answer = - 5°F

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Ir[ Fairbanks, the temperature at 6:00 AM on Sunday is - 48°F. The weather forecast

mestny [16]

Answer:

The answer is 3 days

Step-by-step explanation:

Divide 48/16=3

6 0

3 years ago

— 4x — Зу = = 9<br> 5х = 15<br> Help me now

blsea [12.9K]

Answer:

y=-7

x=3

Step-by-step explanation:

I just used a calculator. Hope this helps. :)

8 0

3 years ago

Determine whether the given vectors are orthogonal, parallel or neither. (a) u=[-3,9,6], v=[4,-12,-8,], (b) u=[1,-1,2] v=[2,-1,1

nevsk [136]

Answer:

a) $u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

b) $u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

Since the angle between the two vectors is not 0 or 180 degrees we can conclude that are either.

c) $u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

Step-by-step explanation:

For each case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

Part a

u=[-3,9,6], v=[4,-12,-8,]

The dot product on this case is:

$u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

And finally we can calculate the angle between the vectors like this:

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

Part b

u=[1,-1,2] v=[2,-1,1]

The dot product on this case is:

$u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

And finally we can calculate the angle between the vectors like this:

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

Since the angle between the two vectors is not 0 or 180 degrees we can conclude that are either.

Part c

u=[a,b,c] v=[-b,a,0]

The dot product on this case is:

$u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

5 0

3 years ago

Read 2 more answers

The angle of depression from a helicopter to a landing pad is 37 degrees. if the helicopter is 1250 feet from the ground. what i

djyliett [7]

The horizontal distance from the helicopter to the landing pad is 1658.81 feet

<em><u>Solution:</u></em>

The figure is attached below

Triangle ABC is a rightangled triangle

A helicopter is flying at point A and landing pad is at point c

Angle of depression of the helicopter is 37 degrees so angle of elevation of this helicopter from landing pad will be same as 37 degrees

The helicopter is 1250 feet from the ground

Therefore, AB = 1250 feet

To find: horizontal distance from the helicopter to the landing pad

BC is the horizontal distance from the helicopter to the landing pad

BC = ?

By the definition of tan,

$tan \theta = \frac{opposite}{adjacent}$

$tan 37 = \frac{AB}{BC}\\\\tan 37 = \frac{1250}{BC}\\\\ 0.75355 = \frac{1250}{BC}\\\\BC = \frac{1250}{0.75355}\\\\BC = 1658.81$

Thus the horizontal distance from the helicopter to the landing pad is 1658.81 feet

5 0

3 years ago

A shoe store advertises that all shoes are on sale for 25% off the regular price. find the sale price of a pair of shoes that ha

DENIUS [597]

100-25=75
find 75% of the 85

percent means parts out of 100
75%=75/100=0.75

'of' means multiply

75% of 85 means
0.75 times 85=63.75

sale price is $63.75

4 0

3 years ago