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frosja888 [35]
3 years ago
14

5x+10=20 what is the answer to this question

Mathematics
2 answers:
slavikrds [6]3 years ago
5 0

Answer:

X=2

Step-by-step explanation:

Snowcat [4.5K]3 years ago
4 0

Answer:

x=2

Step-by-step explanation:

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The dot plot below shows the number of days with different temperatures last month. What is the mean of the data set shown? A. 8
julia-pushkina [17]

Answer:

D. 81

Step-by-step explanation:

I think the attached photo supports for your question to be answered

Here is my answer:

Number of days: 30

Mean of the data set = (68 + 70*2 +74*2 +76*4 +78 +80*5 + 82*3 +84*5 + 88 +88*3 + 92*3) /30 = 81

3 0
3 years ago
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LA plane is trying to travel 250 miles at a bearing of 20° E of S, however, it ends 230 miles away from the
Semmy [17]

Answer:

The wind pushed the plane 65.01 miles in the direction of 45.56 ^{\circ} East of North  with respect to the destination point.

Step-by-step explanation:

Let origin, O, br the starting point and point D be the destination at 250 miles at a bearing of 20° E of S, but due to wind let D' be the actual position of the plane at 230 miles away from the  starting point in the direction of 35° E of South as shown in the figure.

So, we have |OD|=250 miles and |OD'|=230 miles.

Vector \overrightarrow{DD'} is the displacement vector of the plane pushed by the wind.

From figure, the magnitude of the required displacement vector is

|DD'|=\sqrt{|AB|^2+|PQ|^2}\;\cdots(i)

and the direction is \alpha east of north as shown in the figure,

\tan \alpha=\frac{|PQ|}{|AB|}\;\cdots(ii)

From the figure,

|AB|=|OA-OB|

\Rightarrow |AB|=|OD\cos 20 ^{\circ}-OD'\cos 35 ^{\circ}|

\Rightarrow |AB|=|250\cos 20 ^{\circ}-230\cos 35 ^{\circ}|

\Rightarrow |AB|=45.52 miles

Again, |PQ|=|OP-OQ|

\Rightarrow |PQ|=|OD\sin 20 ^{\circ}-OD'\sin 35 ^{\circ}|

\Rightarrow |PQ|=|250\sin 20 ^{\circ}-230\sin 35 ^{\circ}|

\Rightarrow |PQ|=46.42 miles

Now, from equations (i) and (ii), we have

|DD'|=\sqrt{|45.52|^2+|46.42|^2}=65.01 miles, and

\tan \alpha=\frac{|46.42|}{|45.52|}

\alpha=\tan^{-1}\left(\frac{|46.42|}{|45.52|}\right)=45.56 ^{\circ}

Hence, the wind pushed the plane 65.01 miles in the direction of 45.56 ^{\circ} E astof North  with respect to the destination point.

5 0
2 years ago
point q is located in ( -4 , 6) point r is located in ( 8,6) . what is the distance from pint q to point r
Hoochie [10]

Answer: 12

Step-by-step explanation: It’s on the same y axis and -4 to 8 is going to the right on a graph.

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3 years ago
Based on a​ survey, assume that 27​% of consumers are comfortable having drones deliver their purchases. Suppose that we want to
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.3

Step-by-step explanation:

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3 years ago
PLS HELP ILL MARK BRAINLIST !!
andreev551 [17]

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Last option it should be

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3 years ago
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