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miv72 [106K]
3 years ago
6

The inequality represents the spine thickness, in centimeters, of various books on a

Mathematics
1 answer:
SOVA2 [1]3 years ago
6 0
Hdbsbe dhajejr f dhehehdudbd ridehrbr ever rvebe their Bebe
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Ifangle1∠ 1 and angle2∠ 2 are complementary, find the measure of x and the measures of the angles when LaTeX: m\angle1=\left(6x+
Liono4ka [1.6K]

Answer:

<h3>x = 5</h3>

Step-by-step explanation:

The sum of two complementary angle is 90 degrees. Given the following complementary angles;

<1 = 6x + 5

<2 = 8x + 15

<1 + <2 = 90 (complementary angles)

6x + 5 + 8x + 15 = 90

14x + 20 = 90

14x = 90-20

14x = 70

x = 70/14

x = 5

Hence the measure of x is 5

6 0
3 years ago
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
3 years ago
What is the distance between the points
Viefleur [7K]

Answer: its a

Step-by-step explanation:

14.87

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3 years ago
Which of the following would correctly fill in the blank on the following statement? 1.7 x 10__ = 0.00017
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The blank should be the power of -4, so the statement is 1.7 x 10^-4 = 0.00017.
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4 years ago
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