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Fed [463]
3 years ago
13

Please help and explain I'll give u 12pts

Mathematics
1 answer:
White raven [17]3 years ago
7 0

-7x² + 70x - 175

= -7(x² - 10x + 25)

= -7(x - 5)²

= -7(x + -5)²

Answer: -5

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let ∅ be the angle formed in the first quadrant such that its y coordinate is twice its x coordinate. 

The lengths are denoted 2a and a.

Joining the point (0, 0) to the point described, in the first quadrant, we have a right triangle with side lengths (2a, a,  \sqrt{5}a ), where \sqrt{5}a is the hypotenuse, found by the Pythagorean theorem.

sin∅=opposite side/hypotenuse= \frac{2a}{ \sqrt{5}a }= \frac{2}{\sqrt{5}}= \frac{2\sqrt{5}}{5}

Now consider the reflection of the red line segment with respect to the x-axis, the ratio of the distances described still holds. Since here we are in the fourth quadrant, the sine is negative, so sin is -\frac{2\sqrt{5}}{5}.



Answer: {{\frac{2\sqrt{5}}{5}, -\frac{2\sqrt{5}}{5}}}

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The vertical angle to the top of a flagpole from point A on the ground is observed to be 37°11'. The observer walks 17 m directl
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Answer:

22m

Step-by-step explanation:

Let height of flagpole=h

AB==17 m

\angle CAD=37^{\circ}11'=37+\frac{11}{60}=37.183^{\circ}(1 degree= 60 minute)

\angle B=25^{\circ}43'=25+\frac{43}{60}=25.72^{\circ}

We have to find the approximate height of the flagpole.

In triangle CDA,

\frac{CD}{DA}=tan\theta=\frac{Perpendicular\;side}{Base}

\frac{h}{DA}=tan37.183^{\circ}

h=DA(0.759)

In triangle CDB,

tan 25.72^{\circ}=\frac{CD}{DB}

0.482=\frac{h}{DA+17}

0.482DA+8.194=h

Substitute the value

0.482DA+8.194=0.759DA

8.194=0.759DA-0.482DA

8.194=0.277DA

DA=\frac{8.194}{0.277}=29.58

Substitute the value

h=29.58 \times 0.759=22.45 m\approx 22m

Hence, the height of the flagpole=22 m

3 0
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