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elixir [45]
2 years ago
8

Help me please :) 2(2 - x) = -3(x + 1)

Mathematics
1 answer:
kondaur [170]2 years ago
3 0
The answer is x=-7

Explanation:
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Altitudes AA1 and BB1 are drawn in acute △ABC. Prove that A1C·BC=B1C·AC
Sophie [7]

Answer:

See the attached figure which represents the problem.

As shown, AA₁ and BB₁ are the altitudes in acute △ABC.

△AA₁C is a right triangle at A₁

So, Cos x = adjacent/hypotenuse = A₁C/AC ⇒(1)

△BB₁C is a right triangle at B₁

So, Cos x = adjacent/hypotenuse = B₁C/BC ⇒(2)

From (1) and  (2)

∴  A₁C/AC = B₁C/BC

using scissors method

∴ A₁C · BC = B₁C · AC

7 0
3 years ago
Jon has 2/3 of an hour to do some training. It takes him 1/6 of an hour to run one trail. How many times can he run the trail
djverab [1.8K]
Jon will be able to run 2 trails because 1/6 time 2 equals 2/6 you can simplify it to 1/3 and 1/3 + 1/3= 2/3
4 0
3 years ago
Door
maksim [4K]

Answer:

77

Step-by-step explanation:

that is the answer i got i did a quiz with the same question not too long and i got a 100 and i putt 77 aas my answer!!!!!!!!1

6 0
3 years ago
Light in vacuum is incident on the surface of a slab of transparent material. In the vacuum the beam makes an angle of 43.8° wit
REY [17]

Answer:

Index of refraction of transparent is 1.48

Step-by-step explanation:

Snell's law states that;

n1(sinθ1) = n2(sinθ2)

Where;

n1 and n2 represent the indices of refraction for the two media, and θ1 and θ2 are the angles of incidence and refraction that the ray R makes with the normal.

In this question;

n1 = 1;

θ1 = 43.8°

θ2 = 19.3°

n2 is unknown.

Thus using Snell's law, we have;

1 x sin 43.8 = n2 x sin 19.3

n2 = (sin 43.8)/sin 19.3

n2 = 1.48

4 0
3 years ago
The look-out point of a tower is 60 feet above the ground. A man observes a fountain from the look-out point. The angle of depre
Mademuasel [1]
27.98ft

You can get this by using the tangent formula. The answer exactly expressed is 60tan25. 
5 0
3 years ago
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