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Anna71 [15]
4 years ago
6

Explain why it is helpful to understand the identity property of multiplication

Mathematics
1 answer:
kiruha [24]4 years ago
7 0
The identity property, which states that any number multiplied by 1 equals that same number may seem useless and kind of funny now,is more important in higher level maths
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Looking at the top of tower A and base of tower B from points C and D, we find that ∠ACD = 60°, ∠ADC = 75° and ∠ADB = 30°. Let t
katrin2010 [14]

Answer:

\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24

Step-by-step explanation:

We can use the Law of Sines to find segment AD, which happens to be a leg of \triangle ACD and the hypotenuse of \triangle ADB.

The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is \angle CAD. The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}

Now use this value in the Law of Sines to find AD:

\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}

Recall that \sin 45^{\circ}=\frac{\sqrt{2}}{2} and \sin 60^{\circ}=\frac{\sqrt{3}}{2}:

AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}

Now that we have the length of AD, we can find the length of AB. The right triangle \triangle ADB is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio x:x\sqrt{3}:2x, where x is the side opposite to the 30 degree angle and 2x is the length of the hypotenuse.

Since AD is the hypotenuse, it must represent 2x in this ratio and since AB is the side opposite to the 30 degree angle, it must represent x in this ratio (Derive from basic trig for a right triangle and \sin 30^{\circ}=\frac{1}{2}).

Therefore, AB must be exactly half of AD:

AB=\frac{1}{2}AD,\\AB=\frac{1}{2}\cdot 50\sqrt{6},\\AB=\frac{50\sqrt{6}}{2}=\boxed{25\sqrt{6}}\approx 61.24

3 0
3 years ago
Read 2 more answers
Simplify: 2y^2/ 4y^4 • 2y^3<br><br> A) xy^2/16<br> B) 1/y^6<br> C) y^3/6<br> D) 1/ 4y^5
Natali [406]
It is D. if you use a calculator it will tell you that. branliest?
3 0
3 years ago
Given P(A)= 0.95 and P(A∩B)=0.37. Find P(B∣A)
Dmitriy789 [7]

Answer:

P(A)= 0.95  \: and  \: P(A∩B)=0.37. \\  Find  \: P(B∣A) \\ P(B∣A) =  \frac{P(A∩B)}{P(A)}  \\  =  \frac{0.37}{0.95}  \\  \therefore \: P(B∣A) = 0.389

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I need help with these questions. PLEASE HELP
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The answers are MK and PY and AH.
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A man repays a loan of ₹65000 by paying ₹400 in the first month and then increasing the payment by ₹ 300 every month how long it
White raven [17]

dude r u dumb hahahahhahaha lol

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3 years ago
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