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mamaluj [8]
3 years ago
7

Solve the equationlog4x=2

Mathematics
1 answer:
Inessa05 [86]3 years ago
4 0
Change it to exponential form:

log₄(x) = 2   ⇒   4² = x = 16

Final answer: 16
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Find the length of a third side of a right triangle
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The sides of a right triangle can be found using the Pythagorean Theorem: a^2+b^2=c^2

c is the hypotenuse (the longest side) of the right triangle and a and b are the two other legs. Knowing any two sides of the right triangle, you can find the other using this formula.

Hope this helps!
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What is the surface area of a cube that has edges 3 cm long
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9cm^2

Step-by-step explanation:

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Given: -7x<-21 choose the solution set
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Answer:

x > 3

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The only thing you need to do is just divide both sides by -7

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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

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Worth 20 pts
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The answer would be 15 but i am not sure what the problem says her answer is. if it is not 15 then her answer is wrong
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