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GaryK [48]
4 years ago
7

What times 2 equels -108

Mathematics
2 answers:
Igoryamba4 years ago
8 0
-54 i think is the answer 
Inga [223]4 years ago
3 0
Just divide -108 by 2 and it is 54
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Reuben made $168 for 12 hours of work.
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He would have worked 8 hours

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What is the maximum number of consecutive positive integers that can be added together to create a sum less than 400?
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6 0
3 years ago
Evaluate -32 + (2-6)(10).
vitfil [10]

Answer:

-72

Step-by-step explanation:

-32 + (2-6)(10)

PEMDAS

Parenthases first

(-4)(10)

multiply together

(-40)

Then just add the -32 and since a negative plus a negative is always negative the answer is going to be negative

-32 + (-40)

= -72

3 0
3 years ago
Read 2 more answers
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
The length of the hypotenuse of a 45°-45°-90° triangle is 22. Find the length of one leg.
Shkiper50 [21]

Check the picture below.

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