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Vika [28.1K]
3 years ago
8

Sam, Jessie, and Emma go out for burgers. The total cost is $21.39. How much will each person pay if they split the cost equally

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Mathematics
1 answer:
Nadusha1986 [10]3 years ago
4 0

Answer:

$7.13

Step-by-step explanation:

There is 3 people and they need to SPLIT the cost. Split means divide in this situation. So, divide 21.39 by 3 and you should get 7.13.

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deff fn [24]
Answer is E!
12x^2-34x+24
5 0
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How would I solve <br> -6.11 + b = 14.321
Studentka2010 [4]

Answer:

b=20.431

Step-by-step explanation:

you just add  6.11 to both sides. on the left side the -6.11 and +6.11 cancel out and on the right side you will just add 6.11 to 14.321. when you add 14.321 and 6.11 you get 20.431. therefore, b=20.431

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3 years ago
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If f (x)= x-3 and g (x)= 2x-4, find (f+g)(x)
const2013 [10]

(f∘g)(x) is equivalent to f(g(x)). We solve this problem just as we solve f(x). But since it asks us to find out f(g(x)), in f(x), each time we encounter x, we replace it with g(x).

In the above problem, f(x)=x+3.

Therefore, f(g(x))=g(x)+3.

⇒(f∘g)(x)=2x−7+3

⇒(f∘g)(x)=2x−4

Basically, write the g(x) equation where you see the x in the f(x) equation.

f∘g(x)=(g(x))+3 Replace g(x) with the equation

f∘g(x)=(2x−7)+3

f∘g(x)=2x−7+3 we just took away the parentheses

f∘g(x)=2x−4 Because the −7+3=4

This is it

g∘f(x) would be the other way around

g∘f(x)=2(x+3)−7

now you have to multiply what is inside parentheses by 2 because thats whats directly in front of them.

g∘f(x)=2x+6−7

Next, +6−7=−1

g∘f(x)=2x−1

Its a lts easier than you think!

Hope this helped

5 0
3 years ago
If vector u has its initial point at (-7, 3) and its terminal point at (5, -6), u =
attashe74 [19]

First of all, let <span>θθ</span> be some angle in <span><span>(0,π)</span><span>(0,π)</span></span>. Then

<span><span><span>θθ</span> is acute <span>⟺⟺</span> <span><span>θ<<span>π2</span></span><span>θ<<span>π2</span></span></span> <span>⟺⟺</span> <span><span>cosθ>0</span><span>cos⁡θ>0</span></span>.</span><span><span>θθ</span> is right <span>⟺⟺</span> <span><span>θ=<span>π2</span></span><span>θ=<span>π2</span></span></span> <span>⟺⟺</span> <span><span>cosθ=0</span><span>cos⁡θ=0</span></span>.</span><span><span>θθ</span> is obtuse <span>⟺⟺</span> <span><span>θ><span>π2</span></span><span>θ><span>π2</span></span></span> <span>⟺⟺</span> <span><span>cosθ<0</span><span>cos⁡θ<0</span></span>.</span></span>

Now, to see if (say) angle <span>AA</span> of the triangle <span><span>ABC</span><span>ABC</span></span> is acute/right/obtuse, we need to check whether <span><span>cos∠BAC</span><span>cos⁡∠BAC</span></span> is positive/zero/negative. But what is <span><span>cos∠BAC</span><span>cos⁡∠BAC</span></span>? It is the angle made by the vectors <span><span><span>AB</span><span>−→−</span></span><span><span>AB</span>→</span></span> and <span><span><span>AC</span><span>−→−</span></span><span><span>AC</span>→</span></span>. (When you are computing the angle at a particular vertex <span>vv</span>, you should make sure that both the vectors corresponding to the two adjacent sides have that vertex <span>vv</span> as the initial point.) We will first compute these two vectors:

<span><span><span><span>AB</span><span>−→−</span></span>=(0,0,0)−(1,2,0)=(−1,−2,0)</span><span><span><span>AB</span>→</span>=(0,0,0)−(1,2,0)=(−1,−2,0)</span></span><span><span><span><span>AC</span><span>−→−</span></span>=(−2,1,0)−(1,2,0)=(−3,−1,0)</span><span><span><span>AC</span>→</span>=(−2,1,0)−(1,2,0)=(−3,−1,0)</span></span>Therefore, the angle between these vectors is given by:<span><span><span>cos∠BAC=<span><span><span><span>AB</span><span>−→−</span></span>⋅<span><span>AC</span><span>−→−</span></span></span><span>|<span><span>AB</span><span>−→−</span></span>||<span><span>AC</span><span>−→−</span></span>|</span></span>=…</span>(1)</span><span>(1)<span>cos⁡∠BAC=<span><span><span><span>AB</span>→</span>⋅<span><span>AC</span>→</span></span><span>|<span><span>AB</span>→</span>||<span><span>AC</span>→</span>|</span></span>=…</span></span></span>Can you take it from here? From the sign of this value, you should be able to decide if angle <span>AA</span> is acute/right/obtuse.

Now, do the same procedure for the remaining two angles <span>BB</span> and <span>CC</span> as well. That should help you solve the problem.

A shortcut. Since you are not interested in the actual values of the angles, but you need only whether they are acute, obtuse or right, it is enough to compute only the sign of the numerator (the dot product between the vectors) in formula (1). The denominator is always positive.

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3 years ago
Use<br> with numbers. Use<br> with shapes.
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Answer:

Can you clarify?

Step-by-step explanation:

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