How do you solve this problem using substitution or elimination?

nekit [7.7K]

Answer:Do you have a pic of the map? It would make it easier to solve if I could see the map. Its basic area. A=base times height. If the park is 50 m long and is a square, you would do 50 times 50=2500

tep-by-step explanation:

4 0

2 years ago

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Answer this plz I dont know I think its h but IDK WHY. PLZ TELL ME THE ANSWER AND SHOW WORK PLZ. I'll give the brainliest answer

Bas_tet [7]

Your original answer is correct.If you set up a ratio table, it works. For example:
So, they amount of money she's taxed on is $55. We get this by doing 175-20.

$55 is equal to 106% because 6% is tax
Now, you want to find just 100% which is the same as no tax. To get from 106 to 11, you divide by 1.06. So, you also divide 55 by 1.06 to get $51.89
She can't go over this because she won't have enough money.

So, the answer is x≤$51.89
Hope this helps!

5 0

2 years ago

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Does this graph represent a function? Why or why not?

saw5 [17]

No because it does not pass the vertical line test,
meaning that a single vertical line could pass through multiple points on the graph.

7 0

3 years ago

PLS HELP ASAP Graph x+3y=6x

zlopas [31]

Answer:

Not sure what more you need, but here is an example of the graph on Desmos.

3 0

2 years ago

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If vector u has its initial point at (-7, 3) and its terminal point at (5, -6), u =

attashe74 [19]

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

θθ is acute ⟺⟺ θ<π2θ<π2 ⟺⟺ cosθ>0cos⁡θ>0.θθ is right ⟺⟺ θ=π2θ=π2 ⟺⟺ cosθ=0cos⁡θ=0.θθ is obtuse ⟺⟺ θ>π2θ>π2 ⟺⟺ cosθ<0cos⁡θ<0.

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

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

Now, do the same procedure for the remaining two angles BB and CC 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.

6 0

3 years ago