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Vlad [161]
3 years ago
10

How do you know which side is base

Mathematics
2 answers:
Dominik [7]3 years ago
8 0

Answer:

if you're talking about a triangle then it's the bottom side well if you're talking about any shape it's the side the shape is supported on like if you put a box on the ground the face that is touching the ground is the base in this case it is the face that is supporting the rest of the shape

lukranit [14]3 years ago
6 0

Answer:

you have to be more specific on what your talking about

Step-by-step explanation:

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Customers can pick their own apples at well orchards. They pay $6 to enter the orchard and $2 per pound for the apples they pick
Anvisha [2.4K]
Y=6+2x

Since y equals the total cost, it will be on its own in the equation. 6 dollars is a set amount, meaning that no matter how many apples they pick it’ll stay the same. HOWEVER, the cost will change depending on how many pound of apples they get

Every pound of apples is equal to 2 dollars, and since x stands for pounds of apples, it’ll be ( ex: x = 2, so 2x = 2(2) = 4)

So the equation will be y = 6 + 2x

I hope this helps
6 0
3 years ago
A playground is 224 square feet. How many square inches is that?
MAVERICK [17]

Answer:

32256 square inches

Step-by-step explanation:

Formula  

multiply the area value by 144

8 0
3 years ago
Read 2 more answers
Order these decimals greatest to least 0.2 ; -0.4 ; -4.5
Mice21 [21]
0.2,
-0.4,
-4.5
hope this helps
6 0
3 years ago
a tea company surveyed shoppers at two different malls. shoppers were asked whether they prefer sweetened or unsweetened tea. at
FromTheMoon [43]
First, let's create the ratios of sweetened to unsweetened. To do this, you would have to subtract the number of people that preferred unsweetened from the total.
Westside mall: 15:30 or \frac{15}{30}
Eastside mall: 13:26 or \frac{13}{26}

Now, you would divide to both ratios.
15 ÷ 30 = 0.5
13 ÷ 26 = 0.5 

They have the same sum, meaning, they are the same. 

Westside Mall shoppers are just as likely to prefer unsweetened tea as Eastside Mall shoppers. 

I hope this helps!


8 0
3 years ago
Solve the following equation by completing the square. 3x^2-3x-5=13
mr Goodwill [35]

we'll start off by grouping some

\bf 3x^2-3x-5=13\implies (3x^2-3x)-5=13\implies 3(x^2-x)-5=13 \\\\\\ 3(x^2-x)=18\implies (x^2-x)=\cfrac{18}{3}\implies (x^2-x)=6\implies (x^2-x+~?^2)=6

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?

well, let's recall that a perfect square trinomial is

\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²

\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}

6 0
3 years ago
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