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ch4aika [34]
2 years ago
11

One of the questions I don’t get pls has to be turned in by 11

Mathematics
2 answers:
stiv31 [10]2 years ago
7 0

Answer:

6 times n is less than or equal to 30

user100 [1]2 years ago
7 0
6 times a number is less than or equal to 30
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If there is 99 apples and i take one of them how many apples are there now
Finger [1]

Answer:

98

Step-by-step explanation:

99 - 1 = 98

7 0
2 years ago
Read 2 more answers
Which system of equations can be used to find the roots of the equation 12 x 3-5x=2 x 2+x+6
kirill [66]

Answer:

Option (a) is correct.

The system of equation becomes

y=12x^3-5x\\\\ y=2x^2+x+6

Step-by-step explanation:

Given : Equation  12x^3-5x=2x^2+x+6

We have to construct a  system of equations that  can be used to find the roots of the equation 12x^3-5x=2x^2+x+6

Consider the given equation 12x^3-5x=2x^2+x+6

To construct a system of equation put both sides of the given equation equal to a same variable.

Let the variable be "y", Then the equation 12x^3-5x=2x^2+x+6

becomes,

12x^3-5x=y=2x^2+x+6

Thus, The system of equation becomes

y=12x^3-5x\\\\ y=2x^2+x+6

Option (a) is correct.

   

8 0
3 years ago
Read 2 more answers
17+8x=3x+62 if x =16
Anton [14]

X=16 would not be true because it doesn’t work. X would equal 9.

5 0
3 years ago
Subtract. use models if needed (-2x+6)-(13×-16)
galben [10]
-2X+6=-12 negative 12 (-12)
13x-16=-208 negative 208(-208)
-12--208=196
4 0
3 years ago
The owner of a bike shop that produces custom built bike frames has determined that the demand equation for bike frames is given
o-na [289]

Answer:

equilibrium point (10,340)

Step-by-step explanation:

To find the equilibrium point, equal the demand and the supply:

D(q)=S(q)\\\\-6.10q^2-5q+1000=3.2q^2+10q-80

Reorganize the terms in one side and reduce similar terms:

3.2q^2+6.1q^2+5q+10q-80-1000=0\\\\9.3q^2+15q-1080=0

that's a cuadratic equation, solve with the general formula when:

<u><em>a=9.3</em></u>, <u><em>b=15</em></u>, <u><em>c=-1080</em></u>

q_{1}=\frac{-b+\sqrt{b^{2}-4ac} }{2a}\\\\q_{2}=\frac{-b-\sqrt{b^{2}-4ac} }{2a}\\\\q_{1}=\frac{-15+\sqrt{(-15)^{2}-4(9.3)(-1080)} }{2(9.3)}\\\\q_{1}=\frac{-15+201}{18.6}\\\\q_{1}=\frac{186}{18.6}\\\\q_1=10

<em><u>q</u></em> can't be negative because it is the quantity of bike frames, so:

q_{2}=\frac{-b-\sqrt{b^{2}-4ac} }{2a}\\\\q_{2}=\frac{-15-\sqrt{(-15)^{2}-4(9.3)(-1080)} }{2(9.3)}\\\\q_{2}=\frac{-15-201}{18.6}\\\\q_{2}=\frac{-216}{18.6}\\\\

This value of <u><em>q</em></u> can't be considered.

Then substitute the value of <em><u>q</u></em> in <u><em>D(q)</em></u> to find the price <u><em>p</em></u>:

D(10) = -6.10(10)^2-5(10) + 1000\\\\D(10)=340=p

The equilibrium point (q,p) is (10,340).

6 0
3 years ago
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