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2 years ago

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:

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:

a=9.3, b=15, c=-1080

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

q 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 q can't be considered.

Then substitute the value of q in D(q) to find the price p:

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

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

6 0

3 years ago