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andreyandreev [35.5K]
3 years ago
14

Ecuación 2/3 x – x / 6 = 2

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
5 0

Answer:

x=4

Step-by-step explanation:

multiply both sides, colect the terms, divide,  answer!

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PLLLLLLLLLLLLLEASE PLEASE PLEASE PLEASE HELLLLLPPPP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Vika [28.1K]

Answer:

The second one (B)

Step-by-step explanation:

u don't multiply a number by the same number if u know what I mean

hope this helps <3

5 0
3 years ago
Read 2 more answers
Solve 37+z≤54. Graph the solution.
iragen [17]

Answer:

z ≤ 17

Step-by-step explanation:

Step 1: Subtract 37 from both sides.

  • 37 + z - 37\leq 54-37
  • z\leq 17

3 0
3 years ago
Read 2 more answers
PLEASE HELP ME, I DON'T UNDERSTAND THIS! :( Multiply.
photoshop1234 [79]

Hey there! I'm happy to help!

First, let's multiply the numerators. We will put q+5 in parentheses so we can multiply it by 4q.

4q(q+5)

We use the distributive property to undo the parentheses.

First, we multiply 4q by q.

4q×q=4q²

And we multiply 4q and 5.

4q×5=20q

So, our numerator right now is 4q²+20q.

Now, for the denominators.

2(q+4)

We do 2 by q.

2×q=2q

And 2×4, which is 8.

So, our denominator is 2q+8.

Right now, our fraction is \frac{4q^2+20}{2q+8}, and this is your correct answer. However, we can simplify it a bit more. We can divide the 4 by 2, q² by q, and simplify the 20 and the 8.

4/2=2

q²/q=q

20/8=5/2

Now, our final product is (2q+5)/2

But, mark down your answer as \frac{4q^2+20}{2q+8} because that is technically correct.

Have a wonderful day! :D

4 0
3 years ago
Cot^4θ-4cot^2θ-5=0 <br>solve for θ
stich3 [128]
\cot^4\theta-4\cot^2\theta-5=(\cot^2\theta-5)(\cot^2\theta+1)=0

\cot^2\theta-5=0\implies\cot^2\theta=5\implies\cot\theta=\pm\sqrt5

Recall that \cot\theta=\cot(\theta+\pi), which is to say \cot x has period \pi. This in turn means that \cot\theta=0 will have the same solutions as \cot(\theta+n\pi)=0 for any integer n. So the general solution to the first case is

\cot\theta=\pm\sqrt5\implies\theta=\cot^{-1}(\pm\sqrt5)+n\pi

where n is any integer.

On the other hand,

\cot^2\theta+1=0\implies\cot^2\theta=-1

but x^2\ge0 for any value of x, so this equation has no (real) solutions for \theta.
4 0
3 years ago
Can anyone help me with this question please
AlladinOne [14]
QUADRATIC \: \: \: RESOLUTIONS \\ \\ \\ \\Function \: to \: model \: the \: height \: of \: the \\ Grasshopper \: above \: the \: ground \: is \: \\ given \: by \: \: - \\ \\ h(t) \: \: = \: \: - {t}^{2} \: + \: \frac{4}{3} t \: + \: \frac{1}{4} \\ \\ \\ When \: the \: grasshopper \: will \: be \: at \: \\ the \: ground \: , \: \: \\ \\ h(t) \: = \: 0 \\ \\ - {t}^{2} \: + \: \frac{4}{3} t \: + \: \frac{1}{4} \: = \: 0 \\ \\ - 12{t}^{2} \: + \: 16 t \: + 3 \: = \: 0 \\ \\ \: \: \: 12 {t}^{2} \: - \: 16t \: - \: 3 \: = \: 0 \\ \\ \: \: \: 12 {t}^{2} \: + \: 2t \: - \: 18t \: - 3 \: = \: 0 \\ \\ \: \: 2t \: (6t + 1) \: - 3 \: (6t + 1) \: = \: 0 \\ \\ \: (2t - 3) \: (6t + 1) \: \: = \: \: 0 \\ \\ Neglecting \: the \: negative \: value \: \\ as \: time \: cannot \: be \: negative \: \: ,\\ \\ We \: get \: \: - \: \\ \\ t \: \: = \: \: \frac{3}{2} \\ \\ \\ Hence \: \: , \: \: After \: \: || \: \: 1.5 \: seconds \: \: || \: , \: the \: \\ Grasshopper \: will \: land \: on \: the \: ground \: .
4 0
4 years ago
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