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dedylja [7]
3 years ago
6

3.2-4d=2.3d+3 people were in my area but my dog was on a dog in a few weeks lol lol I was so sorry to say that this app has no r

eason why
Mathematics
1 answer:
Keith_Richards [23]3 years ago
4 0
What are you trying to say? do you actually need help or what?
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Joseph tweets 13 times aa day. Define each variable and write an algebraic expression to describe the number of posts after any
Ivahew [28]

Let's say that x is the number of days he tweets. Now, if he tweets any number of days, we can represent all of that with just x. So out expression will look like:

13x

Hope this helps!

5 0
3 years ago
In a normal curve, roughly what percent of all cases fall within plus or minus two standard deviations
Rainbow [258]

Answer:

We know that the total area under the normal curve is 100%. According to the empirical rule of Normal distribution:

Approximately, 68% of data lies within \pm1 standard deviations of mean.

Approximately, 95% of data lies within \pm2 standard deviations of mean.

Approximately, 99.7% of data lies with \pm3 standard deviations of mean,

Therefore, in a normal curve, roughly 95% of all cases fall within plus or minus two standard deviations.



5 0
3 years ago
given T(0, 6, 3) and M(1, 4, -3) find the ordered triple that represents TM then find the magnitude of TM
ozzi

Answer: (1, -2, -6); sqrt41

Step-by-step explanation:

5 0
3 years ago
How do you find the value of 6x-3y
Andrei [34K]
6x- 3y
= 3(2x-y)

Final answer: 3(2x-y)~
7 0
4 years ago
25x^-4-99x^-2-4=0 <br> How do I solve for this?
OlgaM077 [116]

Given

25x^{-4} - 99x^{-2} - 4 = 0

consider substituting y=x^{-2} to get a proper quadratic equation,

25y^2 - 99y - 4 = 0

Solve for y ; we can factorize to get

(25y + 1) (y - 4) = 0

25y+1 = 0 \text{ or } y - 4 = 0

y = -\dfrac1{25} \text{ or }y = 4

Solve for x :

x^{-2} = -\dfrac1{25} \text{ or }x^{-2} = 4

The first equation has no real solution, since x^{-2} = \frac1{x^2} > 0 for all non-zero x. Proceeding with the second equation, we get

x^{-2} = 4 \implies x^2 = \dfrac14 \implies x = \pm\sqrt{\dfrac14} = \boxed{\pm \dfrac12}

If we want to find all complex solutions, we take i=\sqrt{-1} so that the first equation above would have led us to

x^{-2} = -\dfrac1{25} \implies x^2 = -25 \implies x = \pm\sqrt{-25} = \pm5i

8 0
2 years ago
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