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goldenfox [79]
3 years ago
6

Which set of expressions are not equivalent?

Mathematics
1 answer:
IrinaK [193]3 years ago
3 0
Expression,
• 4(3a) - 2(4a) and 4a2
is not equivalent.

You might be interested in
What is the equation of the line that passes through the point (4, 3) and is perpendicular to the line x + y = 4?
zimovet [89]
ANSWER

y - x = - 1

EXPLANATION

The required equation passes through:

(4,3) and is perpendicular to x+y=4.

Rewrite the given equation in slope-intercept form:

y = - x + 4

The slope of this line is -1.

The slope of the required line is perpendicular to this line, so we find the negative reciprocal of this slope.

m= - \frac{1}{ - 1} = 1

The equation of the line can be found using:

y - y_1 = m(x - x_1)

We substitute the slope and point to obtain:

y - 3 = 1(x - 4)

We simplify to get:

y - 3 = x - 4

y - x = - 4 + 3

The required equation is

y - x = - 1
5 0
3 years ago
Please answer!!!! will make brainliest!!!
hammer [34]

Answer:

B

Step-by-step explanation:

1.

  • 3(x+3) - 2 < 4
  • 3x + 9 -2 < 4
  • 3x < 4-7
  • 3x < - 3
  • x < -1
  • x = (-oo, -1)

2.

  • 1 - x ≤ -1
  • - x ≤ -1 - 1
  • -x ≤ -2
  • x ≥ 2
  • x = [2, +oo)

Correct choice is B.

7 0
3 years ago
( 3 1 ​ ) 2 +3 2 =left parenthesis, start fraction, 1, divided by, 3, end fraction, right parenthesis, squared, plus, 3, squared
Fynjy0 [20]

Answer:

I guess that we have the equation:

(1/3)^2 + 3^2 =

And we want to solve it.

Here just remember that:

3^2 = 3*3 = 9

then:

(1/3)^2 = (1/3)*(1/3) = 1/9

replacing these in the expression, we get:

(1/3)^2 + 3^2 = 1/9 + 9

If we want to write this as a single fraction, we can rewrite:

1/9 + 9 = 1/9 + (9/9)*9

           = 1/9 + 81/9 = (1 + 81)/9 = 82/9

4 0
3 years ago
May 26- 1
Dafna11 [192]

Answer:

  neither exponential nor linear

Step-by-step explanation:

A linear relationship will show a common difference between adjacent value. Here, the first differences between the first three values are 5 and 33. They are not the same, so the relation is not linear.

__

An exponential relationship will show first differences that are continuously increasing or decreasing by increasing or decreasing amounts (respectively).

As we showed, the difference between May 28 and May 27 is 33. This is more than the previous difference. However, the difference between May 30 and May 31 is only 28, which is less than a previous difference.

The relation is not exponential.

__

A graph points up the non-linearity of the relation.

4 0
3 years ago
Calculus 2 master needed; stuck on evaluating the integral please show steps <img src="https://tex.z-dn.net/?f=%5Cint%20%7Bsec%2
kakasveta [241]

Answer:

\int \sec(\frac{x}{2})\tan^5({\frac{x}{2}})dx=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)+C

Step-by-step explanation:

So we have the integral:

\int \sec(\frac{x}{2})\tan^5({\frac{x}{2}})dx

First, let's use substitution to get rid of the x/2. I'm going to use the variable y. So, let y be x/2. Thus:

y=\frac{x}{2}\\dy=\frac{1}{2}dx\\2dy=dx

Therefore, the integral is:

=2\int \sec(y)\tan^5(y)dy

Now, as you had done, let's expand the tangent term. However, let's do it to the fourth. Thus:

=2\int \sec(y)\tan^4(y)\tan(y)dy

Now, we can use a variation of the trigonometric identity:

\tan^2(y)+1=\sec^2(y)

So:

\tan^2(y)=\sec^2(y)-1

Substitute this into the integral. Note that tan^4(x) is the same as (tan^2(x))^2. Thus:

=2\int \sec(y)(\tan^2(y))^2\tan(y)dy\\=2\int \sec(y)(\sec^2(y)-1)^2\tan(y)dy

Now, we can use substitution. We will use it for sec(x). Recall what the derivative of secant is. Thus:

u=\sec(y)\\du=\sec(y)\tan(y)dy

Substitute:

2\int (\sec^2(y)-1)^2(\sec(y)\tan(y))dy\\=2\int(u^2-1)^2 du

Expand the binomial:

=2\int u^4-2u^2+1du

Spilt the integral:

=2(\int u^4 du+\int-2u^2du+\int +1du)

Factor out the constant multiple:

=2(\int u^4du-2\int u^2du+\int(1)du)

Reverse Power Rule:

=2(\frac{u^{4+1}}{4+1}-2(\frac{u^{2+1}}{2+1})+(\frac{u^{0+1}}{0+1}}))

Simplify:

=2(\frac{u^5}{5}-\frac{2u^3}{3}+u)

Distribute the 2:

=\frac{2u^5}{5}-\frac{4u^3}{3}+2u

Substitute back secant for u:

=\frac{2\sec^5(y)}{5}-\frac{4\sec^3(y)}{3}+2\sec(y)

And substitute back 1/2x for y. Therefore:

=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)

And, finally, C:

=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)+C

And we're done!

7 0
3 years ago
Read 2 more answers
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