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Evgesh-ka [11]
3 years ago
13

Please help! ! Timed !******

Mathematics
1 answer:
lord [1]3 years ago
3 0

Answer:

(fоg)(x) =  x

Step-by-step explanation:

f(x) = (x-1)/3

g(x)=3x+1

(fоg)(x) = f(g(x)) = ((3x+1)-1) / 3 = 3x / 3 = x

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What is a simplified eCarlotta thinks that 3y + 5y3 is the same as 8y4. Which statement shows that it is NOT the same?Xpression
stira [4]

Questions:

(a) What is a simplified expression for (x + y) – (x – y)?

(b) Carlotta thinks that 3y + 5y^3 is the same as 8y^4. Which statement shows that it is NOT the same?

Answer:

(x + y) - (x - y) =2y

3y + 5y^3 \ne 8y^4

Step-by-step explanation:

Solving (a):

Given

(x + y) - (x - y)

Required

Simplify

(x + y) - (x - y)

Open brackets

(x + y) - (x - y) = x + y - x + y

Collect Like Terms

(x + y) - (x - y) = x - x+ y  + y

(x + y) - (x - y) = y  + y

(x + y) - (x - y) =2y

Solving (b):

Given

3y + 5y^3 and 8y^4

Required

Which expression shows they are not the same

The expression that shows this is:

3y + 5y^3 \ne 8y^4

Take for instance; y=2

Substitute 2 for y in 3y + 5y^3 \ne 8y^4

3*2 + 5*2^3 \ne 8*2^4

Evaluate all exponents

9 + 5*8 \ne 8*16

9 + 40 \ne 128

49 \ne 128

The above expression supports 3y + 5y^3 \ne 8y^4

5 0
3 years ago
√(25 - 10 √(3) + 3)<br> (not as a decimal)
nikitadnepr [17]

Answer:

\huge\boxed{\sqrt{25-10\sqrt3+3}=5-\sqrt3}

Step-by-step explanation:

\sqrt{25-10\sqrt3+3}=(*)\\\\25=5^2\\3=(\sqrt3)^2\\10\sqrt3=(2)(5)(\sqrt3)\\\\(*)=\sqrt{5^2-(2)(5)(\sqrt3)+(\sqrt3)^2}\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=\sqrt{(5-\sqrt3)^2}\\\\\text{use}\ \sqrt{a^2}=|a|\\\\=|5-\sqrt3|=5-\sqrt3\\\\^{\text{because}\ 5>\sqrt3\to5-\sqrt3>0}

6 0
3 years ago
Find the volume.<br> PLEASE HELP I WILL GIVE BRAINLIEST TO THE RIGHT ANSWER
Lana71 [14]
The answer is 1,944. the formula of volume is length times width times height
6 0
3 years ago
PRECAL:<br> Having trouble on this review, need some help.
ra1l [238]

1. As you can tell from the function definition and plot, there's a discontinuity at x = -2. But in the limit from either side of x = -2, f(x) is approaching the value at the empty circle:

\displaystyle \lim_{x\to-2}f(x) = \lim_{x\to-2}(x-2) = -2-2 = \boxed{-4}

Basically, since x is approaching -2, we are talking about values of x such x ≠ 2. Then we can compute the limit by taking the expression from the definition of f(x) using that x ≠ 2.

2. f(x) is continuous at x = -1, so the limit can be computed directly again:

\displaystyle \lim_{x\to-1} f(x) = \lim_{x\to-1}(x-2) = -1-2=\boxed{-3}

3. Using the same reasoning as in (1), the limit would be the value of f(x) at the empty circle in the graph. So

\displaystyle \lim_{x\to-2}f(x) = \boxed{-1}

4. Your answer is correct; the limit doesn't exist because there is a jump discontinuity. f(x) approaches two different values depending on which direction x is approaching 2.

5. It's a bit difficult to see, but it looks like x is approaching 2 from above/from the right, in which case

\displaystyle \lim_{x\to2^+}f(x) = \boxed{0}

When x approaches 2 from above, we assume x > 2. And according to the plot, we have f(x) = 0 whenever x > 2.

6. It should be rather clear from the plot that

\displaystyle \lim_{x\to0}f(x) = \lim_{x\to0}(\sin(x)+3) = \sin(0) + 3 = \boxed{3}

because sin(x) + 3 is continuous at x = 0. On the other hand, the limit at infinity doesn't exist because sin(x) oscillates between -1 and 1 forever, never landing on a single finite value.

For 7-8, divide through each term by the largest power of x in the expression:

7. Divide through by x². Every remaining rational term will converge to 0.

\displaystyle \lim_{x\to\infty}\frac{x^2+x-12}{2x^2-5x-3} = \lim_{x\to\infty}\frac{1+\frac1x-\frac{12}{x^2}}{2-\frac5x-\frac3{x^2}}=\boxed{\frac12}

8. Divide through by x² again:

\displaystyle \lim_{x\to-\infty}\frac{x+3}{x^2+x-12} = \lim_{x\to-\infty}\frac{\frac1x+\frac3{x^2}}{1+\frac1x-\frac{12}{x^2}} = \frac01 = \boxed{0}

9. Factorize the numerator and denominator. Then bearing in mind that "x is approaching 6" means x ≠ 6, we can cancel a factor of x - 6:

\displaystyle \lim_{x\to6}\frac{2x^2-12x}{x^2-4x-12}=\lim_{x\to6}\frac{2x(x-6)}{(x+2)(x-6)} = \lim_{x\to6}\frac{2x}{x+2} = \frac{2\times6}{6+2}=\boxed{\frac32}

10. Factorize the numerator and simplify:

\dfrac{-2x^2+2}{x+1} = -2 \times \dfrac{x^2-1}{x+1} = -2 \times \dfrac{(x+1)(x-1)}{x+1} = -2(x-1) = -2x+2

where the last equality holds because x is approaching +∞, so we can assume x ≠ -1. Then the limit is

\displaystyle \lim_{x\to\infty} \frac{-2x^2+2}{x+1} = \lim_{x\to\infty} (-2x+2) = \boxed{-\infty}

6 0
2 years ago
What is. the fraction1/5 of 80?
Yuliya22 [10]
\frac{1}{5} \times 80=\frac{80}{5}=\frac{16 \times 5}{5}=16
5 0
3 years ago
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