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kifflom [539]
3 years ago
8

A LOT OF POINTS!!!!!!! PLEASE HELP ME BECAUSE I DONT GET!!! Explain why 9(t - 1) is equivalent to 9t - 9 PLEASE TELL ME THE CORR

ECT ANSWER AND WHY STEP BY STEP!!! WHOEVER DOES THIS WILL GET THE BRAINLIEST!!!
Mathematics
1 answer:
Misha Larkins [42]3 years ago
4 0
Easy there on the caps, partner.

The nine, using the distributive property, distributes itself across the brackets, and so 9 times t is 9t and 9 times -1 is -9.

Therefore, 9(t-1) = 9t -9
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A flea is jumping around on the number line. He starts at 3 and jumps 5 units to the left. Where is he now on the number line
weeeeeb [17]

Answer:

Answer. -2

Step-by-step explanation:

hope this helps!!

5 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
Help with this question? thx
Sedbober [7]
EXPLANATION:
1.25 + 0.8 + 2.75 = $4.8

ANSWER: $4.8
6 0
3 years ago
If you start with 85 milligrams of Chromium 51, used to track red blood cells, which
zysi [14]

About 92 days are taken for 90 % of the material to <em>decay</em>.

The mass of radioisotopes (m), measured in milligrams, decreases exponentially in time (t), measured in days. The model that represents such decrease is described below:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} } (1)

Where:

  • m_{o} - Initial mass, in milligrams.
  • m(t) - Current mass, in milligrams.
  • \tau - Time constant, in days.

In addition, the time constant is defined in terms of half-life (t_{1/2}), in days:

\tau = \frac{t_{1/2}}{\ln 2} (2)

If we know that m_{o} = 85\,mg, t_{1/2} = 27.7\,d and m(t) = 8.5\,mg, then the time required for decaying is:

\tau = \frac{t_{1/2}}{\ln 2}

\tau = \frac{27.7\,d}{\ln 2}

\tau \approx 39.963\,d

t = -\tau \cdot \ln \frac{m(t)}{m_{o}}

t = -(39.963\,d)\cdot \ln \frac{8.5\,mg}{85\,mg}

t\approx 92.018\,d

About 92 days are taken for 90 % of the material to <em>decay</em>.

We kindly invite to check this question on half-life: brainly.com/question/24710827

8 0
3 years ago
Help me to solve this questions please.
Yuki888 [10]
I think it is 76 ....... I may be wrong
8 0
3 years ago
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