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3 years ago

Help help help help me , (math question)

Mathematics

1 answer:

insens350 [35]3 years ago

6 0

Answer:

Step-by-step explanation:

1.) Rounded to the nearest millionth, the number is about -

0.000001 <-- This is the millionths place, so the answer would be:

0.000007

Written as the product of a single digit and a power of ten, this number is:

7 x 10^-6

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Can someone help me with this?

Lena [83]

day 5 has the highest visits and day 3 has the lowest visits

4 0

2 years ago

What is the length of segment AB? (4 points)

Sveta_85 [38]

Answer:

10 units

Step-by-step explanation:

$\sqrt{(0-6)^{2} +(10-2)^{2} }\\\\\sqrt{(-6)^{2}+(8)^{2} } \\\\\sqrt{36+64}\\\\\sqrt{100}\\\\10$

3 0

3 years ago

Miguel bikes 23 km per hour and starts at mile 10. Gabby bies 28 km per hour and starts at mile 0. Which system of linear equati

pashok25 [27]

Answer:

y = 23x + 10

y = 28x + 0

Step-by-step explanation:

Given that:

Miguel :

Velocity = 23 km/hr

Starting position = 10 miles

Gabby:

Velocity = 28 km/hr

Starting position = 0 miles

The general form of a linear equation :

y = mx + c

In the scenario above ; the starting points corresponds to the intercept, c

The distance moved per hour (velocity) is the slope (m), which is the change in y per unit change in x

Hence ;

Miguel:

y = 23x + 10

Gabby:

y = 28x + 0

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

PLEASE HELP ME

denis23 [38]

Answer:

(1) The possible outcomes are: X = {0, 1, 2, 3}.

(2) The number of times should Hartley spin a difference of 1 is 36.

(3) The number of times should Hartley spin a difference of 0 is 24.

Step-by-step explanation:

The number of sections on the spinner is 4 labelled as {1, 2, 3, 4}.

The total number of spins for each of the spinner is, <em>n</em> = 96.

(1)

The sample space of spinning both the spinners together are:

S = {(1, 1), (1, 2), (1, 3), (1, 4)

(2, 1), (2, 2), (2, 3), (2, 4)

(3, 1), (3, 2), (3, 3), (3, 4)

(4, 1), (4, 2), (4, 3), (4, 4)}

Total = 16.

The possible outcomes are:

X = {0, 1, 2, 3}.

(2)

The sample space with the difference 1 are:

S₁ = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3)}

n (S₁) = 6

The probability of the difference 1 is:

$P(\text{Diff}=1)=\frac{n(S_{1})}{N}=\frac{6}{16}=\frac{3}{8}$

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 1 is:

$E(\text{Diff}=1)=P(\text{Diff}=1)\times n\\\\=\frac{3}{8}\times 96\\\\=36$

Thus, the number of times should Hartley spin a difference of 1 is 36.

(3)

The sample space with the difference 0 are:

S₂ = {(1, 1), (2, 2), (3, 3), (4, 4)}

n (S₂) = 4

The probability of the difference 0 is:

$P(\text{Diff}=0)=\frac{n(S_{2})}{N}=\frac{4}{16}=\frac{1}{4}$

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 0 is:

$E(\text{Diff}=0)=P(\text{Diff}=0)\times n\\\\=\frac{1}{4}\times 96\\\\=24$

Thus, the number of times should Hartley spin a difference of 0 is 24.

4 0

3 years ago