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3 years ago

Explain why 323.202 is least than 323.21 even though 202 is Grater than 21

2 answers:

7 0

It is less than 324.21, because as you can see, all the numbers are equal until you get to the 0 & 1, and 1 is greater than 0

Serjik [45]3 years ago

3 0

323.21 is greater than 323.202 because of the placement in tenths, hundredths and thousandths, written out as a fraction it would be:
202/1000 versus 21/100
which one is bigger? the 21/100, so its of greater value

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In a right triangle, theta is an acute angle and sin theta = 4/9. What is the

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I’m the dollar menu millionaire

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3 years ago

Where does the circumference of each kind of triangle lie? Question 1 options: Inside the triangle. Outside the triangle. On the

stiks02 [169]

Answer:

Inside the triangle - Acute Triangle

Outside the triangle - Obtuse Triangle

On the hypotenuse - Right Triangle

Step-by-step explanation:

The circumcenter is the point where the perpendicular bisectors of a triangle intersect.

In the special case of a right triangle, the circumcenter lies exactly at the midpoint of the hypotenuse.

The circumcenter of an acute triangle lies inside the triangle.

The circumcenter of an obtuse triangle lies outside the triangle.

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3 years ago

You are visiting a rainforest, but unfortunately,your insect repellent has run out. As a result, at each second, a mosquito land

zmey [24]

Answer:

Step-by-step explanation:

Prob at each second, a mosquito lands on your neck =0.5

Prob it bites/it lands = 0.2

Prob it does not bite/it lands = 0.8

Hence in one second prob for getting a bite = $0.5(0.2) = 0.01$

i.e. 1/100 chance of getting bitten in one second

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6 0

3 years ago

PRECAL: 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

The resultant of two forces acting on a body has a magnitude of 80 pounds. The angles between the resultant and the forces are 2

kifflom [539]

The basic idea is that you form a parallelogram with those two vectors as the two different side lengths another way to see it: start at the tip of one vector and move in the same direction as the other vector (and the same length as the other vector)

With any parallelogram, the adjacent angles are supplementary

180-52-20= 108 degrees

5 0

3 years ago