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lubasha [3.4K]
3 years ago
13

How is 718.012 put in expanded form and written form

Mathematics
1 answer:
adelina 88 [10]3 years ago
5 0

Answer:

567+(6673)- 718)+ 012

Step-by-step explanation:

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What is this one i cant seem to figure it out
pshichka [43]

Answer:

73

Step-by-step explanation:

Since DA is the diameter, this means that you can divide the circle into two half-circles.  You will solve with the right half-circle.  A half-circle has a 180 total degrees.  Add up the values given and solve for x.

(2x) + (7x - 4) + (85) = 180

2x + 7x - 4 + 85 = 180

9x - 4 + 85 = 180

9x + 81 = 180

9x = 99

x = 11

Now that you have the value of x, solve for m∠CZB.

m∠CZB = 7x - 4

m∠CZB = 7(11) - 4

m∠CZB = 77 - 4

m∠CZB = 73

6 0
3 years ago
When i was 18 years old, my sister was half my age. Now I am 82 years. How old is my sister now?​
iVinArrow [24]

Answer:

41 \:  \:  \: years \:  \:  \: old

Step-by-step explanation:

me \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: sis \\ 18 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: 9 \\ 82 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  x \\  \\ use \:  \: cross \: \\  multipaction \\ \\  18x = 82 \times 9 \\ 18x = 738 \\  \frac{18x}{18}  =  \frac{738}{18}  \\ x = 41

4 0
3 years ago
Read 2 more answers
How to do break even analysis​
Ksenya-84 [330]

Answer:

To calculate a break-even point based on units: Divide fixed costs by the revenue per unit minus the variable cost per unit. ...

When determining a break-even point based on sales dollars: Divide the fixed costs by the contribution margin.

7 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
If the original function () = 22 − 1 is shifted to the left 3 units to make the
Rom4ik [11]

A horizontal translation is expressed by transforming

f(x)\mapsto f(x+k)

If k is positive, the function is translated to the left. If k is negative, the function is translated to the right.

So, a 3-units left shift is given by k=3. So you have

f(x)=2x^2-1 \implies f(x+3)=2(x+3)^2-1

4 0
3 years ago
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