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almond37 [142]
3 years ago
6

Use intercepts to graph the linear equation -1/2x+y=-4

Mathematics
2 answers:
ololo11 [35]3 years ago
6 0
Do you need to switch it around or do something else?
timama [110]3 years ago
3 0
In y=mx+b format the equation is y=1/2x-4. you need to mark -4 on your graph then go up one unit then over two units. then continue to go up 1 over 2 until you get your line
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22201+54633333333333
Mariulka [41]
54,633,333,355,534 I hope this helped !
5 0
3 years ago
If < A and < B are supplementary, and < A = 3x - 9 degrees, and < B = 2x + 14 degrees, find x.
Annette [7]

Answer:

Option c, 35

Step-by-step explanation:

<A and <B are supplementary,

so, <A+<B = 180

or, 3x-9+2x+14=180

or, 5x+5=180

or, 5x=175

or, x=35

Answered by GAUTHMATH

7 0
3 years ago
use Taylor's Theorem with integral remainder and the mean-value theorem for integrals to deduce Taylor's Theorem with lagrange r
Vadim26 [7]

Answer:

As consequence of the Taylor theorem with integral remainder we have that

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \int^a_x f^{(n+1)}(t)\frac{(x-t)^n}{n!}dt

If we ask that f has continuous (n+1)th derivative we can apply the mean value theorem for integrals. Then, there exists c between a and x such that

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}dt = \frac{f^{(n+1)}(c)}{n!} \int^a_x (x-t)^n d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{n+1}}{n+1}\Big|_a^x

Hence,

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{(n+1)}}{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} .

Thus,

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}

and the Taylor theorem with Lagrange remainder is

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.

Step-by-step explanation:

5 0
3 years ago
mollys jump rope is 6 1/3 feet long. Gals jump rope 4 2/3 feet lo g. How much longer is Molly's jump rope?
Vlad1618 [11]
6 1/3- 4 2/3 =5 4/3- 4 2/3=1 2/3
You can't subtract fractions unless you have the same denominator.
8 0
3 years ago
Read 2 more answers
The following table shows the estimated populations and annual growth rates for four countries in the year 2000. Find the expect
Liula [17]

Answer:

After 25 years the population will be:

  • Australia: 22271200
  • China: 1580220878
  • Mexico: 157380127
  • Zaire: 112794819

Step-by-step explanation:

Growth rate problem that has a growth rate proportional to the population size can be solved using the equation:

 P(t) = P₀eʳᵗ

  • t is your unit of time. It could be days, or hours, or minutes. It changes depending on each problem. In this problem, t is measured in years because you're jumping from 2000 to 2025. Years just makes the most sense to measure that leap in time.
  • P(t) is the population at time t. An example in this problem could be P(20) would be the population 20 years after the initial count. or maybe P(12) would be the population 12 years after the initial count. or P(0) would be the initial count of the population.
  • P₀ is the initial population at P(0)
  • r is the growth rate.<u><em> Don't forget to convert the percentage to its decimal form</em></u>

Now that everything is set out, lets use the equation to solve for our answer.

P(t) = P₀eʳᵗ

<u>Australia:</u>

P(t)=(19169000)e^{(0.006)t}

after 25 years

P(25)=(19169000)e^{(0.006)(25)}=22271200.6

<u>China:</u>

P(t) = (1261832000)e^{(.009)t}

after 25 years:

P(25)=(1261832000)e^{(.009)(25)}=1580220878

<u>Mexico:</u>

P(t) = (100350000)e^{(0.018)t}

after 25 years:

P(25)=(100350000)e^{(0.018)(25)}=157380127.8

<u>Zaire:</u>

P(t) = (51965000)e^{(0.031)t}

after 25 years:

P(25)=(51965000)e^{(0.031)(25)}=112794819.9

6 0
4 years ago
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