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aalyn [17]
3 years ago
7

Very important!! Solve the inequality 6r + 30 ≥ 12 or -r > 12 (the question is important for high school application c:)

Mathematics
1 answer:
Fofino [41]3 years ago
7 0

First simplify the one inequality, 6r + 30 greater than or equal to 12 just divide by 6 then subtract 30 from both sides of the inequality. You should get r is greater than or equal to -3. For this one you need to reverse the inequality symbol then divide both sides by -1 to get a positive r, and you get r is less than -12. It's not A because there's two solutions, and it's not D for the same reason. It's not C because r isn't greater than -12. In conclusion it's B because it correctly represents the solutions that make the inequality true. I hope this helps you on your high school application :)

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Please help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
lutik1710 [3]

Answer:

What is the question?

Step-by-step explanation:

7 0
3 years ago
An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t
USPshnik [31]
The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

\displaystyle\int_a^bf(x)\,\mathrm dx

Split up the interval [a,b] into n equal subintervals,

[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]

where a=x_0 and b=x_n. Each subinterval has measure (width) \dfrac{a-b}n.

Now denote the left- and right-endpoint approximations by L and R, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are \{x_0,x_1,\cdots,x_{n-1}\}. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, \{x_1,x_2,\cdots,x_n\}.

So, you have

L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)
R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)

Now let T denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)

Factoring out \dfrac12 and regrouping the terms, you have

T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)

which is equivalent to

T=\dfrac12\left(L+R)

and is the average of L and R.

So the trapezoidal approximation for your problem should be \dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2
4 0
3 years ago
Write (-5+8i)+(-5-8i) as a complex number in standard form
gizmo_the_mogwai [7]
(-5+8i)+(-5-8i) =
-5+8i-5-8i=-10
5 0
3 years ago
What is NOT a way to write 52 + 80? 50 + 2 + 80 (50 + 80) + 2 50 + 80 + 2 + 8 50 + 2 + 80 + 0
GuDViN [60]

Answer: 52-80

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
The sum of three consecutive odd numbers is 69. What are the three numbers?
n200080 [17]

Answer:

21, 23, 25

Step-by-step explanation:

Ok, so this one may seem a little tricky, so I'll try to explain it as best as I can. :)

So because they are consecutive odd numbers, that means they are odd numbers that are one after the other. This means each number is 2 apart from the next. So by using this, we know that there is a difference of 6 between the highest and lowest number. Now let's try and make an equation:

3o + 6 = 69

Now in order to get O by itself we have to subtract 6 from both sides:

3o = 63

Now we just have to divide by three on both sides:

o = 21

So now, we know that the lowest number is 21, but we have to add 2 to get 23, and then another 2 to get 25.

Hope this helps :)

5 0
3 years ago
Read 2 more answers
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