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labwork [276]
3 years ago
12

C+1/c-2 =4/7 these are both fractions can you please show your work

Mathematics
1 answer:
vladimir1956 [14]3 years ago
5 0
Yes, I can! If you have any more questions, please ask!

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I need help please ?!!!
andrew11 [14]

Answer:

B

Step-by-step explanation:

8 0
3 years ago
The least common denominator of 3/4 and 1/6 is 24. true or false
Triss [41]

Answer:False


Step-by-step explanation:

4: 4, 8, 12

6: 6, 12

The least common denominator is 12.


3 0
3 years ago
Read 2 more answers
The sum of Cathy's and Tony's ages is 65. Eight years ago, Cathy's age was 2/5 of Tony's age. How old are they now?
Viktor [21]
Let
x------> Cathy's<span> ages
</span>y------> Tony's ages

we know that
x+y=65-----> x=65-y-----> equation 1
y-8=(2/5)*(x-8)-----> equation 2

substitute equation 1 in equation 2
y-8=(2/5)*(65-y-8)---> y-8=22.8-0.4y----> y+0.4y=22.8+8
1.4y=30.8-----> y=22
x=65-y-----> x=65-22----> x=43

the answer is
Cathy's ages is 43 years 
Tony's ages is 22 years

7 0
3 years ago
Which definite integral approximation formula is this: the integral from a to b of f(x)dx ≈ (b-a)/n * [<img src="https://tex.z-d
Stella [2.4K]

The answer is most likely A.

The integration interval [<em>a</em>, <em>b</em>] is split up into <em>n</em> subintervals of equal length (so each subinterval has width (<em>b</em> - <em>a</em>)/<em>n</em>, same as the coefficient of the sum of <em>y</em> terms) and approximated by the area of <em>n</em> rectangles with base (<em>b</em> - <em>a</em>)/<em>n</em> and height <em>y</em>.

<em>n</em> subintervals require <em>n</em> + 1 points, with

<em>x</em>₀ = <em>a</em>

<em>x</em>₁ = <em>a</em> + (<em>b</em> - <em>a</em>)/<em>n</em>

<em>x</em>₂ = <em>a</em> + 2(<em>b</em> - <em>a</em>)/<em>n</em>

and so on up to the last point <em>x</em> = <em>b</em>. The right endpoints are <em>x</em>₁, <em>x</em>₂, … etc. and the height of each rectangle are the corresponding <em>y </em>'s at these endpoints. Then you get the formula as given in the photo.

• "Average rate of change" isn't really relevant here. The AROC of a function <em>G(x)</em> continuous* over an interval [<em>a</em>, <em>b</em>] is equal to the slope of the secant line through <em>x</em> = <em>a</em> and <em>x</em> = <em>b</em>, i.e. the value of the difference quotient

(<em>G(b)</em> - <em>G(a)</em> ) / (<em>b</em> - <em>a</em>)

If <em>G(x)</em> happens to be the antiderivative of a function <em>g(x)</em>, then this is the same as the average value of <em>g(x)</em> on the same interval,

g_{\rm ave}=\dfrac{G(b)-G(a)}{b-a}=\dfrac1{b-a}\displaystyle\int_a^b g(x)\,\mathrm dx

(* I'm actually not totally sure that continuity is necessary for the AROC to exist; I've asked this question before without getting a particularly satisfying answer.)

• "Trapezoidal rule" doesn't apply here. Split up [<em>a</em>, <em>b</em>] into <em>n</em> subintervals of equal width (<em>b</em> - <em>a</em>)/<em>n</em>. Over the first subinterval, the area of a trapezoid with "bases" <em>y</em>₀ and <em>y</em>₁ and "height" (<em>b</em> - <em>a</em>)/<em>n</em> is

(<em>y</em>₀ + <em>y</em>₁) (<em>b</em> - <em>a</em>)/<em>n</em>

but <em>y</em>₀ is clearly missing in the sum, and also the next term in the sum would be

(<em>y</em>₁ + <em>y</em>₂) (<em>b</em> - <em>a</em>)/<em>n</em>

the sum of these two areas would reduce to

(<em>b</em> - <em>a</em>)/<em>n</em> = (<em>y</em>₀ + <u>2</u> <em>y</em>₁ + <em>y</em>₂)

which would mean all the terms in-between would need to be doubled as well to get

\displaystyle\int_a^b f(x)\,\mathrm dx\approx\frac{b-a}n\left(y_0+2y_1+2y_2+\cdots+2y_{n-1}+y_n\right)

7 0
3 years ago
What is the solution to the system of equations?<br> O (-8,-4)<br> O(-4,-8)<br> O (44)<br> 0 (4-4)
Novay_Z [31]

Answer:

Step-by-step explanation:

I'm pretty sure the answer is C sorry if I'm wrong

3 0
3 years ago
Read 2 more answers
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