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gtnhenbr [62]
2 years ago
14

If x2 mx m is a perfect-square trinomial, which equation must be true? x2 mx m = (x – 1)2 x2 mx m = (x 1)2 x2 mx m = (x 2)2 x2 m

x m = (x 4)2
Mathematics
1 answer:
Leno4ka [110]2 years ago
6 0

An expression has numbers, variables, and mathematical operations. The equation that must be true so that x²+mx+m is a perfect square trinomial is x²+mx+m=(x+2)².

<h3>What is an Expression?</h3>

In mathematics, an expression is defined as a set of numbers, variables, and mathematical operations formed according to rules dependent on the context.

A perfect square trinomial is in the form (a+b)²=a²+b²+2ab. If we compare the perfect square trinomial x²+mx+m, we will get that the value of m should be such that it satisfies the equation m^2+2m. Since there is only one value that can satisfy this equation that is 2.

Therefore, the equation that must be true so that x²+mx+m is a perfect square trinomial is x²+mx+m=(x+2)².

Learn more about Expression:

brainly.com/question/13947055

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JulijaS [17]

Answer:

f_{avg}=\frac{1}{e-1}

Step-by-step explanation:

We are given that a function

f(x)=2lnx

We have to find the average value of function on the given interval [1,e]

Average value of function on interval [a,b] is given by

\frac{1}{b-a}\int_{a}^{b}f(x)dx

Using the formula

f_{avg}=\frac{1}{e-1}\int_{1}^{e}lnx dx

By Parts integration formula

\int(uv)dx=u\int vdx-\int(\frac{du}{dx}\int vdx)dx

u=ln x and v=dx

Apply by parts integration

f_{avg}=\frac{1}{e-1}([xlnx]^{e}_{1}-\int_{1}^{e}(\frac{1}{x}\times xdx))

f_{avg}=\frac{1}{e-1}(elne-ln1-[x]^{e}_{1})

f_{avg}=\frac{1}{e-1}(e-0-e+1)=\frac{1}{e-1}

By using property lne=1,ln 1=0

f_{avg}=\frac{1}{e-1}

8 0
3 years ago
If you borrow $225 for eight years at an interest rate of 6% how much interest will you pay ?
weeeeeb [17]

Answer:$58.55

Step-by-step explanation:


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4 years ago
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A Norman window is a window with a semi-circle on top of regular rectangular window. (See the picture.) What should be the dimen
Vikki [24]

Answer:

bottom side (a) = 3.36 ft

lateral side (b) = 4.68 ft

Step-by-step explanation:

We have to maximize the area of the window, subject to a constraint in the perimeter of the window.

If we defined a as the bottom side, and b as the lateral side, we have the area defined as:

A=A_r+A_c/2=a\cdot b+\dfrac{\pi r^2}{2}=ab+\dfrac{\pi}{2}\left (\dfrac{a}{2}\right)^2=ab+\dfrac{\pi a^2}{8}

The restriction is that the perimeter have to be 12 ft at most:

P=(a+2b)+\dfrac{\pi a}{2}=2b+a+(\dfrac{\pi}{2}) a=2b+(1+\dfrac{\pi}{2})a=12

We can express b in function of a as:

2b+(1+\dfrac{\pi}{2})a=12\\\\\\2b=12-(1+\dfrac{\pi}{2})a\\\\\\b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a

Then, the area become:

A=ab+\dfrac{\pi a^2}{8}=a(6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a)+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a^2+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}-\dfrac{\pi}{8}\right)a^2\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right)a^2

To maximize the area, we derive and equal to zero:

\dfrac{dA}{da}=6-2\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right )a=0\\\\\\6-(1-\pi/4)a=0\\\\a=\dfrac{6}{(1+\pi/4)}\approx6/1.78\approx 3.36

Then, b is:

b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a\\\\\\b=6-0.393*3.36=6-1.32\\\\b=4.68

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3 years ago
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vladimir1956 [14]
1.. Plug in f... 4 - (2 x 1).
2.. Multiply 2 x 1... 4 - 2
3.. Subtract... 4 - 2 = 2
The answer is 2
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a whole, fraction wise, is always 1, so in this case we're looking at a whole of 5/5 = 1, and she withdrew 3/5, so what remained is 2/5

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8 0
2 years ago
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