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3 years ago

Add the product l(m-n) , m (m-n) , n(n-l)

Mathematics

1 answer:

DochEvi [55]3 years ago

4 0

Step-by-step explanation:

l(m-n)+m(m-n)+n(n-l)

lm-ln+m²-mn+n²-nl

lm-2ln+m²+n²-mn

You might be interested in

F(x)=2x^(2)-5x-20 find f(-9)

Fed [463]

Answer:

Step-by-step explanation:

$f(x) = 2 {x}^{2} - 5x - 20$

To find f(-9) , substitute the value of x that's - 9 into f (x). That is for every x in f(x) replace it with - 9

We have

$f( - 9) = 2 ({ - 9})^{2} - 5( - 9) - 20 \\ = 2(81) + 45 - 20 \\ = 162 + 25 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ = 187 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

We have the final answer as

Hope this helps you

7 0

3 years ago

Read 2 more answers

Numbers 1, and 3 please

Ronch [10]

Answer:

1. shifts the graph right 2 units

2. y = -2(x -3)² +7

Step-by-step explanation:

1) Replacing x with x-h in any function shifts the graph h units to the right. Here, you have replaced x with (x-2), so the graph will be shifted 2 units to the right.

3) The vertex form of the equation of a parabola is ...

y = a(x -h)² +k . . . . . . . . for vertex (h, k) and vertical scale factor 'a'

Here, the vertex is (h, k) = (3, 7), and the parabola opens downward. This tells us the sign of 'a' is negative.

The graph is not so clear that it is easy to read the value of 'a' directly from it, but there are several clues.

The zeros of the above function are found at h±√(k/a). This graph shows the zeros to be located such that √(7/a) is slightly less than 2. This means the magnitude of 'a' will be slightly more than 7/2² = 1.75. The y-intercept of the function is 7-9a. It is less than -7, but probably more than -14. This puts bounds on 'a':

-14 < 7-9a < -7

-21/9 < -a < -14/9 ⇒ -2.33 < -a < -1.56

If we assume that 'a' is an integer value, we have bounded its magnitude as being between 1.75 and 2.33, so a=-2 is a reasonable choice.

The equation of the graph may be ...

y = -2(x -3)² +7

8 0

3 years ago

A scale for a scale drawing is 10cm:1mm. which is larger the actual object or drawing explain

Bogdan [553]

If I'm understanding your question, the drawing is 10cm and for every 10cm, the drawing is etched 1mm. This means that no matter how big the drawing gets, the actual object will always be bigger since it's 10cm larger always than the drawing. Hope this helped!

4 0

3 years ago

If the integral of the product of x squared and e raised to the negative 4 times x power, dx equals the product of negative 1 ov

Nataly_w [17]

Answer:

$A + B + E = 32$

Step-by-step explanation:

Given

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

Required

Find $A +B + E$

We have:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

Using integration by parts

$\int {u} \, dv = uv - \int vdu$

Where

$u = x^2$ and $dv = e^{-4x}dx$

Solve for du (differentiate u)

$du = 2x\ dx$

Solve for v (integrate dv)

$v = -\frac{1}{4}e^{-4x}$

So, we have:

$\int {u} \, dv = uv - \int vdu$

$\int\limits {x^2\cdot e^{-4x}} \, dx = x^2 *-\frac{1}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x} 2xdx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} - \int -\frac{1}{2}e^{-4x} xdx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx$

-----------------------------------------------------------------------

Solving

$\int xe^{-4x} dx$

Integration by parts

$u = x$ ---- $du = dx$

$dv = e^{-4x}dx$ ---------- $v = -\frac{1}{4}e^{-4x}$

So:

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x}\ dx$

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} + \int e^{-4x}\ dx$

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}$

So, we have:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} [ -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}]$

Open bracket

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} -\frac{x}{8}e^{-4x} -\frac{1}{8}e^{-4x}$

Factor out $e^{-4x}$

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{x^2}{4} -\frac{x}{8} -\frac{1}{8}]e^{-4x}$

Rewrite as:

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{4}x^2 -\frac{1}{8}x -\frac{1}{8}]e^{-4x}$

Recall that:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{64}Ax^2 -\frac{1}{64} Bx -\frac{1}{64} E]Ce^{-4x}$

By comparison:

$-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2$

$-\frac{1}{8}x = -\frac{1}{64}Bx$

$-\frac{1}{8} = -\frac{1}{64}E$

Solve A, B and C

$-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2$

Divide by $-x^2$

$\frac{1}{4} = \frac{1}{64}A$

Multiply by 64

$64 * \frac{1}{4} = A$

$A =16$

$-\frac{1}{8}x = -\frac{1}{64}Bx$

Divide by $-x$

$\frac{1}{8} = \frac{1}{64}B$

Multiply by 64

$64 * \frac{1}{8} = \frac{1}{64}B*64$

$B = 8$

$-\frac{1}{8} = -\frac{1}{64}E$

Multiply by -64

$-64 * -\frac{1}{8} = -\frac{1}{64}E * -64$

$E = 8$

So:

$A + B + E = 16 +8+8$

$A + B + E = 32$

4 0

3 years ago

The number of bacteria in a certain population increases according to an exponential growth model, with a growth rate of 3.5% pe

uysha [10]

Generic exponential growth model: y = Ao[1+r]^t

In this case: r = 3.5% = 0.035

y = 2Ao .....[the double of the initial value]

Then: 2Ao =Ao (1 + 0.035)^t

(1.035)^t =2

Take logarithm to both sides

t ln(1.035) = ln(2)

t = ln(2) / ln(1.035) = 0.693 / 0.0344 = 20.15

Answer: 20.15 hours.

3 0

4 years ago

Add the product l(m-n) , m (m-n) , n(n-l)​

Add the product l(m-n) , m (m-n) , n(n-l)