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rusak2 [61]
2 years ago
12

F(x) = x2 – 2x + 5 . What is the value of f(12) ? f(12) =

Mathematics
1 answer:
V125BC [204]2 years ago
5 0
The value of f(12) is 5, f(12)=5
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Please answer quickly<br>expand and simplify: 4(5x-3y) (x-4y) -(3x-4y)(2x+3y)​
Gekata [30.6K]

Answer: 14x^2-93xy+60y^2 Hope that helps!

Step-by-step explanation:

1. Expand by distributing terms

(20x-12y)(x-4y)-(3x-4y)(2x+3y)

2. Use the Foil method:(a+b)(c+d)= ac+ad+bc+bd

20x^2-80xy-12yx+48y^2-(3x-4y)(2x+3y)

3. Use the Foil method : (a+b)(c+d)= ac+ad+bc+bd

20x^2-80xy-12yx+48y^2-(6x^2+9xy-8yx-12y^2)

4. Remove parentheses 20x^2-80xy-12yx+48y^2-6x^2-9xy+ 8yx+12y^2

5. Collect like terms (20x^2-6x^2)+(-80xy-12xy-9xy+8xy)+(48y^2+12y^2)

6. Simplify.

And your answer would be 14x^2-93xy+60y^2

5 0
3 years ago
Solve the inequality for w. w-7&lt;21
Radda [10]
Add 7 to each side of the inequality.

w < 28.
8 0
3 years ago
Look at image. reflection across x=1​
elena-14-01-66 [18.8K]

Answer:

A' would be at (0,-4)

B' would be at (-2,1)

C' would be at (-2,-3)

Step-by-step explanation:

A' would be at (0,-4)

B' would be at (-2,1)

C' would be at (-2,-3)

5 0
3 years ago
Read 2 more answers
Consider the differential equation:
Wewaii [24]

(a) Take the Laplace transform of both sides:

2y''(t)+ty'(t)-2y(t)=14

\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s

where the transform of ty'(t) comes from

L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)

This yields the linear ODE,

-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s

Divides both sides by -s:

Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}

Find the integrating factor:

\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C

Multiply both sides of the ODE by e^{3\ln|s|-s^2}=s^3e^{-s^2}:

s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}

The left side condenses into the derivative of a product:

\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}

Integrate both sides and solve for Y(s):

s^3e^{-s^2}Y(s)=7e^{-s^2}+C

Y(s)=\dfrac{7+Ce^{s^2}}{s^3}

(b) Taking the inverse transform of both sides gives

y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that \frac{7t^2}2 is one solution to the original ODE.

y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7

Substitute these into the ODE to see everything checks out:

2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14

5 0
3 years ago
A storage container for oil is in the shape of a cylinder with a diameter of 10 feet and a height of 17 feet. Which measurement
Archy [21]

Answer:

Step-by-step explanation:

use  Area=pi * r^{2}

diameter is 2 * r

so r = 5

pi * 5^{2} = 78.5398

Area * height  = volume

78.5398 * 17 = 1335.18

the cylinder holds about 1335 ft^{3}

5 0
3 years ago
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