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Art [367]
3 years ago
8

Factorise 28ab^2-21ab^2

Mathematics
2 answers:
DanielleElmas [232]3 years ago
4 0

Answer:

7ab^2

Step-by-step explanation:

28ab^2-21ab^2=7ab^2

ra1l [238]3 years ago
4 0

The Answer is :

7ab^2

I hope u get what ur looking for and I wish u give me brainlist But I know u won't cuz everyone say that . But thank you so much if u put for me and It would be a very appreciated from you and again thank you so much

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If 5-3(2a+1)=-4a+10, what is the value of a+2​
AlekseyPX

Answer:

-2

<em>BRAINLIEST, PLEASE!</em>

Step-by-step explanation:

5 - 3(2a + 1) = -4a + 10

5 - 6a - 3 = -4a + 10

2 - 6a = -4a + 10

-2a = 8

a = -4

-4 + 2 = -2

5 0
3 years ago
Read 2 more answers
What is the distance between the points (–3, 4) and (–7, 4)?
uysha [10]

Answer:

d =  \sqrt{( - 7  + 3) {}^{2}  + (4 - 4) {}^{2} }  \\ d =  \sqrt{16}  \\ d = 4

Good luck!

Intelligent Muslim,

From Uzbekistan.

6 0
4 years ago
Consider this sphere with a diameter of 5 m.
abruzzese [7]

Answer: only C and E

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 11z on the curve of intersection of the plane x − y + z =
Taya2010 [7]

Answer:

\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}

<em>Maximum value of f=2.41</em>

Step-by-step explanation:

<u>Lagrange Multipliers</u>

It's a method to optimize (maximize or minimize) functions of more than one variable subject to equality restrictions.

Given a function of three variables f(x,y,z) and a restriction in the form of an equality g(x,y,z)=0, then we are interested in finding the values of x,y,z where both gradients are parallel, i.e.

\bigtriangledown  f=\lambda \bigtriangledown  g

for some scalar \lambda called the Lagrange multiplier.

For more than one restriction, say g(x,y,z)=0 and h(x,y,z)=0, the Lagrange condition is

\bigtriangledown  f=\lambda \bigtriangledown  g+\mu \bigtriangledown  h

The gradient of f is

\bigtriangledown  f=

Considering each variable as independent we have three equations right from the Lagrange condition, plus one for each restriction, to form a 5x5 system of equations in x,y,z,\lambda,\mu.

We have

f(x, y, z) = x + 2y + 11z\\g(x, y, z) = x - y + z -1=0\\h(x, y, z) = x^2 + y^2 -1= 0

Let's compute the partial derivatives

f_x=1\ ,f_y=2\ ,f_z=11\ \\g_x=1\ ,g_y=-1\ ,g_z=1\\h_x=2x\ ,h_y=2y\ ,h_z=0

The Lagrange condition leads to

1=\lambda (1)+\mu (2x)\\2=\lambda (-1)+\mu (2y)\\11=\lambda (1)+\mu (0)

Operating and simplifying

1=\lambda+2x\mu\\2=-\lambda +2y\mu \\\lambda=11

Replacing the value of \lambda in the two first equations, we get

1=11+2x\mu\\2=-11 +2y\mu

From the first equation

\displaystyle 2\mu=\frac{-10}{x}

Replacing into the second

\displaystyle 13=y\frac{-10}{x}

Or, equivalently

13x=-10y

Squaring

169x^2=100y^2

To solve, we use the restriction h

x^2 + y^2 = 1

Multiplying by 100

100x^2 + 100y^2 = 100

Replacing the above condition

100x^2 + 169x^2 = 100

Solving for x

\displaystyle x=\pm \frac{10}{\sqrt{269}}

We compute the values of y by solving

13x=-10y

\displaystyle y=-\frac{13x}{10}

For

\displaystyle x= \frac{10}{\sqrt{269}}

\displaystyle y= -\frac{13}{\sqrt{269}}

And for

\displaystyle x= -\frac{10}{\sqrt{269}}

\displaystyle y= \frac{13}{\sqrt{269}}

Finally, we get z using the other restriction

x - y + z = 1

Or:

z = 1-x+y

The first solution yields to

\displaystyle z = 1-\frac{10}{\sqrt{269}}-\frac{13}{\sqrt{269}}

\displaystyle z = \frac{-23\sqrt{269}+269}{269}

And the second solution gives us

\displaystyle z = 1+\frac{10}{\sqrt{269}}+\frac{13}{\sqrt{269}}

\displaystyle z = \frac{23\sqrt{269}+269}{269}

Complete first solution:

\displaystyle x= \frac{10}{\sqrt{269}}\\\\\displaystyle y= -\frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{-23\sqrt{269}+269}{269}

Replacing into f, we get

f(x,y,z)=-0.4

Complete second solution:

\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}

Replacing into f, we get

f(x,y,z)=2.4

The second solution maximizes f to 2.4

5 0
3 years ago
You deposit $3000 in an account with an APR 7.5% and continuous compounding. How much will you have after 20 years?
BartSMP [9]

Answer:

I = $4,500

Step-by-step explanation:

Simple Interest (I) = PRT / 100

Where:

P= Principal (Amount deposited) = $3000

R= Rate = 7.5%

T= Time = 20 years.

Simple Interest (I) = PRT / 100

I = ($3000 * 7.5 * 20) / 100

I = $450,000 / 100

I = $4,500.

8 0
3 years ago
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