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Anika [276]
3 years ago
15

How do you find the HCF of 15t and 6s?

Mathematics
1 answer:
earnstyle [38]3 years ago
4 0

Answer:

HCF=3

Step-by-step explanation:

Factors of 15t= 3*5*t

Factors of 6s= 2*3*s

HCF= Common factors

common factor = 3

HCF=3

You might be interested in
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
A. 0.27 and 0.50
defon

Answer:

Part 1

a. .27 < .50

b. .33 < .75

c. 4.60 > 3.89

d. .76 > .08

e. .09 < .11

f. 3.33 > 3.30

Part 2

.27(2)= .54

.50(1)= .50

.27(2)= .54 is larger than .50 because if it was money, .54 is greater than .50

Step-by-step explanation:

3 0
3 years ago
Which expression can be used to convert 80 dollars (USD) to Australian dollars (aud) 1 USD=1.0343 AUD 1 AUD 0.9668 USD
lesya692 [45]

Answer: 1\ \text{USD}\equiv 1.35\ \text{AUD}

Step-by-step explanation:

1 USD is equivalent to 1.35 AUD

Therefore, 80 USD is equivalent to

\Rightarrow 80\ \text{USD}\equiv 80\times 1.35\\\\\Rightarrow 108\ \text{AUD}

4 0
3 years ago
Order these decimals 4.8,4.4082,2.933,4.8
Ainat [17]

Answer:

acending order: 2.933,4.8,4.8,4082

descending order: 4082,4.8,4.8,2.933

6 0
3 years ago
What is the area of a hexagonal prism with a side length of 11, a height of 20, and apothem of 9.5? Note: Answer must not be fra
katrin [286]
I assume the solid is a "regular hexagonal prism", by which it means that the cross section is a regular hexagon with all sides and angles equal.
Also, the exact apothem is 9.5262... but we will adopt the given value of 9.5.

First lateral surface area
A1=6*side length*height=6*11*20=1320 units
Then the end areas
A2=2 faces * 6 triangles/face * [base * height (apothem) /2]
=2*6*11*9.5/2
=627 units

Total area=1320+627=1947 units
8 0
3 years ago
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