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pentagon [3]
3 years ago
7

60% of what number is 21

Mathematics
2 answers:
trapecia [35]3 years ago
4 0
'of' in mathematics means multiplication
'is' in mathematics means equal

Translating 60% of what number is 21 into an equation, we get:

60% * x = 21

60%, or 60 percent, is 60 over 100 which is 60/100. This can be simplified to 3/5.

3/5 * x = 21

Multiply both sides by 5.

3/5 * 5 * x = 21 * 5

3x = 105

Divide 3 on both sides.

3x/3 = 105/3

x = 35

60% of 35 is 21.
Lerok [7]3 years ago
3 0
60% of 35 is 21 

Change the percentage (60) into a decimal by dividing it over 100:
\frac{60}{100} = 0.6

Divide 21 by the decimal:
<span>\frac{21}{0.6} = 35</span>
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Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .
kogti [31]

Answer:

The family of all prime numbers such that a^{2} + b^{2} + c^{2} -1 is a perfect square is represented by the following solution:

a is an arbitrary prime number. (1)

b = \sqrt{1 + 2\cdot a \cdot c} (2)

c is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if (x+y)^{2} = x^{2} + 2\cdot x\cdot y  + y^{2}. From statement, we must fulfill the following identity:

a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}

By Associative and Commutative properties, we can reorganize the expression as follows:

a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2} (1)

Then, we have the following system of equations:

x = a (2)

(b^{2}-1) = 2\cdot x\cdot y (3)

y = c (4)

By (2) and (4) in (3), we have the following expression:

(b^{2} - 1) = 2\cdot a \cdot c

b^{2} = 1 + 2\cdot a \cdot c

b = \sqrt{1 + 2\cdot a\cdot c}

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, a, b, c > 1. If a, b and c are prime numbers, then  2\cdot a\cdot c must be an even composite number, which means that a and c can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, b must be a natural number, which means that:

1 + 2\cdot a\cdot c \ge 4

2\cdot a \cdot c \ge 3

a\cdot c \ge \frac{3}{2}

But the lowest possible product made by two prime numbers is 2^{2} = 4. Hence, a\cdot c \ge 4.

The family of all prime numbers such that a^{2} + b^{2} + c^{2} -1 is a perfect square is represented by the following solution:

a is an arbitrary prime number. (1)

b = \sqrt{1 + 2\cdot a \cdot c} (2)

c is another arbitrary prime number. (3)

Example: a = 2, c = 2

b = \sqrt{1 + 2\cdot (2)\cdot (2)}

b = 3

4 0
3 years ago
This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex
noname [10]

L(x,y,z,\lambda)=10x+10y+2z+\lambda(5x^2+5y^2+2z^2-42)

L_x=10+10\lambda x=0\implies1+\lambda x=0

L_y=10+10\lambda y=0\implies1+\lambda y=0

L_z=2+4\lambda z=0\implies1+2\lambda z=0

L_\lambda=5x^2+5y^2+2z^2-42=0

\begin{cases}L_x=0\\L_y=0\\L_z=0\end{cases}\implies1+\lambda x=1+\lambda y=1+2\lambda z=0\implies x=y=2z

5x^2+5y^2+2z^2=5(2z)^2+5(2z)^2+2z^2=42z^2=42\implies z^2=1

z^2=1\implies z=\pm1\implies x=y=\pm2

There are two critical points, at which we have

f\left(2,2,1\right)=42\text{ (a maximum value)}

f\left(-2,-2,-1\right)=-42\text{ (a minimum value)}

3 0
3 years ago
Please help me with 1-4 and 6 and 7
Tom [10]
I could only get number one which was; 1.75
3 0
3 years ago
Order these numbers from least to greatest
pantera1 [17]

Answer:

8 5/6 -> 8.839 -> 177/20 -> 8.99

Step-by-step explanation:

8 5/6 = 8.333

8.839

177/20 = 8.85

8.99

8 0
2 years ago
Which pair of complex factors results in a real number product? 15(–15i) 3i(1 – 3i) (8 + 20i)(–8 – 20i) (4 + 7i)(4 – 7i)
photoshop1234 [79]

Answer:

(4 + 7i)(4 – 7i)

Step-by-step explanation:

This pair will produce a real answer because they are complex conjugates.

Complex conjugates ( a+bi) (a-bi)  when multiplied together form a real number ( a^2 + b^2)

6 0
3 years ago
Read 2 more answers
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