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2 years ago

Express the prime number 19 as the difference of two squares? 19 =

Mathematics

2 answers:

Lemur [1.5K]2 years ago

6 0

Step-by-step explanation:

Hey there!

Let's take out the square of 1st ten natural numbers to find out the exact answer.

1² = 1

2² = 4

3² = 9

4² = 16

5² = 25

6² = 36

7² = 49

8² = 64

9² = 81

10² = 100

Now; You can take any of these numbers and subtract from their squares.

So, let's check 6² and 5²

36-25 = 11 which is not equal to 19

Again check for 10²-9²

100 - 81 = 19 (True value)

Therefore, 19 can be expressed as the difference of square of 10 and 9.

Hope it helps!

o-na [289]2 years ago

5 0

Answer:

The two squares are 81 and 100, and their difference is 19

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Dy/dx = 2xy^2 and y(-1) = 2 find y(2)

Anarel [89]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2887301

—————

Solve the initial value problem:

dy
——— = 2xy², y = 2, when x = – 1.
dx

Separate the variables in the equation above:

$\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}$

Integrate both sides:

$\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\
\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\
\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{y}=x^2+C_1}$

$\mathsf{\dfrac{1}{y}=-(x^2+C_1)}$

Take the reciprocal of both sides, and then you have

$\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}$

In order to find the value of C₁ , just plug in the equation above those known values for x and y, then solve it for C₁:

y = 2, when x = – 1. So,

$\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}$

$\mathsf{C_1=-\,\dfrac{3}{2}}$

Substitute that for C₁ into (i), and you have

$\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}$

So y(– 2) is

$\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: ordinary differential equation ode integration separable variables initial value problem differential integral calculus

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3 years ago

A(c−b)=d solve for c

Rainbow [258]

Answer:

The solution to this equation could not be determined.

Step-by-step explanation:

Simplifying

a(c + -1b) = d

Reorder the terms:

a(-1b + c) = d

(-1b * a + c * a) = d

(-1ab + ac) = d

Solving

-1ab + ac = d

Solving for variable 'a'.

Move all terms containing a to the left, all other terms to the right.

Combine like terms: d + -1d = 0

-1ab + ac + -1d = 0

The solution to this equation could not be determined.

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2 years ago

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joja [24]

Answer:

A) 5 + 2 =7 x 4= 28

B) 8

Step-by-step explanation:

A) 5 x 4 = 20 and 2 x 4 = 8 so you add 20 + 8 and you get 28

B) so 5 x 4 is 20 and there is 8 left over and if you do 2 x 4 you get 8

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2 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Mr Matt spent 1/4 of a day at his office on Monday and 7/10 of a day at the office on Tuesday what fraction of a day did he spen

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Answer:

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Step-by-step explanation:

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8 months ago

Express the prime number 19 as the difference of two squares? 19 =​

Express the prime number 19 as the difference of two squares? 19 =