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2 years ago

Five times the sum of a number and 3 equals 9

Mathematics

1 answer:

jasenka [17]2 years ago

5 0

Answer:

5(x+3)=9

5x+15=9

5x=-6

x=-6/5

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64/72 in simplest form

Stells [14]

64=2^6, 72=2^3*3^2

gcd (64,72)=8
Therefore 64/8 =8
72/8=9
gcd (8,9) =1
8,9 are relatively prime.
Thus, 64/72=8/9

4 0

2 years ago

Prove: cot(x) sec^4(x) = cot(x) +2tan(x) + tan^3(x)

masha68 [24]

$\bf cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\qquad\qquad 
sec(\theta)=\cfrac{1}{cos(\theta)}\quad\qquad 1+tan^2(\theta)=sec^2(\theta)\\\\\\
tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}
\qquad \qquad cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\\\\
-------------------------------\\\\
cot(x)sec^4(x)=cot(x)+2tan(x)+tan^3(x)\\\\
-------------------------------\\\\$

$\bf \textit{so, let's do the left-hand-side}\\\\
cot(x)sec^2(x)sec^2(x)\implies cot(x)[1+tan^2(x)][1+tan^2(x)]
\\\\\\
cot(x)[1^2+2tan^2(x)+tan^4(x)]
\\\\\\
cot(x)+2tan^2(x)cot(x)+tan^4(x)cot(x)
\\\\\\
cot(x)+2\cdot \cfrac{sin^2(x)}{cos^2(x)}\cdot \cfrac{cos(x)}{sin(x)}+\cfrac{sin^4(x)}{cos^4(x)}\cdot \cfrac{cos(x)}{sin(x)}
\\\\\\
cot(x)+2\cdot \cfrac{sin(x)}{cos(x)}+\cfrac{sin^3(x)}{cos^3(x)}\implies \boxed{cot(x)+2tan(x)+tan^3(x)}$

8 0

2 years ago

In the era just prior to the Reformation, which of the following is true about the Catholic Church?

vovikov84 [41]

The answer is That it had a peaceful relationship with the protestants

3 0

2 years ago

Read 2 more answers

Evaluate.

mafiozo [28]

Answer:

-539.25

Step-by-step explanation:

(w^2x−3)÷10⋅z

w=−9, x = 2.7, and z=−25

((-9)^2*2.7−3)÷10⋅(-25)

Parentheses first

The exponent in the parentheses

(81*2.7−3)÷10⋅(-25)

Then multiply

(218.7−3)÷10⋅(-25)

Then subtract

(215.7)÷10⋅(-25)

Now multiply and divide from left to right

21.57*(-25)

-539.25

8 0

2 years ago

Read 2 more answers

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: <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>

7 0

3 years ago