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2 years ago

9

a family spends 1/3 of its weekly budget on rent 1/4 on food 1/8 on clothes 1/12 on entertainment and the remainder is banked ho

w much is banked if the weekly income is €846.72

1 answer:

Serjik [45]2 years ago

8 0

1/3+1/8+1/12+1/4=19/24
19/24=791.66
846.72-791.66=55.06

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Square root of 2tanxcosx-tanx=0

kobusy [5.1K]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3242555

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Solve the trigonometric equation:

$\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}$

Restriction for the solution:

$\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.$

Square both sides of (i):

$\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}$

$\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}$

Let

$\mathsf{sin\,x=t\qquad (0\le t$

So the equation becomes

$\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}$

Solving the quadratic equation:

$\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.$

$\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}$

$\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}$

You can discard the negative value for t. So the solution for (ii) is

$\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}$

Substitute back for t = sin x. Remember the restriction for x:

$\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}$

where k is an integer.

I hope this helps. =)

3 0

2 years ago

What are the solutions to the equation?<br> 7x^3=28x

11Alexandr11 [23.1K]

7x³ = 28x is our equation. We want its solutions.

When you have x and different powers, set the whole thing equal to zero.

7x³ = 28x

7x³ - 28x = 0

Now notice there's a common x in both terms. Let's factor it out.

x (7x² - 28) = 0

As 7 is a factor of 7 and 28, it too can be factored out.

x (7) (x² - 4) = 0

We can further factor x² - 4. We want a pair of numbers that multiply to 4 and whose sum is zero. The pairs are 1 and 4, 2 and 2. If we add 2 and -2 we get zero.

x (7) (x - 2) (x + 2) = 0

Now we use the Zero Product Property - if some product multiplies to zero, so do its pieces.

x = 0 -----> so x = 0

7 = 0 -----> no solution

x - 2 = 0 ----> so x = 2 after adding 2 to both sides

x + 2 = 0 ---> so = x - 2 after subtracting 2 to both sides

Thus the solutions are x = 0, x = 2, x = -2.

6 0

2 years ago

What's 7/12 minus 3/12 simplified? BTW ONLY IN 4TH GRADE MAKE IT CLEAR

dusya [7]

7/12 - 3/12 = 1/3
Simplify each fraction if possible.
Same denominator? YES!
7 - 3 = 4
Since they have the same denominator, you just subtract the numerators.
Then simplify 4/12
4/4= 1 12/4= 3
Answer: 1/3
Hope that helps!

4 0

2 years ago

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2 years ago

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