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3 years ago

6

Check my answers please?? I have three questions here , just wanted to make sure I got them right , thanks (:

2 answers:

vodka [1.7K]3 years ago

6 0

Q.13 is no solution since the y intercept of the 2 equation differ but gradient is the same therefore if you were to draw a graph there wont be any intersection between the equations

the next question's answer is no solution

jekas [21]3 years ago

4 0

13 It's no solution

14. infiniteley many

15 y >= 4x + 7

only first one is wrong

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Sin^2(x/2)=sin^2x <br> Find the value of x

Veronika [31]

$\sin^2\dfrac x2=\sin^2x$
$\dfrac{1-\cos x}2=1-\cos^2x$
$1-\cos x=2-2\cos^2x$
$2\cos^2x-\cos x-1=0$
$(2\cos x+1)(\cos x-1)=0$
$\implies\begin{cases}2\cos x+1=0\\\cos x-1=0\end{cases}$ $\implies\begin{cases}\cos x=-\frac12\\\cos x=1\end{cases}$

The first case occurs in $0\le x$ for $x=\dfrac{2\pi}3$ and $x=\dfrac{4\pi}3$ . Extending the domain to account for all real $x$ , we have this happening for $x=\dfrac{2\pi}3+2n\pi$ and $\dfrac{4\pi}3+2n\pi$ , where $n\in\mathbb Z$ .

The second case occurs in $0\le x$ when $x=0$ , and extending to all reals we have $x=2n\pi$ for $n\in\mathbb Z$ , i.e. any even multiple of $\pi$ .

6 0

3 years ago

Help me please nobody seems to help me :(

podryga [215]

Answer:

A: (4, -2)

B: (-3, -3)

C: (-2, 5)

D: (-3, -3)

E: (1, -4)

F: (0, 3)

6. Option B

Step-by-step explanation:

5 0

3 years ago

Using the method of completing the square, put each circle into the form

tatiyna

Answer:

Standard form: $(x-\frac{1}{2})^2 + (y-0)^2 = 15$

Center: $(\frac{1}{2}, 0)$

Radius: $r =\sqrt{15}$

Step-by-step explanation:

The equation of a circle in the standard form is

$(x-h)^{2} + (y-k)^{2} = r^{2}$

Where the point (h, k) is the center of the circle

To transform this equation $4x^{2} -4x + 4y^{2} - 59 = 0$ this equation in the standard form we use the method of square.

First, we group similar variables

$(4x^{2} -4x) + (4y^{2}) - 59 = 0$

Divide both sides of equality by 4

$(x^{2} -x) + (y^{2}) - 14.75 = 0$

Now we complete square for variable x.

Take the coefficient "b" that accompanies the variable x and divide by 2. Then, elevate the result to the square:

$b =-1\\\\\frac{b}{2}= \frac{-1}{2}= -\frac{1}{2}\\\\(\frac{b}{2})^2= (-\frac{1}{2})^2 = \frac{1}{4}$

Now add $(\frac{b}{2})^2$ on both sides of the equality

$(x^{2} -x +\frac{1}{4}) + (y^{2}) - 14.75 = (\frac{1}{4})$

Factor the expression and simplify the independent terms

$(x-\frac{1}{2})^2 + (y^{2}) = 15$

$(x-\frac{1}{2})^2 + (y-0)^2 = 15$

Then

$h =\frac{1}{2}\\\\k=0$

and the center is $(\frac{1}{2}, 0)$

radius $r =\sqrt{15}$

3 0

3 years ago

Put the following list of numbers in ascending order. 1.2, 0, -0.4, -2.6, 0.8

NeX [460]

Answer:

-2.6, -0.4, 0, 0.8, 1.2

Step-by-step explanation:

ascending just means smallest to biggest, so start with the lowest negative and work up

5 0

3 years ago

Read 2 more answers

Find the height of the pyramid.<br> Volume

MA_775_DIABLO [31]

Answer:

5

Step-by-step explanation:

like my answer please...

3 0

3 years ago