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2 years ago

Use the relationship among the three angles of any triangle to solve. Two angles of a triangle have the same

Mathematics

1 answer:

borishaifa [10]2 years ago

7 0

Answer:

51°,51°,78°

Step-by-step explanation:

The sum of angles in a triangle add up to 180°

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yaroslaw [1]

Answer:

n= 11 & n = 13 is the answer

8 0

2 years ago

Read 2 more answers

Rewrite x2 − 6x + 7 = 0 in the form (x − a)2 = b, where a and b are integers, to determine the a and b values.

Setler79 [48]

Answer:

Therefore values of a and b are

$a=3\ and\ b = 2$

Step-by-step explanation:

Rewrite $x^{2}-6x+7=0$ in the form

$(x-a)^{2}=b$

where a and b are integers,

To Find:

a = ?

b = ?

Solution:

$x^{2}-6x+7=0$ ..............Given

Which can be written as

$x^{2}-6x=-7$

$(\frac{1}{2} coefficient\ of\ x)^{2}=(\frac{1}{2}\times -6)^{2}=9$

Adding half coefficient of X square on both the side we get

$x^{2}-6x+9=-7+9=2$ ...................( 1 )

By identity we have (A - B)² =A² - 2AB + B²

Therefore,

$x^{2}-6x+9=x^{2}-2\times 3\times x+3^{2}=(x-3)^{2}$

Substituting in equation 1 we get

$(x-3)^{2}=2$

Which is in the form of

$(x-a)^{2}=b$

On comparing we get

a = 3 and b = 2

Therefore values of a and b are

$a=3\ and\ b = 2$

4 0

3 years ago

Help please help I don’t know this

Ganezh [65]

Answer:

option d is the correct answer.

Step-by-step explanation:

3 0

3 years ago

General solutions of sin(x-90)+cos(x+270)=-1<br> {both 90 and 270 are in degrees}

mixer [17]

Answer:

$\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

Step-by-step explanation:

Given:

$\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1$

First, note that

$\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x$

So, the equation is

$-\cos x+\sin x= -1$

Multiply this equation by $\frac{\sqrt{2}}{2}:$

$-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}$

The general solution is

$x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

4 0

3 years ago

Wrong question???????

melomori [17]

Answer:

what do you mean ???????????

4 0

2 years ago