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USPshnik [31]
2 years ago
9

Please solve for x please use statement reason

Mathematics
1 answer:
nataly862011 [7]2 years ago
8 0

AE || BC

===》 DCB angle = DAE angle

AC /

__________________________________

Thus :

DCB angle = x

And also we know that the submission of interior angles of any triangle must equals to 180 degrees so :

In BCD triangle :

B + C + D = 180

x + x + D = 180

D = 180 - 2x

__________________________________

In the other hand we know that :

BDC angle + BDA angle = 180

because they make a straight line together so ;

180 - 2x + 50 = 180

- 2x = 180 - 230

- 2x = - 50

x = 25

And we're done...

Have a great time ♡

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Write a polynomial function with rational coefficients so that P(x) = 0 has the given roots. 4, 16, and 1 + 9i
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I remember dis, yey

so if a polynomial has roots r_1, r_2, r_3, it can be factored into
f(x)=a(x-r_1)^b(x-r_2)^c(x-r_3)^d where a,b,c,d are constants

also, if a polynomial has rational coefients and a+bi is a root, then a-bi must also be a root


so our roots we need are
4,16, 1+9i and 1-9i

so assuming multiplity 1 (that means we have something like [/tex]f(x)=a(x-r_1)^1(x-r_2)^1(x-r_3)^1[/tex])

we get that your function is
P(x)=(x-4)(x-16)(x-(1+9i))(x-(1-9i)) which simplifies to
P(x)=(x-4)(x-16)(x-1-9i)(x-1+9i) which expands to
P(x)=x^4-22x^3+186x^2-1768x+5248
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Eight times the sum of a number and 2 equals 4
Scrat [10]
8 x n + 2 = 4
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5x – 2= 8<br> Please help me
Marta_Voda [28]

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Step-by-step explanation:

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A shirt is on sale for 25% off and you paid $21.75. What was the original cost of the shirt?
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Simplify: cos2x-cos4 all over sin2x + sin 4x
GrogVix [38]

Answer:

\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)

Step-by-step explanation:

\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}

Apply formula:

\cos\left(A\right)-\cos\left(B\right)=-2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right) and

\sin\left(A\right)+\sin\left(B\right)=2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)

We get:

=\frac{-2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\sin\left(\frac{2x-4x}{2}\right)}{2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\cos\left(\frac{2x-4x}{2}\right)}

=\frac{-\sin\left(\frac{2x-4x}{2}\right)}{\cos\left(\frac{2x-4x}{2}\right)}

=\frac{-\sin\left(\frac{-2x}{2}\right)}{\cos\left(\frac{-2x}{2}\right)}

=\frac{-\sin\left(-x\right)}{\cos\left(-x\right)}

=\frac{-\cdot-\sin\left(x\right)}{\cos\left(x\right)}

=\frac{\sin\left(x\right)}{\cos\left(x\right)}

=\tan\left(x\right)

Hence final answer is

\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)

6 0
3 years ago
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