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Savatey [412]
2 years ago
8

Find the length here asap

Mathematics
2 answers:
SOVA2 [1]2 years ago
6 0

Answer:

A,  40.16

Step-by-step explanation:

cosθ = adjacent (BC) / hypotenuse (AB)

cos 17° = BC / 42

BC = 40.16

bezimeni [28]2 years ago
6 0

Answer:

\boxed{BC = 40.16}

Step-by-step explanation:

Cos 17 = \frac{adjacent}{hypotenuse}

Where adjacent = BC, hypotenuse = 42

0.959 = \frac{BC}{42}

BC = 0.959 * 42

BC = 40.16

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If α and β are the zeros of the quadratic polynomial f(x) = 3x2–4x + 5, find a polynomialwhose zeros are 2α + 3β and 3α + 2β.
Lelechka [254]

Answer:

\boxed{\sf \ \ \ 3x^2-20x+37\ \ \ }

Step-by-step explanation:

Hello,

a and b are the zeros, we can say that

f(x)=3(x^2-\dfrac{4}{3}x+\dfrac{5}{3}) = 3(x-a)(x-b)=3(x-(a+b)x+ab)

So we can say that

a+b=\dfrac{4}{3}\\ab=\dfrac{5}{3}

Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b

for instance we can write

(x-2a-3b)(x-3a-2b)=x^2-(2a+3b+3a+2b)x+(2a+3b)(3a+2b)\\= x^2-5(a+b)x+6a^2+6b^2+9ab+4ab

and we can notice that

a^2+b^2=(a+b)^2-2ab so

(x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab

it comes

x^2-5*\dfrac{4}{3}x+6(\dfrac{4}{3})^2+\dfrac{5}{3}

multiply by 3

3x^2-20x+2*16+5=3x^2-20x+37

4 0
3 years ago
1) solve for w: 3w - 10 = 4w + 5<br> 2) solve for t:6 -2t &gt;18
Gre4nikov [31]
Problem # 1
 3w-10=4w+5
(3w-10) + (10-4w) = (4w+5) + (10-4w)
(3w-10) + (-4w+10) = (4w+5) + (-4w +10)
3w-10-4w+10 = 4w +5 -4w +10
3w-4w=15
-w=15
w= -15 (w is negative 15)

problem #2.
6-2t > 18
6-2t-6 > 18-6
-2t > 12 
(-1) (-2t) >  12 (-1)
2t > -12
next divide by 2
2t/2 > -12/2
 t >  -6     (t is greater then negative 6)

 
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