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Colt1911 [192]
2 years ago
7

Someone help with these its due tomorrow. Ahhhh

Mathematics
1 answer:
Oksana_A [137]2 years ago
4 0
I'll help with 1.) 
9m:3m = 6m:x
cross multiply
9x=3*6
9x=18
divide both sides 9
x = 2
the missing side is 2m

The missing angle is 30°

Question 2.)
find the ratios to solve for missing side
24:6 = b:4
cross multiply
24*4 = 6b
96 = 6b
divide both sides by 6
16 = b

The missing angle is 32°

Question 3.)
Find missing side d
make a ratio
5:d = 3:9
cross multiply
5*9 = 3d
45 = 3d 
divide both sides by 3
15 = d

the missing angle is 90°

Question 4.)
Find missing side s
write as proportion or ratio
36:6 = s:2
cross multiply
36*2 = 6s
72 = 6s
divide both sides by 6
s = 12

The missing angle t is 71°

Hope this helps :)
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Find the quotient of these Complex Numbers. <br><br><br> (5-i) / (3+2i)
JulsSmile [24]

Let the given complex number

z = x + ix = \dfrac{5-i}{3+2i}

We have to find the standard form of complex number.

Solution:

∴ x + iy = \dfrac{5-i}{3+2i}

Rationalising numerator part of complex number, we get

x + iy = \dfrac{5-i}{3+2i}\times \dfrac{3-2i}{3-2i}

⇒ x + iy = \dfrac{(5-i)(3-2i)}{3^2-(2i)^2}

Using the algebraic identity:

(a + b)(a - b) = a^{2} - b^{2}

⇒ x + iy = \dfrac{15-10i-3i+2i^2}{9-4i^2}

⇒ x + iy = \dfrac{15-13i+2(-1)}{9-4(-1)} [ ∵ i^{2} =-1]

⇒ x + iy = \dfrac{15-2-13i}{9+4}

⇒ x + iy = \dfrac{13-13i}{13}

⇒ x + iy = \dfrac{13(1-i)}{13}

⇒ x + iy = 1 - i

Thus, the given complex number in standard form as "1 - i".

5 0
3 years ago
2/3 : 1 in its simplest form
slamgirl [31]

Answer:

2 : 3

Step-by-step explanation:

You need to show the ratio using the smallest whole numbers possible.

Multiply both numbers by 3.

2/3 : 1

2/3 × 3 : 1 × 3

2 : 3

5 0
2 years ago
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