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Natali5045456 [20]
3 years ago
9

Find the area of the figure

Mathematics
2 answers:
olya-2409 [2.1K]3 years ago
7 0
It will be 47 I hope this helps
Aleonysh [2.5K]3 years ago
6 0

the correct answer is 3249 sorry i took so long

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Find two complex numbers that have a sum of i10 a different of -4and a product of -29​
lianna [129]
<h2>-2+5i and 2+5i</h2>

Step-by-step explanation:

   Let the complex numbers be a+ib\textrm{ and }c+id.

Given, sum is 10i, difference is -4 and product is -29.

(a+c)+i(b+d)=10i ⇒ a+c=0,b+d=10

(a-c)+i(b-d)=-4 ⇒ a-c=-4,b-d=0

a=-2,c=2,b=5,d=5

(a+ib)(c+id)=(-2+5i)(2+5i)=-4-25=-29

Hence, all three equations are consistent yielding the complex numbers -2+5i\textrm{ and }2+5i.

8 0
3 years ago
Find the inequality represented by the graph
bezimeni [28]
The answer is y ≥ 1/2x
4 0
3 years ago
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Dan borrowed money from his parents and he pays back his parents the same amount of money every month He records his payment sch
timurjin [86]

Answer:

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Step-by-step explanation:

Jmfsxbbihswyvbn

3 0
3 years ago
The formula for finding the horsepower of a vehicle is: Horsepower = weight × ( velocity 234 ) 3 Weight is the total weight of t
Luden [163]
The answer is option 2
 Power = 175 hp
 The correct formula to calculate the power is:
 Power = weight * (speed / 234) ³
 Where the weight is that of the vehicle plus the driver's
 Therefore the calculation is as follows:
 Power = (2200 + 180) * (98/234) ³
 Power = 174.8266 hp
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8 0
3 years ago
What is simplest form of sqrt 2 + sqrt 5 / sqrt 2 - sqrt -5? √2 + √5 / √2 - √-5
Leni [432]

Answer:

\frac{-7-2\sqrt{10}}{3}

Step-by-step explanation:

we have

\frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}-\sqrt{5}}

Simplify

Multiply the expression by  \frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}}

(\frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}-\sqrt{5}})(\frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}})

Apply difference of squares in the denominator

\frac{(\sqrt{2}+\sqrt{5})^2}{(\sqrt{2})^2-(\sqrt{5})^2}

\frac{2+2\sqrt{10}+5}{2-5}

\frac{7+2\sqrt{10}}{-3}

\frac{-7-2\sqrt{10}}{3}

6 0
3 years ago
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