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mojhsa [17]
2 years ago
12

Can someone please help me please

Mathematics
1 answer:
qwelly [4]2 years ago
4 0

Answer:

1) \dfrac{12+\sqrt{-16}}{4}=\dfrac{12+\sqrt{16i^{2}}}{4}\\\\\\=\dfrac{12+4i}{4}=\dfrac{12}{4}+\dfrac{4i}{4}\\\\\\= 3 + i

Real part = 3  & imaginary part = 1

2) 4 + 7i - 8 = 4 -8 + 7i

                   = -4 + 7i

Real part = -4 & imaginary part = 7

3)\dfrac{\sqrt{9}+\sqrt{-4}}{5}=\dfrac{3+\sqrt{4i^{2}}}{5}\\\\\\=\dfrac{3}{5}+\dfrac{2i}{5}\\\\\\Real \ part = \dfrac{3}{5} \ & \ imaginary \ part = \dfrac{2}{5}

4) -2-\sqrt{-25}=-2-\sqrt{25i^{2}}=-2-5i

Real part = -2  & imaginary part = -5

5)\dfrac{15-13i}{3}=\dfrac{15}{3}-\dfrac{13i}{3}\\\\\\Real \ part = \dfrac{15}{3} \ & \ imaginary \ part = \dfrac{-13}{3}

b) 1) 6 + 4i

2) -12+ 5i

3) 5-14i

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Rampur Sarpanch requested one of his villager to donate a 6m wide land adjusted to his 132.8m long side of his right triangular
Ronch [10]

Answer:

Area of the remaining triangle with the villager is 1243.13 m²

Step-by-step explanation:

Triangle ABC is the triangular plot of a villager shown in the figure attached.

Sarpanch requested the villager to donate land which is 6 m wide and along the side AC which measures 132.8m.

Other sides of the plot has been given as AB = 50m and BC = 123 m.

Now area of this land before donation = \frac{1}{2}\times {\text{Height}}\times \taxt{Base}

= \frac{1}{2}\times (123)\times (50)

= 3075 square meter

After donation of the land the triangle formed is ΔDBE.

In ΔABC,

tan(ABC)=(\frac{AB}{BC})

tan(∠ABC) = \frac{50}{123}

                 = 0.4065

∠ABC = tan^{-1}(0.4065)

          = 22.12°

In ΔEFC,

tanC = \frac{EF}{CF}

0.4065 = \frac{6}{CF}

CF = \frac{6}{0.4065}

CF = 14.76 m

Since DE = AC - (CF + AG)

               = 132.8 - (2×14.76)

               = 132.8 - 29.52

               = 102.48 m

Now in ΔDBE,

sin(∠DEB) = \frac{BE}{DE}

sin(22.12) = \frac{BE}{102.48}

DB = 102.48×0.3765

     = 38.59 m

Similarly, cos(22.12) = \frac{BE}{DE}

0.9264 = \frac{BE}{102.48}

BE = 102.48×0.9264

     = 94.94m

Now area of ΔDBE = \frac{1}{2}(DB)(BE)

                                = \frac{1}{2}(38.59)(94.94)

                                = 1831.87 square meter

Area of remaining triangle with the villager = Area of ΔABC - Area of ΔDBE

= 3075 - 1831.87

= 1243.13 square meter

3 0
3 years ago
Plato Necessities produces camera cases. they have found that the cost, c (x), of making x camera cases is a quadratic function
Phoenix [80]
This answer has been deleted.
3 0
3 years ago
In the figure, GK bisects FGH
bija089 [108]

Answer:

w = 7

Step-by-step explanation:

Given:

m<FGK = (7w + 3)°

m<FGH = 104°

angle bisector of <FGH = GK

Required:

Value of w

SOLUTION:

Since GK bisects angle FGH, it divides the angle into two equal parts. Therefore, the following equation can be generated to find the value of w:

m<FGH = 2*m<FGK

104 = 2*(7w + 3) (substitution)

Divide both sides by 2

\frac{104}{2} = \frac{2*(7w + 3)}{2}

52 = 7w + 3

Subtract 3 from each side

52 - 3 = 7w + 3 - 3

49 = 7w

Divide both sides by 7

\frac{49}{7} = \frac{7w}{7}

7 = w

w = 7

8 0
3 years ago
Solve 4x^2 - 36x + 32 = 0<br><br>Sorry! It's suppose to be + 32.<br>Please show all work.
mylen [45]

Answer:

x=1, x=8

Step-by-step explanation:

4x2−36x+32=0

Step 1: Factor left side of equation.

4(x−1)(x−8)=0

Step 2: Set factors equal to 0.

x−1=0 or x−8=0

 (+1)        (+8)

x=1 or x=8




6 0
3 years ago
Which ordered pairs make both inequalities true? Check all that apply.
Thepotemich [5.8K]
All others did not satisfy the inequality simultaneously except #4 and #6
For #4:
2 < 5(1) + 2 i.e. 2 < 7 (true)
2 >= 1/2(1) + 1 i.e. 2 >= 3/2  (true)
For #6
2 < 5(2) + 2 i.e. 2 < 12 (true)
2 >= 1/2(2) + 1 i.e. 2 >= 2

7 0
3 years ago
Read 2 more answers
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