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Alex73 [517]
3 years ago
8

Simplify the expression and write the result in standard form, a+bi. 6 - 8i / -2

Mathematics
1 answer:
gregori [183]3 years ago
4 0

Answer:

\frac{6 - 8i}{ - 2}  =  \frac{6}{ - 2}  -  \frac{8i}{ - 2} (splitting \: denomenator) \\ =  -  3 -  - 4i \\ =   - 3 + 4i \\  thank \: you

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Semmy [17]
The equation of a circle:
(x-h)^2+(y-k)^2=r^2
(h,k) - the coordinates of the centre
r - the radius

A(2,1) - \hbox{the centre} \\
h=2 \\ k=1 \\ \\
(x-2)^2+(y-1)^2=r^2 \\ \\
\hbox{passes through B(10,7)} \\
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7 0
3 years ago
2
Zigmanuir [339]

Answer:

The weight of the water in the pool is approximately 60,000 lb·f

Step-by-step explanation:

The details of the swimming pool are;

The dimensions of the rectangular cross-section of the swimming pool = 10 feet × 20 feet

The depth of the pool = 5 feet

The density of the water  in the pool = 60 pounds per cubic foot

From the question, we have;

The weight of the water in Pound force = W = The volume of water in the pool given in ft.³ × The density of water in the pool given in lb/ft.³ × Acceleration due to gravity, g

The volume of water in the pool = Cross-sectional area × Depth

∴ The volume of water in the pool = 10 ft. × 20 ft. × 5 ft. = 1,000 ft.³

Acceleration due to gravity, g ≈ 32.09 ft./s²

∴ W = 1,000 ft.³ × 60 lb/ft.³ × 32.09 ft./s² = 266,196.089 N

266,196.089 N ≈ 60,000 lb·f

The weight of the water in the pool ≈ 60,000 lb·f

6 0
2 years ago
A book distributor is trying to divide an order of textbooks into equally sized groups for shipping in cartons. The textbooks ca
Phoenix [80]
Calculate least common factor of 12, 15 & 18.
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3 0
3 years ago
A horizontal trough is 16 m long, and its end are isosceles trapezoids with an altitude of 4 m, a lower base of 4 m, and an uppe
Ganezh [65]

Answer:

0.28cm/min

Step-by-step explanation:

Given the horizontal trough whose ends are isosceles trapezoid  

Volume of the Trough =Base Area X Height

=Area of the Trapezoid X Height of the Trough (H)

The length of the base of the trough is constant but as water leaves the trough, the length of the top of the trough at any height h is 4+2x (See the Diagram)

The Volume of water in the trough at any time

Volume=\frac{1}{2} (b_{1}+4+2x)h X H

Volume=\frac{1}{2} (4+4+2x)h X 16

=8h(8+2x)

V=64h+16hx

We are not given a value for x, however we can express x in terms of h from Figure 3 using Similar Triangles

x/h=1/4

4x=h

x=h/4

Substituting x=h/4 into the Volume, V

V=64h+16h(\frac{h}{4})

V=64h+4h^2\\\frac{dV}{dt}= 64\frac{dh}{dt}+8h \frac{dh}{dt}

h=3m,

dV/dt=25cm/min=0.25 m/min

0.25= (64+8*3) \frac{dh}{dt}\\0.25=88\frac{dh}{dt}\\\frac{dh}{dt}=\frac{0.25}{88}

=0.002841m/min =0.28cm/min

The rate is the water being drawn from the trough is 0.28cm/min.

3 0
3 years ago
Help me pleaseee.....
Lana71 [14]

Answer:

titutex=cos\alp,\alp∈[0:;π]

\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+

1−x

2

∣=

2

(2x

2

−1)\Leftright∣cos\alp+sin\alp∣=

2

(2cos

2

\alp−1)

\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N

2

cos(\alp−

4

π

)∣=N

2

cos(2\alp)\Right\alp∈[0;

4

π

]∪[

4

3π

;π]

1) \displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;

4

π

]

\displaystyle cos(\alp -\frac{\pi}{4})=cos(2\alp )\dotscos(\alp−

4

π

)=cos(2\alp)…

2. \displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[

4

3π

;π]

\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−cos(\alp−

4

π

)=cos(2\alp)…

1

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Display

6 0
3 years ago
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