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agasfer [191]
3 years ago
8

16 OT 26

Mathematics
1 answer:
hram777 [196]3 years ago
5 0

Answer:

Total = 166cm^2

Step-by-step explanation:

Given

See attachment

Required

Determine the surface area

The figure is made up of two cuboids and the surface area of a cuboid is:

Area = 2LB + 2LH + 2BH

Because the bottom part of the upper piece is not visible, the area is calculated as:

Area = 2LB + 2LH + BH

This gives:

Area = 2 * 5*4 + 2 * 4 * 2 + 5 * 2

Area = 66

For the bottom piece:

We calculate its area as Area = 2LB + 2LH + 2BH. However, the area of the invisible part will be subtracted.

So, we have:

Area = 2*3*5 + 2*3*5 + 2*5*5 - 2*5

Area = 100

The total surface area is:

Total = 66 + 100

Total = 166cm^2

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(-11x^2 +6) - (14x^2 +2)
Nataly [62]

Answer:

−1(5−2)(5+2)

Step-by-step explanation:

Eliminate redundant parentheses

(−11²+6)−1(14² +2)

Distribute

−11² + 6 -- 1(14² +2)

−11² + 6 - 14² − 2

8 0
2 years ago
Find the real numbers x and y if -3+ix^2y and x^2+y+4i are conjugate of each other. Pls solve with the steps
Firdavs [7]
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4

Therefore, for the two expressions to be conjugates, we must satisfy the two conditions. 

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the 

   x²y = -4 ... (I)

Condition 2: Real parts are the same

   x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

   x²y = -4 ... (I)
   x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

   y = -3 - x² ... (II)

Substituting into equation (I)

   x²y = -4 ... (I)
   x²(-3 - x²) = -4
   -3x² - x⁴ = -4
   x⁴ + 3x² - 4 = 0
   (x² + 4)(x² - 1) = 0
   (x² + 4)(x-1)(x+1) = 0

Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

   y = -3 - x² ... (II)
   y = -3 - (±1)²
   y = -3 - 1
   y = -4

So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:

   -3 + ix²y 
   = -3 + i(±1)²(-4)
   = -3 - 4i

   x² + y + 4i
   = (±1)² - 4 + 4i
   = 1 - 4 + 4i
   = -3 + 4i

They result in conjugates
4 0
3 years ago
Read 2 more answers
6x8 = nx 16<br> What is the value of the unknown number?
BigorU [14]
Um I think the answer is 3
3 0
3 years ago
Simplify <br> 2x x y x 3
nikitadnepr [17]
Which is it
2x times y times 3 = 6xy
or,
2xxyx3= 6x^3y

7 0
3 years ago
in equivalent ratios , if the numerator of the first ratio is greater than the denominator of the first ratio , then the numerat
Murljashka [212]
less than the denominator in the second ratio
8 0
3 years ago
Read 2 more answers
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