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emmasim [6.3K]
3 years ago
14

Find the value of below algebriac expression for x=2,y=-1 and z=2 i. 4x2−3y2+5z24x2−3y2+5z2 ii. 3x3−2x(4yz+5x2)3x3−2x(4yz+5x2) i

ii. 1−4x(yz+3xy)1−4x(yz+3xy)
Mathematics
1 answer:
Advocard [28]3 years ago
8 0

Answer:

i. 33

ii. 1488

iii. 65

Step-by-step explanation:

i. 4x² − 3y² + 5z²

Putting the values x=2, y=-1 and z=2

4(2)² − 3(-1)² + 5(2)²

4×4 − 3×(1)² + 5×(4)

(16) -  (3×1) + (20)

16 -  3+ 20

36 - 3

33

ii. 3x³ − 2x(4yz+5x²)

Putting the values x=2, y=-1 and z=2

[3(2)³ − 2×2(4×(-1)×2 + 5×(2)²)]

[3(8)³ − 2×2(-8 + 5×4)]

[3 ×512 -  2×2(-8 +20)]

[1536 - 4(12)]

[1536 - 48]

1488

iii. 1−4x(yz+3xy)

Putting the values x=2, y=-1 and z=2

[1 - 4 × 2((-1)×2 + 3×2×(-1))]

[1- 8(-2 + (-6))]

[1- 8 (-2 -6)]

[1 - 8(-8)]

[1 +64]

65

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7/8m+9/10-2m-3/5<br> Combine like terms to create an equivalent expression.
stich3 [128]

Answer:

-1.123m+0.3

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Step-by-step explanation:

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Parentheses first:

9-[42-5]

Brackets next:

9- 37

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pav-90 [236]

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The total area of the blue sectors is the sum of 98 pi/15 and 49 pi/4 (units^2):


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