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mart [117]
4 years ago
7

Any1 can help me wif q12

Mathematics
1 answer:
xenn [34]4 years ago
8 0
I) HCF - use the smallest powers of each common factors
HCF (A,B) = 2^2 × 3^4 × 5^2

LCM - use the highest powers of each factors
LCM (A,B) = 2^4 × 3^6 × 5^2 × 7^2 × 11^16

ii) Add powers together.
A×B = 2^6 × 3^10 × 5^4 × 7^2 × 11^16
sqrt(A × B)
Divide powers by 2.
sqrt(A × B) = 2^3 × 3^5 × 5^2 × 7 × 11^8

iii) C = 3^7 × 5^2 × 7
Ck = (3^7 × 5^2 × 7) × k
B/c Ck should be a product that is a perfect cube, the powers of the products should be divisible by 3.
(3^7 × 5^2 × 7) × k = 3^9 × 5^3 × 7^3

k = (3^9 × 5^3 × 7^3) / (3^7 × 5^2 × 7)
k = 3^(9-7) × 5^(3-2) × 7^(3-1)
k = 3^2 × 5 × 7^2
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Nookie1986 [14]

Hello mother fu*k

Step-by-step explanation:

3 0
3 years ago
A principal of $3500 is invested at 7.5% interest, compounded annually. How much will the investment be worth after 13 years? Us
Anna71 [15]
The equation is for when you compound annually is
A=P(1+r)^t

A=future amount
P=initial amount or principal
r=interest rate in decimal form
t=time in years



so
A=A
P=3500
r=7.5%=0.075
t=13

A=3500(1+0.075)^{13}
A=3500(1.075)^{13}
use your calculator
A=8961.4457
round to nearest dollar
A=$8961

that's how much the investment will be worth after 13 years
7 0
4 years ago
Multiply Conjugates Using the Product of Conjugates Pattern
SpyIntel [72]

Answer:

The product is the difference of squares is $$\left(11-b\right)\left(11+b\right)=121-{{b}^2}$$

Step-by-step explanation:

Explanation

  • The given expression is (11-b)(11+b).
  • We have to multiply the given expression.
  • Square the first term 11. Square the last term b.

$$\begin{aligned}&(11-b)(11+b)=(11)^{2}-(b)^{2} \\&(11-b)(11+b)=121-b^{2}\end{aligned}$$

5 0
2 years ago
2r + 3s = 14<br><br> 2( )+ 3s = 14<br><br> +3s = 14<br><br> 3s=<br> S =
Soloha48 [4]

Answer:

4

Step-by-step explanation:

3s=14-2

3s=12

3s/3=12/3

s=4

5 0
2 years ago
Carolyn was asked to solve the following system of equations. Her work is shown. What is the solution to the system of linear eq
Natalka [10]

Answer:

(11, 13)

Step-by-step explanation:

Carolyn's work is incorrect. Below is the correct solution:

Given:

3x - 2y = 7 ----› Equation 1

y = x + 2 -----› Equation 2

Substitute y = (x + 2) in equation 1

3x - 2y = 7 ----› Equation 1

3x - 2(x + 2) = 7 (substitution)

3x - 2(x) -2(+2) = 7 (distributive property)

[Note: this is where Carolyn made a mistake]

3x - 2x - 4 = 7

Collect like terms

x - 4 = 7

x = 7 + 4 (addition property of equality).

x = 11

Substitute x = 11 in equation 2

y = x + 2 -----› Equation 2

y = 11 + 2 (substitution)

y = 13

✅The solution to the system of equations would be:

(11, 13)

8 0
3 years ago
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