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3 years ago

Whats the LCM of 40 and 5

Mathematics

1 answer:

Simora [160]3 years ago

3 0

Answer:

The LCM of 40 and 5 is simply 40.

Step-by-step explanation:

To find the lowest common multiple of any 2 numbers, we will first list the prime factors of each number.

40: 2 × 2 × 2 × 5

5: 5

Then we need to multiply each factor the most amount of times it appears in either number.

2: 3 times in the first number

5: 1 time in either number

Now we multiply these.

2 × 2 × 2 × 5 = 40

The LCM of 40 and 5 is simply 40.

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Both questions plzzz thanks

andrew11 [14]

Answer:

Step-by-step explanation:

The formula for determining simple interest is expressed as

I = PRT/100

Where

I represents interest paid on the loan.

P represents the principal or amount taken as loan

R represents interest rate

T represents the duration of the loan in years.

1) From the information given,

P = cost of gym = $2200

R = 6%

T = 3 years

I = (2200 × 6 × 3)/100 = $395

The total amount that he would pay after 3 years is

2200 + 395 = $2596

If he decides to continue going to the gym, the total amount that the would pay in 3 years(3 × 12 = 36 is

80 × 36 = 2880

Therefore, paying for the gym would cost more. He should take the loan.

2) if she takes the loan, the interest would be

I = (18000 × 7.5 × 5)/100 = $6750

The total amount that she would pay back after 5 years is

6750 + 18000 = $24750

Since the value that would be added to her house is $25000, therefore, she should buy it.

6 0

3 years ago

Simplify each exponential expression using the properties of exponents and match it to the correct answer.

saveliy_v [14]

1) (2 $\times$ 3^-2)^3 (5 $\times$ 3^2)^2 / (3^-2)(5 $\times$ 2)^2 = 2

2) (3^3) (4^0)^2 (3 $\times$ 2)^-3 (2^2) = 1/2

3) (3^7 $\times$ 4^7) (2 $\times$ 5)^-3 (5)^2 / (12^7) (5^-1) (2^-4) = 2

4) (2.3)^-1 (2^0) / (2.3)^-1 = 1

<u>Step-by-step explanation</u>:

Step 1 :

(2 $\times$ 3^-2)^3 (5 $\times$ 3^2)^2 / (3^-2)(5 $\times$ 2)^2

⇒ (2^3) (3^-6) (5^2) (3^4) / (3^-2) (10^2)

⇒ (2.2^2.5^2) (3^-6.3^4) / (3^-2) (10^2)

⇒ (2)(10^2) (3^-2) / (3^-2) (10^2)

⇒ 2

Step 2 :

(3^3) (4^0)^2 (3 $\times$ 2)^-3 (2^2)

Any number with power 'zero' is 1.

⇒ (3^3) (1^2) (3^-3) (2^-3) (2^2)

⇒ (2^(-3+2))

⇒ 2^-1 = 1/2

Step 3 :

(3^7 $\times$ 4^7) (2 $\times$ 5)^-3 (5)^2 / (12^7) (5^-1) (2^-4)

⇒ (12^7) (2^-3) (5^-3) (5^2) / (12^7) (5^-1) (2^-4)

⇒ (2^-3) (5^(-3+2)) / (5^-1) (2^-4)

⇒ (2^-3) (5^-1) / (5^-1) (2^-4)

⇒ 1 / (2^-1)

⇒ 2

Step 4 :

(2.3)^-1 (2^0) / (2.3)^-1

⇒ (2^0)

⇒ Any number with power 'zero' is 1

⇒ 1

8 0

3 years ago

Bryson’s dad is 38 years old. If this is four years less than three times Bryson’s age, how old is Bryson?

klemol [59]

Answer:

Bryson’s age is 14 years

Step-by-step explanation:

Let

x ----> Bryson’s age

we know that

Bryson’s age multiplied by 3 minus 4 must be equal to 38

The linear equation that represent this situation is

$3x-4=38$

solve for x

Adds 4 both sides

$3x=38+4$

$3x=42$

Divide by 3 both sides

$x=14$

therefore

Bryson’s age is 14 years

7 0

3 years ago

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$

$=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}$

What we use?

We use that

$e^{i\pi n}=cos(\pi n)+i sin(\pi n)$

and

$\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}$

6 0

3 years ago

F(x) = 4x2 – 7x + 7<br> Find f (7)

Colt1911 [192]

Answer:

4×2÷8

so 8+7=15

=7x=15

so,x=15/7

5 0

2 years ago

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