3 years ago

Help,??????????????????

Mathematics

1 answer:

abruzzese [7]3 years ago

8 0

So you would find out what number in the 9 times tables goes into 25 and there's your answer x

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Help please! Thankyou

Soloha48 [4]

Answer:

Thus, option A is correct.

Explanation:

<u>match the proportions</u>

$240,000 → 60%

x → 100%

<u>set up a equation:</u>

$\sf \dfrac{60}{100} = \dfrac{240,000}{x}$

$\sf x = \$400,000$

4 0

3 years ago

A $60 jacket is on sale for 20% off. Explain the steps you would take to find the total cost of the jacket. Identify the new cos

Pie

I think it would be $48

8 0

3 years ago

Lindsey owes $5,381 on a credit card with a 24.9% interest rate compounded monthly. What is the monthly payment she should make

Digiron [165]

If you multiply .249*5381= 1339.87

3 0

3 years ago

Approximate the sum of the convergent infinite series the least usingnumber of terms so that the error is less than .0001. What

____ [38]

Solution :

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ $a_n= \frac{(-1)^n}{n ! 2^n} \ \ \ \ \ a_{n+1}= \frac{(-1)^{n+1}}{(n+1) ! 2^{n+1}}$

Error = $|a_{n+1}|$

Error ≤ 0.0001

$|\frac{(-1)^{n+1}}{(n+1)!2^{n+1}}| \leq 0.0001$

$|\frac{1}{(n+1)!2^{n+1}}| \leq 10^{-4}$

$(n+1)! 2^{n+1} \geq 10000$

Now try, n ≥ 5

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = $\sum_{n=1 }^{5 } \frac{(-1)^n}{n ! 2^n}$ (with error 0.0001)

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = $\frac{-1}{1!2}+\frac{1}{2!2^2}-\frac{1}{3! 2^3}+\frac{1}{4! 2^4}-\frac{1}{5! 2^5}$

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = 0.6065104

3 0

3 years ago

Can someone help me with this problem please :)

WARRIOR [948]

The correct answer is d

6 0

4 years ago

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