All categories

3 years ago

Whats the greatest ten you can multiply by 19 to get close to but not go over 684

Mathematics

1 answer:

USPshnik [31]3 years ago

4 0

Answer:

The desired expression is 19 x (3 tens) = 19 x 30 = 570

So, the greatest ten = 3 (tens)

Step-by-step explanation:

Let us check each tens.

19 x (1 ten) = 19 x 10 = 190

19 x (2 tens) = 19 x 20 = 380

19 x (3 tens) = 19 x 30 = 570

19 x (4 tens) = 19 x 40 = 760

But, here 760 is OVER the given term 684.

and 570 is the nearest value to the value 684.

So, the desired expression is 19 x (3 tens) = 19 x 30 = 570

You might be interested in

Help me :,)

IrinaK [193]

Answer:

The rule of the arithmetic sequence is 13 - 2n

The 30th term is -47

Step-by-step explanation:

∵ f(n) = 11 and g(n) = -2(n - 1) = -2n + 2

∴ f(n) + g(n) = 11 + -2n + 2 = 13 - 2n

Use n = 1 , 2 , 3 , 4 to check the type of the sequence

∵ n = 1 ⇒ 13 - 2(1) = 11

∵ n = 2 ⇒ 13 - 2(2) = 13 - 4 = 9

∵ n = 3 ⇒ 13 - 2(3) = 13 - 6 = 7

∵ n = 4 ⇒ 13 - 2(4) = 13 - 8 = 5

∵ 11 , 9 , 7 , 5 is an arithmetic sequence with difference -2

∴ The rule of the arithmetic sequence is 13 - 2n

∴ The 30th term = 13 - 2(30) = -47

5 0

3 years ago

Read 2 more answers

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

3 years ago

The temperature was -5 ℉ in Denver and then fell 8 ℉ by the morning.<br> Answer now please

allochka39001 [22]

Answer:

it was -5 and then it fell 8 degrees

-5 -8 = -13 degrees F

Step-by-step explanation:

8 0

3 years ago

Please help!!!<br> will pick brainliest

katrin2010 [14]

Answer:

122 and 75 i think

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Please solve with work much appreciated

mamaluj [8]

Answer:

0.046

Step-by-step explanation:

$\text{BAC} = \dfrac{10N - 7.5H}{6.8 M}$

Data:

N = 4

H = 2

M = 80

Calculation:

$\text{BAC} = \dfrac{10N - 7.5H}{6.8M} = \dfrac{10\times4 - 7.5\times2}{6.8\times80} = \dfrac{40 - 15}{544} = \dfrac{25}{544} = 0.046$

6 0

3 years ago