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2 years ago

In a school election, 2/5 of the 30 candidates are girls and the rest are boys. How many candidates are boys?

Mathematics

2 answers:

Marianna [84]2 years ago

7 0

This should solve this problem

MrMuchimi2 years ago

3 0

Answer: 18

explanation: 30 split into 5 groups have 6 people in each. And 3 groups of 6 is 18. :) hope this helped

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Find the first five terms of each arithmetetic sequence describe a1 = -12 d = 5

Fudgin [204]

Answer:

- 12, - 7, - 2, 3, 8

Step-by-step explanation:

The general terms of an arithmetic sequence are

a, a + d, a + 2d + ....... + a + (n - 1 )d

To obtain consecutive terms in the sequence add d = 5 to the previous term

$a_{1}$ = - 12

$a_{2}$ = - 12 + 5 = - 7

$a_{3}$ = - 7 + 5 = - 2

$a_{4}$ = - 2 + 5 = 3

$a_{5}$ = 3 + 5 = 8

4 0

2 years ago

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

2 years ago

Can someone help me which are correct?

taurus [48]

Answer:

Between the first option and the last option.

Step-by-step explanation:

first choice: c, a, d, b, e

last/fourth choice: a,c,d,b,e

7 0

3 years ago

My grade is a 77%, if i get a 0% on a paper worth 30% of my grade, what will my grade be now??

blsea [12.9K]

$Grade=Paper+Other$
$Grade=(Grade_{paper}\times Value_{paper})+(Grade_{other}\times Value_{other})$
$Grade=(0 \times 30\%)+(77\% \times (1-30\%))$
$Grade=(0 \times 30\%)+(77\% \times 70\%)$
$Grade=0+53.9\%$
$Grade=53.9\%$

6 0

3 years ago

The perimeters of two squares are in the ratio 2 : 4. The perimeter of

Bess [88]

Answer:

Step-by-step explanation:

2 : 4

8 : 16

8 0

2 years ago