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3 years ago

10

An alloy consists of nickel, zinc, and copper in the ratio 2:7:9.

1 answer:

Alexus [3.1K]3 years ago

5 0

Answer:

0.4kg, 1.4kg and 1.8kg

Step-by-step explanation:

Given parameters:

Mass of alloy = 3.6kg

Ratio of elements in the alloy = 2:7:9

Unknown:

Mass of each element in the alloy = ?

Solution:

To find the mass of each element in the alloy, first find the sum of the ratio;

2+7+9 = 18

Mass of nickel = $\frac{2}{18}$ x 3.6 = 0.4kg

Mass of zinc = $\frac{7}{18}$ x 3.6 = 1.4kg

Mass of copper = $\frac{9}{18}$ x 3.6 = 1.8kg

The mass of each metal is 0.4kg, 1.4kg and 1.8kg

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JulijaS [17]

Answer:

The signa notation to represent the first five ten f(x) is given by $f(x)=\sum\limits^{5}_{n=1}a_n$

Step-by-step explanation:

Given sequence is -5, -9, 13...

Let f(x) be the given sequence and is denoted by

$f(x)=\{-5,-9,-13,...\}$

Let the first term be $a_1, 2^{\textrm{nd}}$ term be $a_3,...$

ie, $a_1=-5,a_2=-9, a_3=-13,...$

To find the common difference d:

$d=a_2-a_1$

$=-9-(-5)$

$=-9-(+5)$

$d=-4$

$d=a_3-a_2$

$=-13-(-9)$

$=-13+9$

$d=-4$

Therefore the common difference d is -4 for given sequence f(x) with $a_1=-5$ and d=-4, the seqence f(x) is an arithmetic sequence

By defintion of arithmetic sequence

$a_n=a+(n-1)d\hfill(1)$

Now to find $a_4, a_5$ :

put n=4 in equation (1)

$a_4=a+(4-1)d$

$a_4=-5+(3)(-4)$ [since a=-5, d=-4]

$=-5-12$

$a_4=-17$

in equation (1)

$a_5=a+(5-1)d$

$a_5=-5+(4)(-4)$ [since a=-5, d=-4]

$=-5-16$

$a_5=-21$

Therefore $a_4=-17$ and $a_5=-21$

Therefore $f(x)=\{-5,-9,-13,-17,-21,...\}$

Now to represent the sum of the first five terms of f(x) using sigma notation as below

$f(x)=\sum_{n=1}^5 a_n$

where $\sum_{n=1}^5 a_n=a_1+a_2+a_3+a_4+a_5$

$-5-9-13-17-21$

$\sum_{n=1}^5 a_n=-65$

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4 years ago