1) help me please!!!!

1. an alloy contains zinc and copper in the ratio of 7:9 find weight of copper of it had 31.5 kgs of zinc.

m_a_m_a [10]

Answer:

Step-by-step explanation:

Question (1). An alloy contains zinc and copper in the ratio of 7 : 9.

If the weight of an alloy = x kgs

Then weight of copper = $\frac{9}{7+9}\times (x)$

= $\frac{9}{16}\times (x)$

And the weight of zinc = $\frac{7}{7+9}\times (x)$

= $\frac{7}{16}\times (x)$

If the weight of zinc = 31.5 kg

31.5 = $\frac{7}{16}\times (x)$

x = $\frac{16\times 31.5}{7}$

x = 72 kgs

Therefore, weight of copper = $\frac{9}{16}\times (72)$

= 40.5 kgs

2). i). 2 : 3 = $\frac{2}{3}$

4 : 5 = $\frac{4}{5}$

Now we will equalize the denominators of each fraction to compare the ratios.

$\frac{2}{3}\times \frac{5}{5}$ = $\frac{10}{15}$

$\frac{4}{5}\times \frac{3}{3}=\frac{12}{15}$

Since, $\frac{12}{15}>\frac{10}{15}$

Therefore, 4 : 5 > 2 : 3

ii). 11 : 19 = $\frac{11}{19}$

19 : 21 = $\frac{19}{21}$

By equalizing denominators of the given fractions,

$\frac{11}{19}\times \frac{21}{21}=\frac{231}{399}$

And $\frac{19}{21}\times \frac{19}{19}=\frac{361}{399}$

Since, $\frac{361}{399}>\frac{231}{399}$

Therefore, 19 : 21 > 11 : 19

iii). $\frac{1}{2}:\frac{1}{3}=\frac{1}{2}\times \frac{3}{1}$

$=\frac{3}{2}$

$\frac{1}{3}:\frac{1}{4}=\frac{1}{3}\times \frac{4}{1}$

= $\frac{4}{3}$

Now we equalize the denominators of the fractions,

$\frac{3}{2}\times \frac{3}{3}=\frac{9}{6}$

And $\frac{4}{3}\times \frac{2}{2}=\frac{8}{6}$

Since $\frac{9}{6}>\frac{8}{6}$

Therefore, $\frac{1}{2}:\frac{1}{3}>\frac{1}{3}:\frac{1}{4}$ will be the answer.

IV). $1\frac{1}{5}:1\frac{1}{3}=\frac{6}{5}:\frac{4}{3}$

$=\frac{6}{5}\times \frac{3}{4}$

$=\frac{18}{20}$

$=\frac{9}{10}$

Similarly, $\frac{2}{5}:\frac{3}{2}=\frac{2}{5}\times \frac{2}{3}$

$=\frac{4}{15}$

By equalizing the denominators,

$\frac{9}{10}\times \frac{30}{30}=\frac{270}{300}$

Similarly, $\frac{4}{15}\times \frac{20}{20}=\frac{80}{300}$

Since $\frac{270}{300}>\frac{80}{300}$

Therefore, $1\frac{1}{5}:1\frac{1}{3}>\frac{2}{5}:\frac{3}{2}$

V). If a : b = 6 : 5

$\frac{a}{b}=\frac{6}{5}$

$=\frac{6}{5}\times \frac{2}{2}$

$=\frac{12}{10}$

And b : c = 10 : 9

$\frac{b}{c}=\frac{10}{9}$

Since a : b = 12 : 10

And b : c = 10 : 9

Since b = 10 is common in both the ratios,

Therefore, combined form of the ratios will be,

a : b : c = 12 : 10 : 9

7 0

3 years ago

If t(n) 2n+3 what is the 3rd term

german

The third term is 9

t(3)=2(3)+3
6+3 = 9

3 0

3 years ago

What is the solution of log2x + 6 144 = 2?

anzhelika [568]

The answer is x = 3.

Hope this helps (:

4 0

3 years ago

Read 2 more answers

PLEASE HELP!!

sp2606 [1]

Standard form is just putting (in this case) variables in alphabetical order. First we simplify- 2x+4=4y would become 1/2x+1=y. This is already in standard form, as numbers w/o variables come at the end. simplifying the next one is more tricky. first you get a variable/number alone-
9-2x-2y=4x+3
9-2y=6x+3
6-2y=6x
1-1/3y=x
1=x+1/3y
3=x+y
x+y=3
sorry if I was wrong and not of any help, But I do believe this is correct.

5 0

3 years ago

Read 2 more answers

What is the area of this ^2

Ad libitum [116K]

<h2>Question:- Area of the given diagram</h2>

<h2>Answer :- </h2>

Divide the diagram as shown in attachment.

Now first calculate the area of the rectangle having side 1 and 12

So area = length ×breadth

Area = 1× 12 = 12 unit²

Now find the Area of the remaining triangle with sides 12,13,5 units

So

$Area = \frac{1}{2} \times base \times height \:$

Now put the value as -> base= 5 and height= 12

$Area = \frac{1}{2} \times base \times height \: \\ Area = \frac{1}{2} \times 5 \times 12 \\ Area = \frac{1}{ \cancel2} \times 5 \times \cancel{12}^{ \: \: 6} \\ Area = 5 \times 6 \\ Area = 30 \: units^{2}$

Now add both areas for your answer

$Area{ \tiny{1}} = 12 {units}^{2} \\ Area{ \tiny2} = 30 {units}^{2} \\ Area{ \tiny{total}} = Area{ \tiny{1}} + Area{ \tiny{2}} \\ Area{ \tiny{total}} = 12 + 30 = 42 {units}^{2}$

8 0

3 years ago