MU Grade 8 Pr Algebra V16, Kemi O'Brien (4028/8) -

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

2 years ago

Joanna is paid $14 per hour is it linear

Dafna1 [17]

Well me myself and I would say yes!!!

8 0

3 years ago

How can u find the radius wit the height nd volume

lesantik [10]

Answer:

The radius of a cylinder(r) = √(V / π × h), where V is the volume of a cylinder, h is the height of the cylinder, and π(Pi) is a mathematical constant with an approximate value of 3.14.

Step-by-step explanation:

hope this helps.

5 0

2 years ago

Find m<AED<br> help me with this hurry!!!

Nataly_w [17]

First lets find the value of x. We can do this by making m∠AEB and m∠DEC equal to each other in an equation because they are vertical angles (vertical angles are equal to each other).

Your equation should look like this: m∠AEB = m∠DEC

Plug in the values of m∠AEB and m∠DEC into the equation. Now your equation should look like this:

(3x + 21) = (2x + 26)

Subtract 2x from both sides.

x + 21 = 26

Subtract 21 from both sides.

x = 5

Now plug 5 for x in either ∠AEB or ∠DEC; I will plug it into ∠AEB.

m∠AEB = 3(5) + 21

15 + 21 = 36

m∠AEB = 36°, now since ∠AEB and ∠AED are forming a straight line, this means they are supplementary so they must add up to 180 degrees.

Make m∠AEB and m∠AED add up to 180 in an equation and solve for m∠AED.

36 + m∠AED = 180

Subtract 36 from both sides.

m∠AED = 144°

4 0

2 years ago

Please help! i think its ratio between boys and girls for it

Jlenok [28]

Answer:

1.

Step-by-step explanation:

I am 100% positive.

If you look at the bottom of the graph you see number from 0-14 counting by 2's. Those numbers stand for how many girls/boys there are exactly.

The colored lines represent boys/girls and you can determine which is which by looking at the two boxes on the right and reading the labels. Looking at the cookies and cream two lines you see that the girl one is longer. If you look back at the bottom and see which number both lines are lined up to you can determine which is on a higher number. That would be your answer.

Hope this helped lots and lots,

sincerely, me

6 0

3 years ago

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