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5 months ago

10

Given triangle JKL,write the coordinates of the vertices of a triangle J’K’L’ that result from a translation of triangle JKL two

units to the left

1 answer:

sweet [91]5 months ago

7 0

Step-by-step explanation:

since I don't have any data about the coordinates, I am simply calling them xj, yj, and xj' and yj'.

J = (xj, yj)

K = (xk, yk)

L = = (xl, yl)

a translation (shift) 2 units to the left just means, that the y coordinates regain the same, but the x coordinates are decreased ("to the left" = getting smaller) by 2.

J' = (xj - 2, yj)

K' = (xk - 2, yk)

L' = (xl - 2, yl)

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2 years ago

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 = $\frac{9}{16}\times (72)$

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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

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