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2 years ago

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HELP ME PLEASE!!!

1 answer:

Pavel [41]2 years ago

7 0

Answer:

Lo siento amigo mío pero ha llegado el momento. Necesitas hacer tu propia tarea porque la profunda y oscura verdad es ... nadie te ayudará cuando falles en la prueba. Por eso estoy aquí para llevarte. Deja de pedir respuestas y pruébalas tú mismo. Pídele ayuda a tu maestro. Pero deja de pedirle respuestas a la gente. Eso es definitivamente una trampa.

Step-by-step explanation:

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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

Find each product.Is it an enlargement or reduction?

allsm [11]

Yessss that should be correct

4 0

2 years ago

Solve for x 5x - 10 = -10

Zigmanuir [339]

Answer:

the answer is x = 0

Step-by-step explanation:

Cancel equal terms on both sides of the equation:

$5x = 0$

Divide both side of the equation by 5.

$x = 0$

Therefore, the answer is x = 0

3 0

3 years ago

Read 2 more answers

Match the three graphs to the function they represent ?

Mama L [17]

Suppose that a>b>1, then

$a^x>b^x$ and $\log_ax$

Therefore, since 2<3<7, $\log_2x>\log_3x>\log_7x.$

Choose an arbitrary x>1. You have that a takes the greatest values at x, c takes the smallest value at x. Thus,

a>b>c and

$a=\log_2x;$
$b=\log_3x;$
$c=\log_7x.$

Answer: correct option is B.

3 0

2 years ago

Given that IK is a midsegment of AFGH, find GH.

guajiro [1.7K]

It’s either C or B but I’m not sure

8 0

3 years ago

Read 2 more answers