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2 years ago

Could someone help me

Mathematics

1 answer:

Strike441 [17]2 years ago

5 0

1. The area of the rectangle is 336. 14 times 24= 336 or width times length= 336.
2. The perimeter of the rectangle is 76. Because, 2 times(14+24)=76
3. If x= 10, the area would still be 336

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PLSSS HELP ME WITH MATH

labwork [276]

Step-by-step explanation:

always remember, equations and inequalities are like a scale with 2 pans (left and right).

a variable is like a placeholder, an empty spot on the pans. and the expressions on both sides are kind of the conditions for the things to put in these empty spots so that the status of the scale stays unchanged.

for an equation the scale has to stay balanced all the time.

and in an inequality one side is heavier that the other (it normally does not matter how much heavier) with "<" or ">" signs, or the scale can be also balanced with "<=" or ">=" signs. just the other side must never be heavier.

so, in our case here

2x - 5 = 3

yes, the solution (the ONLY solution, actually) is x = 4. no other value of x allows the scale to be balanced.

2x - 5 >= 3

well, since it is a ">=" sign, we can be lazy and treat it like the equation above, and x = 4 is therefore a valid solution for the inequality too.

but so is every other value of x that makes the left side "heavier". what about e.g. 5 ?

2×5 - 5 >= 3

10 - 5 >= 3

5 >= 3

true, great ! so, e.g. x = 5 is also a valid solution.

what else ?

let's simplify the inequality

2x - 5 >= 3

2x >= 8

x >= 4

so, really every value of x that is greater or equal to 4 is a valid solution for the inequality.

-2x - 5 = 3

yes, the solution (the ONLY solution, actually) is x = -4. no other value of x allows the scale to be balanced.

-2x - 5 >= 3

well, since it is a ">=" sign, we can be lazy and treat it like the equation above, and x = -4 is therefore a valid solution for the inequality too.

but so is every other value of x that makes the left side "heavier". what about e.g. -5 ?

-2×-5 - 5 >= 3

remember, -×- = +

10 - 5 >= 3

5 >= 3

true, great ! so, e.g. x = -5 is also a valid solution.

what else ?

let's simplify the inequality

-2x - 5 >= 3

-2x >= 8

x <= -4

a multiplication or division by a negative value flips the inequality sign, because such an operation makes a light weight heavy and a heavy weight light.

so, really every value of x that is less or equal to -4 is a valid solution for the inequality.

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

= 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

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3 years ago

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miskamm [114]

Answer:

it 152.3

Step-by-step explanation:

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3 years ago

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