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3 years ago

PLS PLS PLS PLSSSSS HELPPPP

Mathematics

2 answers:

Ulleksa [173]3 years ago

8 0

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Ration and Percentage,

To get the Adams rake,

let's calculate 5% of 357 ltrs.

therefore, (5/100)* 357 = 17.85 ltrs

let's subtract the 5 % value from Tapiwa rake,

==> 357 - 17.85 == 339.15

hence the Adams rake is 339.15 ==> round off ==> 339 litres.

Vaselesa [24]3 years ago

8 0

Answer: 340

Step-by-step explanation:

Okay So the T person raked 5% more than the A person.

The T person raked a total of 357 leaves.

So if A is X, then T is 1.05x since T raked 5% more.

so if 1.05x = 357, x = 357/1.05 = 340

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If a > b > c > d, then which is larger, a+c or b+d ? Can we tell from a > b > c > d which of a+d and b+c is la

victus00 [196]

Answer:

1. a+c is larger than b+d

2. No way to tell whether a+d or b+c is larger.

Step-by-step explanation:

1. Which is larger, a+c or b+d?

Let a, b, c, and d be any numbers such that $a > b > c > d$ .

Specifically, note that $a > b$ , and subtracting b from both sides of the inequality, observe that $a-b > 0$ .

Similarly, $c > d$ , and subtracting d from both sides of the inequality, observe that $c-d > 0$ .

From this, add "a-b" (a positive number, as proven above) to both sides of the inequality.

$(a-b)+(c-d) > (a-b)+0$

Addition by zero (the additive identity) doesn't change anything, so the right side remains "a-b"...

$(a-b)+(c-d) > a-b$

... and "a-b" is positive...

$(a-b)+(c-d) > a-b > 0$

... so, by the transitive property of inequality...

$(a-b)+(c-d) > 0$

Recall that subtraction is addition by a negative number...
$a+(-b)+c+(-d) > 0$

...and that addition is associative and commutative, so things can be added in any order, so the middle two terms on the left side can be rearranged...

$a+c+(-b)+(-d) > 0$

Adding b + d to both sides of the inequality

$(a+c+(-b)+(-d))+(b+d) > 0+(b+d)$

... and simplifying

$a+c > b+d$

So, a+c is larger than b+d.

2. Which is larger, a+d or b+c?

Consider the following two examples:

Example 1

Suppose a=10; b=3; c=2; d=1.

Note that $a > b > c > d$ ( $10 > 3 > 2 > 1$ ) and, also observe that $a+d=(10)+(1)=11$ , and $b+c=(3)+(2)=5$ , so a+d is larger than b+c.

Example 2

However, suppose a=10; b=9; c=8; d=1.

Note that $a > b > c > d$ ( $10 > 9 > 8 > 1$ ) but that $a+d=(10)+(1)=11$ , and $b+c=(9)+(8)=17$ , so a+d is smaller than b+c.

So, in one example, a+d is bigger, and in the other, a+d is smaller. Therefore, there is no way to tell which of a+d or b+c is larger from only the given information.

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