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3 years ago

The smaller the absolute value of a negative number, the larger, smaller or farther from zero the number is ???

Mathematics

1 answer:

Mama L [17]3 years ago

6 0

It doesn't matter. It's the same distance from zero

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Help asap will give brainliest

ArbitrLikvidat [17]

<u>Answer:</u>

• Equation = $2(\frac{7}{4}w + w) = 176$

• length = 56 ft

• width = 32 ft

<u>Step-by-step explanation:</u>

The ratio of the length to width is 7:4.

∴ $\frac{l}{w} = \frac{7}{4}$

⇒ $l = \frac{7}{4} w$ [Equation 1]

We know that the perimeter of the garden is 176 feet.

∴ $2(l + w) = 176$

⇒ $2(\frac{7}{4}w + w) = 176$

⇒ $\frac{7}{4} w + w = 88$ [From Equation 1]

⇒ $\frac{7 w + 4w}{4} = 88$

⇒ $\frac{11w}{4} = 88$

⇒ $11w = 352$

⇒ $w = \bf 32 \space\ ft$

We know from Equation 1 that:

$l = \frac{7}{4} w$

∴ $l = \frac{7}{4} (32)$

⇒ $l = \bf 56 \space\ ft$

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2 years ago

2*10^3 in scientific notation

Natali5045456 [20]

2000, you just need to put three zeros at the end of the number.

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4 years ago

The table represents a linear equation . which equation correctly uses point(-2,-6) to write the equation of this line in point-

Minchanka [31]

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3 years ago

Read 2 more answers

5(2z-2)=9(z+2) how to solve it

IceJOKER [234]

Answer:

z=28

Step-by-step explanation:

10z-10=9z+18

-9z -9z

z-10=18

+10 +10

z=28

7 0

3 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

3 years ago