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3 years ago

How’s many times does 49 go into 53?

Mathematics

2 answers:

alina1380 [7]3 years ago

4 0

Answer:

1.082.

You will have to divide 53 by 49, 53/49.

Hope that helped :)

ra1l [238]3 years ago

3 0

Answer:

26.5 times

Step-by-step explanation:

just divide it

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Given that |-6| > 5, which statement is not true?

ollegr [7]

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3 years ago

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Find the value of y for the given value of x. Round to the nearest hundredth. y = 15 + 3 ln x; x = 7.2

Nikolay [14]

Answer:

y = 20.922

Step-by-step explanation:

Simply plug in <em>x</em> and evaluate:

y = 15 + 3ln(7.2)

Only way to calculate ln is to plug into a calc:

We should get 20.9222 as our answer.

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3 years ago

What set of refections would carry trapezoid ABCD onto itself?

BaLLatris [955]

Hello!

Trapezoid ABCD is shown. A is at negative 5, 1. B is at negative 4, 3. C is at negative 2, 3. D is at negative 1, 1.
A) x-axis, y=x, y-axis, x-axis
b) x-axis, y-axis, x-axis
c) y=x, x-axis, x-axis
d) y-axis, x-axis, y-axis, x-axis
The best answer is C180 rotation wud take that point to 4th quadrant
reflection in x-axis takes that to 1st quadrant
<span>reflection in y-axis brings it back to 2nd quadrant again. So, the sequence of transformations will bring A back to where it started
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3 years ago

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What is 100×3857×29473

marshall27 [118]

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3 years ago

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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.

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3 years ago