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2 years ago

Forty-two less than a number z

Mathematics

2 answers:

Ann [662]2 years ago

6 0

42 less than a number z, means that z is greater than the number by 42

z - 42 is your expression

hope this helps

Tasya [4]2 years ago

4 0

42 less then a number z....

z - 42 <== ur expression

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Kathy measured her finger using a ruler with centimeters. Then she measured the same finger with a ruler using millimeters. Can

Stells [14]

Answer:

Yes, 10 millimeters is the same as 1 centimeter.

Step-by-step explanation:

Kathy can compare the two measurement because 10 millimeters equals 1 centimeter

This means

10 millimeters of her finger = 1 centimeter of the same finger

100 millimeters of her finger = 10 centimeters of her finger

1000 millimeters of her finger = 100 centimeters of her finger

Therefore, Kathy can compare both measurement ( millimeters and centimeters).

6 0

2 years ago

Read 2 more answers

Find the sum Sn of the first n terms from the given sequence.

VMariaS [17]

The Sequence:

1 , 1+2 , 1+2+3 , 1+2+3+4, ...

here, every term is an AP. finding the general formula for the sum of the elements of each term is:

Sₙ = $\frac{x}{2}(2a + (x-1)d)$ <em> [where x = number of term, a = first term and d = common difference]</em>

here, the first term is always 1 and so is the common difference.

Sₙ = $\frac{x}{2}(2 +x-1)$

Sₙ = $\frac{x}{2}(1 +x)$ = $\frac{1}{2}(x + x^{2})$

which is the formula for a general term in our series

now, we need to find the sum of the first n terms of this series

$\displaystyle\sum_{x=1}^{n} [\frac{1}{2}(x + x^{2})]$

$\displaystyle\frac{1}{2} [\sum_{x=1}^{n} (x) + \sum_{x=1}^{n}(x^{2})]$

in this formula, for the first term, it's just an AP from x = 1 to x = n

for the second term, we have a general formula $\frac{n(n+1)(2n+1)}{6}$

$\frac{1}{2}[\frac{n}{2}(2a + (n-1)d)+ \frac{n(n+1)(2n+1)}{6} ]$

in this AP (first term), the first term and the common difference is 1 as well

$\frac{1}{2}[\frac{n}{2}(2 + n-1)+ \frac{n(n+1)(2n+1)}{6} ]$

$\frac{1}{2}[\frac{n}{2}(n+1)+ \frac{n(n+1)(2n+1)}{6} ]$

$[\frac{n}{4}(n+1)+ \frac{n(n+1)(2n+1)}{12} ]$

$\frac{n}{4}(n+1) [1+\frac{(2n+1)}{3} ]$

$\frac{n}{4}(n+1) [\frac{(3+2n+1)}{3} ]$

$\frac{n}{4}(n+1) [\frac{(2n+4)}{3} ]$

$\frac{n}{2}(n+1) [\frac{(n+2)}{3} ]$

$\frac{n(n+1)(n+2)}{6}$

which is the sum of n terms of the given sequence

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