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Maslowich
3 years ago
13

Find two consecutive odd integers whose product is 323

Mathematics
1 answer:
Harrizon [31]3 years ago
4 0
Use a let statement 
first
let x and x + 2 be the number so you write it like this
<u>let x = the first consecutive integer
</u><u>let x + 2 = the second consecutive integer
</u>
second
x(x+2)=323
x^2 + 2x = 323
        -323  -323
x^2 + 2x -323 = 0

third
try to factor -323 so it is 19 and -17
(x + 19) (x - 17) = 0
x = 19 
x = -17

hope this help

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MatroZZZ [7]

For this case we must simplify the following expression:

\frac {6-3 \sqrt [3] {6}} {\sqrt [3] {9}}

Multiplying the numerator and denominator by(\sqrt [3] {9}) ^ 2

\frac {6-3 \sqrt [3] {6}} {\sqrt [3] {9}} * \frac {(\sqrt [3] {9}) ^ 2} {(\sqrt [3] { 9}) ^ 2} =

We rewrite:

\frac {\frac {6-3 \sqrt [3] {6}} * (\sqrt [3] {9}) ^ 2} {\sqrt [3] {9} * (\sqrt [3] {9 }) ^ 2} =

By properties of powers we have that:

a ^ m * a ^ n = a ^ {m + n}\\\frac {(6-3 \sqrt [3] {6}) * (\sqrt [3] {9}) ^ 2} {(\sqrt [3] {9}) ^ 3} =\\\frac {(6-3 \sqrt [3] {6}) * (\sqrt [3] {9}) ^ 2} {9} =

We rewrite, moving the exponent within the radical:

\frac {(6-3 \sqrt [3] {6}) * \sqrt [3] {9 ^ 2}} {9} =\\\frac {(6-3 \sqrt [3] {6}) * \sqrt [3] {81}} {9} =

We can rewrite3 * 3 ^ 3 = 81

\frac {(6-3 \sqrt [3] {6}) * \sqrt [3] {3 * 3 ^ 3}} {9} =

We simplify:

\frac {(6-3 \sqrt [3] {6}) * 3 \sqrt [3] {3}} {9} =

We apply distributive property:

\frac {18 \sqrt [3] {3} -9 \sqrt [3] {18}} {9} =

Simplifying we finally have:

2 \sqrt [3] {3} - \sqrt [3] {18}

Answer:

2 \sqrt [3] {3} - \sqrt [3] {18}

5 0
3 years ago
In the geometric pattern below, n represents the number of blocks in the bottom row of each figure.
MrMuchimi
Figure 1
n=1→f(n)=f(1)=1

Options 2 or 3

Figure 2
n=2→f(n)=f(2)=5

With Option 2:
f(n)=f(n-1)+2n
n=2→f(2)=f(2-1)+2(2)
f(2)=f(1)+4=1+4→f(2)=5  ok

With Option 3
f(n)=f(n-1)+n
n=2→f(2)=f(2-1)+2
f(2)=f(1)+2=1+2→f(2)=3  No

Answer: Second option: f(1)=1; f(n)=f(n-1)+2n; for n>=2




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