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maw [93]
3 years ago
5

find two consecutive odd intergers that twice the larger is fifteen more than three times the smaller

Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
6 0

Answer:

11, 13

Step-by-step explanation:

Let n and n+2 be the consecutive odd integers

2(n+2) = 3n + 15

2n + 4 = 3n + 15

n = 11

n + 2 = 13

Svetach [21]3 years ago
4 0

Answer:

11, 13

Step-by-step explanation:

an odd number can be represented by 2n+1

since they are consecutive, the larger odd number will be 2n + 1 + 2

now,

2(2n+3) = 2n + 1 +15

solving this eqn, we get n = 5

so the two numbers are (2n+1) = 11 and 11+2= 13

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3 years ago
Is 9/10 larger than 13/14
Reptile [31]

Answer:

13/14 is greater

Step-by-step explanation:

Find the lcm for both:

<u>9 x 7 </u>     = <u>63</u>

10 x 7        70

<u>13 x 5 </u>    = <u>65</u>

14 x 5        70

so therefore 13/14 is greater since it has a greater value when finding the lcm

5 0
3 years ago
Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!!
LUCKY_DIMON [66]

Answer:

=8\sqrt{15}b^{\frac{7}{2}}

Step-by-step explanation:

\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}

=\sqrt{40}\sqrt{b^2}\sqrt{b^2}\sqrt{24b^3}

=\sqrt{40}b^2\sqrt{24b^3}

\sqrt{24b^3}

=\sqrt{24}\sqrt{b^3}

=\sqrt{24}b^{\frac{3}{2}}

=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2

=\sqrt{2^3\cdot \:3}b^{\frac{3}{2}}\sqrt{40}b^2

=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

\sqrt{2^3}

=2^{3\cdot \frac{1}{2}

=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3}\sqrt{5}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}

2^{\frac{3}{2}+\frac{3}{2}}

=2^3

=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}

b^{\frac{3}{2}+2}

=b^{\frac{7}{2}}

=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}

=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}

=8\sqrt{15}b^{\frac{7}{2}}

4 0
3 years ago
Which two expressions are equivalent?
Mnenie [13.5K]

Answer:

I believe it's the first one

4 0
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