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Sladkaya [172]
3 years ago
5

Adrian has already earned $66 washing and waxing cars. He earns $9 per car. He wants to save at least $250 for a new

Mathematics
1 answer:
pav-90 [236]3 years ago
5 0
175

66+9=75
250-75=175

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Write the first five terms of the sequence defined by the explicit formula
denis-greek [22]

<u>Answer:</u>

The correct answer option is: 19, 13, 3, -11, -29.

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

We are given the following explicit formula for an arithmetic sequence:

a_n=21-2n^2

So to find the first five terms of this sequence, we will substitute the values of n (num here in the given formula:

a_1=21-2(1)^2\\\\a_1=19


a_2=21-2(2)^2\\\\a_2=13


a_3=21-2(3)^2\\\\a_3=3


a_4=21-2(4)^2\\\\a_4=-11


a_5=21-2(5)^2\\\\a_5=-29

Therefore, the first five terms of the sequence defined by the explicit formula a_n=21-2n^2 are 19, 13, 3, -11, -29.


5 0
3 years ago
Read 2 more answers
If 3x is one factor of 3x^2-9x, what is the other factor?
Rudik [331]
The answer is:  " (x² − 3) " .
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Explanation:
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Given:  3x² <span>− 9x ;  factor out a "3x" ; 
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   </span>→  3x (x² − 3) ;
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     The answer is:  " (x² − 3) " .
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7 0
3 years ago
How do u solve 6x-(3+8x)-11 applying the distributive property?
Irina-Kira [14]
Distribute the negative:
6x-3-8x-11
Combine like terms:
-2x-14
Simplify by factoring:
-2(x+7)
5 0
3 years ago
If (a,b)=(4,6) then find the value of b.
MAVERICK [17]

Answer:

Concepts: Mathematical Understanding

  1. (a,b) is an ordered pair just like (x,y)
  2. Hence b= 6
8 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
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