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Troyanec [42]
3 years ago
7

Simplify the expression 3a + b, when a=1.4 -2x and b=-0.2x + 1.7*

Mathematics
2 answers:
Aleksandr-060686 [28]3 years ago
7 0

Answer:

Step-by-step explanation:

3a + b

a = 1.4 - 2x

b = -0.2x + 1.7

3 ( 1.4 - 2x ) +  -0.2x + 1.7

= 4.2 - 6x - 0.2x + 1.7

= 4.2 + 1.7 - 6x - 0.2x ( by rearranging the like terms )

= <u><em>5.9 - 6.2 x</em></u>

hope this helps

plz mark as brainliest!!!!!!!!

kenny6666 [7]3 years ago
3 0

Answer:5.9-6.2x

Step-by-step explanation:

a=1.4-2x

b=-0.2x+1.7

 3a+b

=3(1.4-2x)+(-0.2x+1.7)

=4.2-6x-0.2x+1.7

=4.2+1.7-6x-0.2x

=5.9-6.2x

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Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​
taurus [48]

Answer:

12942 is the sum of positive integers between 1 (inclusive) and 199 (inclusive) that are not multiples of 4 and not multiples 7.

Step-by-step explanation:

For an arithmetic series with:

  • a_1 as the first term,
  • a_n as the last term, and
  • d as the common difference,

there would be \displaystyle \left(\frac{a_n - a_1}{d} + 1\right) terms, where as the sum would be \displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n).

Positive integers between 1 (inclusive) and 199 (inclusive) include:

1,\, 2,\, \dots,\, 199.

The common difference of this arithmetic series is 1. There would be (199 - 1) + 1 = 199 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}.

Similarly, positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 4 include:

4,\, 8,\, \dots,\, 196.

The common difference of this arithmetic series is 4. There would be (196 - 4) / 4 + 1 = 49 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 7 include:

7,\, 14,\, \dots,\, 196.

The common difference of this arithmetic series is 7. There would be (196 - 7) / 7 + 1 = 28 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 28 (integers that are both multiples of 4 and multiples of 7) include:

28,\, 56,\, \dots,\, 196.

The common difference of this arithmetic series is 28. There would be (196 - 28) / 28 + 1 = 7 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}.

The requested sum will be equal to:

  • the sum of all integers from 1 to 199,
  • minus the sum of all integer multiples of 4 between 1\! and 199\!, and the sum integer multiples of 7 between 1 and 199,
  • plus the sum of all integer multiples of 28 between 1 and 199- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

19900 - 4900 - 2842 + 784 = 12942.

8 0
3 years ago
..1+4=5. 2+5=12. 3+6=21. 8+11=40 or this could be 96 which is right and why
andre [41]
Multiply the numbers being added then add the first number. 4X1=4, 4+1=5.
2X5=10, 10+2=12. 11X8=88, 88+8=96. I'm not sure how to get 40, though. 
Hope I could help.
7 0
2 years ago
Match each verbal description to its corresponding expression.
Aliun [14]
I need points thank you
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3 years ago
Which is a true statement?
myrzilka [38]
B is correct because -6.75 is smaller than -6.5
8 0
3 years ago
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Zigmanuir [339]

Answer:

the system of equation has infinite solution

Step-by-step explanation:

5x_1 + x_2 = 0 , 25x_1 + 5x_2 = 0

Solve the first equation for x_2

Subtract 5x1 on both sides

5x_1 + x_2 = 0

x_2 =-5x_1

Now substitute -5x1 on second equation

25x_1 + 5x_2 = 0

25x_1 + 5(-5x_1)= 0

25x_1-25x_1= 0

0=0

So the system of equation has infinite solution

5 0
3 years ago
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