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Ugo [173]
3 years ago
6

Find the common difference in the arithmetic sequence where p35 =-9p + 4 and p28 = 5p - 17?

Mathematics
1 answer:
Sphinxa [80]3 years ago
6 0
P(35) = -9p + 4 = -9(35) + 4 = -315 + 4 = -311
p(28) = 5p - 17 = 5(28) - 17 = 140 - 17 = 123

a + 34d = -311
a + 27d = 123

7d = -434
d = -62

Therefore, the commpm difference is -62.
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Can someone help<br> Me please?
Sindrei [870]

9514 1404 393

Answer:

  • minimum: 2 at x=0
  • maximum: 10 at x=10

Step-by-step explanation:

When looking for extremes, one must consider both the turning points and the ends of the interval. Here, there is a relative minimum at x=7, and a relative maximum at x=3. However, the values at the ends of the interval are more extreme than these.

The absolute minimum on the interval is 2 at x=0.

The absolute maximum on the interval is 10 at x=10.

8 0
3 years ago
How do I solve<br> -3x+y=24<br> -2x=+18+5y
ExtremeBDS [4]

Answer:

Let's solve by eliminating x. Since 18 is the lowest common multiple , let's multiply the first equation by 3 and the second by -2.

18x−21y=24

−18x+10y=−2

Add the two equations.

−11y=22

Divide by -11 to isolate y.

y=−2

Substitute y=−2 into either of the 2 equations, I'll just use the second one.

9x+10=1

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x=−1

Step-by-step explanation: that is your anwser

3 0
3 years ago
HELP will give branliest!
amm1812

Answer:

perimeter 140

Step-by-step explanation:

60L x 10W = 600

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8 0
3 years ago
A x + b y = c what are the variables?
Murljashka [212]
Hey!
If A, B, C are representing numbers then they are not the variables. The variables are x and y.
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7 0
3 years ago
Read 2 more answers
How many different strings can be formed by rearranging the letters in the word DCDCD?
Natali [406]
<h2>Answer:</h2>

The different strings which can be formed by rearranging the letters in the word DCDCD are:

                                     10

<h2>Step-by-step explanation:</h2>

We are asked to find the different number of arrangements that may be me made in the word:

                               DCDCD

There are a total of 5 words such that D is repeated 3 times and C is repeated two times.

Hence, the total number of ways of arrangement is given by:

=\dfrac{5!}{3!\times 2!}\\\\=\dfrac{5\times 4\times 3!}{3!\times 2!}\\\\=\dfrac{5\times 4}{2!}\\\\=\dfrac{5\times 4}{2}\\\\=10

Hence, the total number of ways are:  10

3 0
3 years ago
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