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3 years ago

What is the common difference for this arithmetic sequence? 54, 50, 46, 42, 38, A. 34 B. 4 C. 54 D. -4

Mathematics

2 answers:

Vaselesa [24]3 years ago

8 0

Answer:

-4

Step-by-step explanation:

Take the second term and subtract the first term

50 -54 = -4

We can check by taking the third term and subtracting the second term

46-50 = -4

The common difference is -4

olganol [36]3 years ago

8 0

If we will analyze, we will notice a standard:

From 54 to 50 we subtracted 4 --> (-4);
From 50 to 46 we subtracted 4 --> (-4);

This is repeated until last term of sequence.

Terefore, the correct alternative is letter d).

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Write the following as a fraction in simple form <br><br>1.6

mestny [16]

Hi!
im not really sure what the answer is but i got 1 3/5
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7 0

3 years ago

Read 2 more answers

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

a hat contains 3 blue and 5 black tickets. if one ticket is chosen at random from the hat what is the probability that it is blu

Nikitich [7]

Answer:

3/8

Step-by-step explanation:

3/(3 + 5) = 3/8

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3 years ago

If 15 baseballs weighs 75 ounces how many baseballs with 15 ounces

igor_vitrenko [27]

Answer:

Step-by-step explanation:

75/15=5

Bam

5 0

3 years ago

Jermiah works at an appliance store. He reacantly sold 12 clothes dryers and 36 other appliances. Of the next 100 appliances he

Viefleur [7K]

12/36 = x/100
Then cross multiply
1200 = 36x
Divide by 36
x = 33.33

Can’t sell 0.33 of an appliance, so your answer is 33.

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3 years ago