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Yuliya22 [10]
3 years ago
15

Please help me asap

Mathematics
1 answer:
Arada [10]3 years ago
6 0

Answer:

See explanation

Step-by-step explanation:

Car A: Started at 0 and ended at 300, thus, car A travels 300 miles.

It travels 6 hours, so car A speed is \frac{300}{6}=50 mph.

Car B: Started at 100 and ended at 300, thus, car B travels 300-100=200 miles.

It travels 5 hours, so car B speed is \frac{200}{5}=40 mph.

Since 50>40, car A traveled faster than car B.

The graph for the car A is steeper than the graph for the car B.

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Find the four arithmetic means between -21 and -36.
lara31 [8.8K]
The numbers given in the problem above are part of an arithmetic sequence with first and sixth terms equal to -21 and -36, respectively. Firstly, calculate for the common difference (d).
 
                                    d = (-36 - -21) / (6 - 1) = -3

The arithmetic mean is calculated by adding -3 to the term prior to it. 

            a2 = -21 + -3 = -24               a3 = -24 + -3 = -27
            a4 = -27 + -3 = -30               a5 = -30 + -3 = -33

Thus the four arithmetic means are -24, -27, -30, and -33.
4 0
3 years ago
This math my graduation depends on it
In-s [12.5K]

In an arithmetic sequence, consecutive terms have a fixed distance d between them. If a₁ is the first term, then

2nd term = a₂ = a₁ + d

3rd term = a₃ = a₂ + d = a₁ + 2d

4th term = a₄ = a₃ + d = a₁ + 3d

and so on, up to

nth term = a_n = a_{n-1} + d = a_{n-2} + 2d = a_{n-3} + 3d = \cdots = a_1 + (n-1)d

so that every term in the sequence can be expressed in terms of a₁ and d.

6. It's kind of hard to tell, but it looks like you're given a₁₃ = -53 and a₃₅ = -163.

We have

a₁₃ = a₁ + 12d = -53

a₃₅ = a₁ + 34d = -163

Solve for a₁ and d. Eliminating a₁ and solving for d gives

(a₁ + 12d) - (a₁ + 34d) = -53 - (-163)

-22d = 110

d = -5

and solving for a₁, we get

a₁ + 12•(-5) = -53

a₁ - 60 = -53

a₁ = 7

Then the nth term is recursively given by

a_n = a_{n-1}-5

and explicitly by

a_n = 7 + (n-1)(-5) = 12 - 5n

7. We do the same thing here. Use the known terms to find a₁ and d :

a₁₉ = a₁ + 18d = 15

a₃₈ = a₁ + 37d = 72

⇒   (a₁ + 18d) - (a₁ + 37d) = 15 - 72

⇒   -19d = -57

⇒   d = 3

⇒   a₁ + 18•3 = 15

⇒   a₁ = -39

Then the nth term is recursively obtained by

a_n = a_{n-1}+3

and explicitly by

a_n = -39 + (n-1)\cdot3 = 3n-42

8. I won't both reproducing the info I included in my answer to your other question about geometric sequences.

We're given that the 1st term is 3 and the 2nd term is 12, so the ratio is r = 12/3 = 4.

Then the next three terms in the sequence are

192 • 4 = 768

768 • 4 = 3072

3072 • 4 = 12,288

The recursive rule with a₁ = 3 and r = 4 is

a_n = 4a_{n-1}

and the explicit rule would be

a_n = 3\cdot4^{n-1}

7 0
2 years ago
Mr.Davis's students were surveyed on their favorite subject. One fifth prefer math, 0.35 prefer social studies, and 45% prefer r
lara [203]

its already in order


7 0
2 years ago
Read 2 more answers
1. Find 3 prime factorization of the number 68.
n200080 [17]
1x68
2x34
4x17
Hope that helps
8 0
2 years ago
Read 2 more answers
A used electronics store pays $30 for each cell phone, spends $20 refurbishing the phone, and sells the refurbished phone to a c
Amiraneli [1.4K]


I am subtracting the amount it took to make the phone, from the amount they sell the phone to a customer for.
115-(30+20)=x


115-50=x

65=x





8 0
3 years ago
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