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4 years ago

What is the 7th term in the sequence?

2 answers:

5 0

To determine the 7th term of a given sequence, we use the formula given above which is

an = 30 - 4n

where an is the value at the nth term of a sequence

Therefore,

an = 30 - 4n
an = 30 - 4(7)
an = 2

Andreas93 [3]4 years ago

5 0

Hello there.

What is the 7th term in the sequence?

Note: Use the explicit rule.

an=30−4nan=30−4n

Enter your answer in the box.
a7=a7=

an = 2

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Round 776644 to the nearest thousand

prohojiy [21]

Answer:

776644 rounds to 777000

Step-by-step explanation:

hundred thousand ten thousand thousand hundred tens ones

7 7 6 6 4 4

We are rounding to the nearest thousand so we are rounding the first 6. We need to look at the next 6 (the one in the hundreds place) 6>=5 so we will round the 6 in the thousands place up to 7

776644 rounds to 777000

6 0

4 years ago

Can somebody please help me!!

Free_Kalibri [48]

Answer:

C≈15.71

Step-by-step explanation: use the circumference formula

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3 years ago

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If a drug's recommended dose is 5 mg/kg, and a solution contains a concentration of 2 mg/mL of the drug, what volume of the solu

Luba_88 [7]

Answer:

Option B. 9 ml.

Step-by-step explanation:

Recommended dose of a drug = 5 mg/kg body weight.

weight of the cat = 8 lb

1 lb = 0.4536 kg.

8 lb = 0.4536 × 8 = 3.6288 kg. ≈ 3.6 kg

A solution contains a concentration of 2 mg/ml

so recommended dose for 3.63 kg = 5mg × 3.6 = 18 mg

∵ 1 ml. contains a concentration = 2 mg

∴ for 18 mg concentration dose will be = 18/2 = 9 ml

9 ml of the solution should be given to a cat.

Option B. 9 ml.

4 0

3 years ago

Explain why each of the following integrals is improper. (a) 4 x x − 3 dx 3 Since the integral has an infinite interval of integ

erma4kov [3.2K]

Answer:

Since the integral has an infinite discontinuity, it is a Type 2 improper integral

Since the integral has an infinite interval of integration, it is a Type 1 improper integral

Since the integral has an infinite discontinuity, it is a Type 2 improper integral

Step-by-step explanation:

Considering a

$\int\limits^4_3 {\frac{x}{x- 3} } \, dx$

Looking at this we that at x = 3 this integral will be infinitely discontinuous

Considering b

$\int\limits^{\infty}_0 {\frac{1}{1 + x^3} } \, dx$

Looking at this integral we see that the interval is between $0 \ and \ \infty$ which means that the integral has an infinite interval of integration , hence it is a Type 1 improper integral

Considering c

$\int\limits^{\infty}_{- \infty} {x^2 e^{-x^2}} \, dx$

Looking at this integral we see that the interval is between $-\infty \ and \ \infty$ which means that the integral has an infinite interval of integration , hence it is a Type 1 improper integral

Considering d

$\int\limits^{\frac{\pi}{4} }_0 {cot (x)} \, dx$

Looking at the integral we see that at x = 0 cot (0) will be infinity hence the integral has an infinite discontinuity , so it is a Type 2 improper integral

7 0

3 years ago

2+7347 and much is it

Lana71 [14]

7349 that's the answer and really? That was easy

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3 years ago

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