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3 years ago

If the pattern continues, what would be the seventh number in the pattern?

Mathematics

2 answers:

trasher [3.6K]3 years ago

5 0

Answer: The answer is A. 76

Luda [366]3 years ago

3 0

The answer would be 76

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An arithmetic sequence has the following properties:

WINSTONCH [101]

Answer:

Easy peasy

Step-by-step explanation:

7+(n-1)5

3 0

3 years ago

2x + 4y = 22<br> What is the answer?

Goryan [66]

Answer:

Step-by-step explanation:

y=-\frac{1}{2}x+\frac{11}{2}

5 0

2 years ago

Read 2 more answers

Beatrice built about 1/3 of a sandcastle. Linda built 4/7 of the same castle. What fraction of the sandcastle did they build tog

wlad13 [49]

Answer

Find out the what fraction of the sandcastle did they build together .

To prove

Let us assume that the fraction of the sandcastle they build together be x .

As given

$Beatrice\ built\ about\ \frac{1}{3}\ of\ a\ sandcastle.$

$Linda\ built\ \frac{4}{7}\ of\ the\ same\ castle.$

Than the equation becomes

$x = \frac{1}{3} +\frac{4}{7}$

L.C .M of (3,7) = 21

solving

$x = \frac{7 + 3\times4}{21}$

$x = \frac{19}{21}$

Therefore the fraction of the sandcastle did they build together be

$x = \frac{19}{21}$

Hence proved

4 0

3 years ago

A production process fills containers by weight. Weights of containers are approximately normally distributed. Historically, the

forsale [732]

Answer:

option (c) n = 201

Step-by-step explanation:

Data provided in the question:

Standard deviation, s = 5.5 ounce

Confidence level = 99%

Length of confidence interval = 2 ounces

Therefore,

margin of error, E = (Length of confidence interval ) ÷ 2

= 2 ÷ 2

= 1 ounce

Now,

E = $\frac{zs}{\sqrt n}$

here,

z = 2.58 for 99% confidence interval

n = sample size

thus,

1 = $\frac{2.58\times5.5}{\sqrt n}$

n = (2.58 × 5.5)²

n = 201.3561 ≈ 201

Hence,

option (c) n = 201

5 0

3 years ago

Evaluate the integral. (remember to use absolute values where appropriate. Use c for the constant of integration.) 5 cot5(θ) sin

ozzi

$I=5\int \frac{cos^{4}\theta }{sin\theta }\times cos\theta d\theta \\\\I=5\int \left ( 1-sin^{2}\theta \right )^{2}\times \frac{cos\theta }{sin\theta }d\theta \\put\ \sin\theta =t\\\\dt=cos\theta d\theta \\\\I=5\int\frac{t^{4}+1-2t^{2}}{t}dt\ \ \ \ \ \ \ \ \ \ \because (a-b)^2=a^2+b^2-2ab\\\\I=5\left ( \int t^{3}dt + \int \frac{1}{t} -2\int t \right )dt$

by using the integration formula

we get,

$\\I=5\left ( \frac{t^{4}}{4} +logt -t^{2}\right )\\\\I=\frac{5}{4}t^{4}+5\log t-5t^{2}+c$

now put the value of t=\sin\theta in the above equation

we get,

$\int 5\cot^5\theta \sin^4\theta d\theta=\frac{5}{4}sin^{4}\theta+5\log \sin\theta - 5sin^{2} \theta+c$

hence proved

7 0

2 years ago