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3 years ago

Integrate x^2+5 please i need it fast

Mathematics

1 answer:

Vinvika [58]3 years ago

6 0

Answer:

$\frac{1}{3}$ x³ + 5x + c

Step-by-step explanation:

Using the power rule

∫ a $x^{n}$ dx = $\frac{a}{n+1}$ $x^{n+1}$

Thus

∫ x² + 5 dx

= $\frac{1}{3}$ x³ + 5x + c ← c is the constant of integration

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Select the correct answer.

Nat2105 [25]

Answer:

cndjxnn

Step-by-step explanation:

vncnfxjdncnnxxjjxnxxnjxxjdnxnxjxj, x

7 0

3 years ago

A man sells a type of nut for $7 per pound and a different one for $4.20 per pound, how much of each type should be used to make

qwelly [4]

Answer:

a type nut is 10 pounds

a different one is 14 pounds

Step-by-step explanation:

let a type of the nut be represented by t

Let a different one be represented by d

a type of nut cost $7 per pound

a different one cost $4.20 per pound

The cost of the mixture for 24 pounds = 5.37 * 24

= $128.88

t + d = 24 ........(1)

7t + 4.2d = 128.88 ..........(2)

From equation (1), t = 24 - d

Put t = 24 - d in equation 2

7(24 - d) + 4.2d = 128.88

168 - 7d + 4.2d = 128.88

168 - 2.8d = 128.88

-2.8d = 128.88 - 168

-2.8d = -39.12

d = -39.12 / -2.8

d= 13.97

d = 14 pounds

t = 24 - d

t = 24 - 14

t = 10 pounds

A type nut is 10 pounds. A different one is 14 pounds

5 0

3 years ago

Read 2 more answers

What is the simplest form for the fractions 12/30,12/20 and 21/35

Yuki888 [10]

Simplest form of:
12/30=2/5
12/20=3/5
21/35=3/5

6 0

3 years ago

A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0

3 years ago

HELP

san4es73 [151]

Answer:

$300 debit to Advertising Expense

Step-by-step explanation:

7 0

2 years ago