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3 years ago

Seema used compatible numbers to estimate the product of (–25.31)(9.61). What was her estimate?

Mathematics

2 answers:

zepelin [54]3 years ago

8 0

If she used rounding properly, she would have gotten -25 x 10 = -250

Ratling [72]3 years ago

3 0

Answer:

-250

Step-by-step explanation:

just did the test

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Find the percent increase. Round to the nearest percent.

lys-0071 [83]

Answer:

32%

Step-by-step explanation:

at least twenty caters long

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3 years ago

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GrogVix [38]

Answer:

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Step-by-step explanation:

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1 year 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

Simplify 2a+2−3a−5+a−4+7a.

zhuklara [117]

Answer:

7a-7

Step-by-step explanation:

2a+2−3a−5+a−4+7a.

Combine like terms

2a-3a+a+7a+2-5-4

7a-7

3 0

3 years ago

Read 2 more answers

How to determine if a function is even or odd calculus?

aalyn [17]

A function $f(x)$ is even if $f(-x)=f(x)$ , and is odd if $f(-x)=-f(x)$ .

Example:

$f(x)=x^2$ is even, since $f(-x)=(-x)^2=x^2=f(x)$ .

$g(x)=x^3-x$ is odd, since $g(-x)=(-x)^3-(-x)=-x^3+x=-(x^3-x)=-g(x)$ .

6 0

3 years ago