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3 years ago

What number is 76% of 97

Mathematics

2 answers:

IrinaVladis [17]3 years ago

7 0

Answer:

73.72

Step-by-step explanation:

76x96/100=73.72

alexandr1967 [171]3 years ago

6 0

76%*97
=73.72

I hope this helps.

Comment if you need further explanation.

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Answer:

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2 years ago

Need help this is so confusing and please show work.

AlexFokin [52]

Answer:

Please see attached picture for full solution.

Step-by-step explanation:

You should multiply the number outside the bracket with each term, with their symbols (plus/minus), in the bracket.

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3 years ago

The area of a circle is 49 cm2. Find the circumference of the circle.

Nataly_w [17]

~Hello There!~

The formula to find the area of a circle is:

$\pi r^{2}$

We already know this answer, so we write it as:

$\pi r^{2} =49$

We need to calculate the radius (r) from this so divide both sides by $\pi$

$r^{2} = 15.59718442$

Square root this to get r

r = 3.949327085

The formula for circumference is $\pi *d$

(d is the diameter )

Double the radius to get the diameter:

d = 7.89865417

Multiply this value by $\pi$ to get the answer

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4 0

3 years ago

Read 2 more answers

If a = –2 – 5i and b = –i, then find the value of the ab^3 in fully simplified form.

Kitty [74]

Answer:

5 + 2i

We were given the complex numbers;

a = - 2 - 5i

and

b = - i

We want to find the product;

ab^3 ^ = Exponent

We substitute the complex numbers into the expression and simplify

( - 2 - 5i ) (i)^3 ^ = Exponent

This is rewritten as:

( - 2 - 5i) (i)^2 x i ^ = Exponent

Note that

i^2 = - 1 ^ = Exponent

We substitute to obtain:

( - 2 - 5i ) x - i

Let us expand to get:

- 2 x - i + 5i x - i

This simplifies to:

2i - 5i^2 ^ = Exponent

This gives:

2i - 5 ( - 1 ) = 2i + 5

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8 0

3 years ago

Perform long division on the integrand, write the proper fraction as a sum of partial fractions, and then evaluate the integra

worty [1.4K]

Answer:

$x^2+4x -\frac{1}{2} lnx + \frac{2}{13} ln(x-2) + C$

Step-by-step explanation:

Given the integrand $\int\limits{\dfrac{2x^3 - 2x + 1}{x^2-2x} } \, dx$ , before evaluating the integral function, we will need to simplify the function first by applying long division as shown in the attachment.

Hence the partial form of the function $\dfrac{2x^3 - 2x + 1}{x^2-2x} } = 2x+4 + \frac{6x+1}{x^2-2x}$

Integrating its partial sum

$\int\limits \dfrac{2x^3 - 2x + 1}{x^2-2x} }dx = \int\limits (2x+4 + \frac{6x+1}{x^2-2x})\ dx\\\\= \int\limits {2x} \, dx + \int\limits {4} \, dx + \int\limits {\frac{6x+1}{x^2-2x} \, dx\\ = \frac{2x^2}{2}+4x -\frac{1}{2} \int\limits{\frac{1}{x} } \, dx + \frac{2}{13} \int\limits{\frac{1}{x-2} } \, dx$

$= \frac{2x^2}{2}+4x -\frac{1}{2} lnx + \frac{2}{13} ln(x-2) + C$

$= x^2+4x -\frac{1}{2} lnx + \frac{2}{13} ln(x-2) + C$

<em>NB: Find the partial sum calculation also in the attachment. </em>

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3 years ago