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zalisa [80]
2 years ago
8

What is b and c in this expression?

le="\sqrt[4]{15^7} }" alt="\sqrt[4]{15^7} }" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
ozzi2 years ago
8 0

The value of b and  c is 4 and 7.

<h3>What is algebra?</h3>

Algebra is the part of mathematics in which letters and other general symbols are used to represent numbers and quantities in formulae and equations.

Algebra is divided into different sub-branches such as elementary algebra, advanced algebra, abstract algebra, linear algebra, and commutative algebra.

Elementary algebra: elementary algebra, branch of mathematics that deals with the general properties of numbers and the relations between them. Algebra is fundamental not only to all further mathematics and statistics but to the natural sciences, computer science, economics, and business.

From the given question,

\sqrt[4]{15^7} This expression can also be written in the form of a^\frac{b}{c}

We will convert \sqrt[4]{15^7} to the required form:

(15^\frac{7c}{4b} )\left \{ {{a=15^\frac{7c}{4b} } \atop {b=b} , c = c} \right.

Hence,

From the above equation we get the value of b and c is 4 and 7.

Learn more about algebra here:brainly.com/question/17260216

#SPJ1

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In the Midpoint Rule for triple integrals we use a triple Riemann sum to approximate a triple integral over a box B, where f(x,
Tomtit [17]

To approximate the volume with 8 boxes, we have to split up the interval of integration for each variable into 2 subintervals, [0, 1] and [1, 2]. Each box will have midpoint m_{i,j,k} that is one of all the possible 3-tuples with coordinates either 1/2 or 3/2. That is, we're sampling f(x,y,z)=\cos(xyz) at the 8 points,

(1/2, 1/2, 1/2)

(1/2, 1/2, 3/2)

(1/2, 3/2, 1/2)

(3/2, 1/2, 1/2)

(1/2, 3/2, 3/2)

(3/2, 1/2, 3/2)

(3/2, 3/2, 1/2)

(3/2, 3/2, 3/2)

which are captured by the sequence

m_{i,j,k}=\left(\dfrac{2i-1}2,\dfrac{2j-1}2,\dfrac{2k-1}2\right)

with each of i,j,k being either 1 or 2.

Then the integral of f(x,y,z) over B is approximated by the Riemann sum,

\displaystyle\iiint_B\cos(xyz)\,\mathrm dV\approx\sum_{i=1}^2\sum_{j=1}^2\sum_{k=1}^2\cos m_{i,j,k}\left(\frac{2-0}2\right)^2

=\displaystyle\sum_{i=1}^2\sum_{j=1}^2\sum_{k=1}^2\cos\frac{(2i-1)(2j-1)(2k-1)}8

=\cos\dfrac18+3\cos\dfrac38+3\cos\dfrac98+\cos\dfrac{27}8\approx\boxed{4.104}

(compare to the actual value of about 4.159)

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3 years ago
I'm supposed to find the volume of a cube but the only number I am given is 6. Do I multiply 6 three times?
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V= a^3
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