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3 years ago

How to multiply fractions with a mixed number?

1 answer:

4 0

Okay here's an example so if we have 2, 3/2*1/2 we convert the mixed numbers to improper fractions. So in order to solve a mixed number you add the denominator, numerator and whole to get 7/2*1.2
So Multiple
7/2*1/2
Refine the fractions
7/2*2
Multiple the numbers 2*2=4 to get
7/4
and that equals 1, 3/4

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Find the simplified product. 2V5x^3(-3/10x^2)

mariarad [96]

Answer:

Step-by-step explanation:

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3 years ago

Clara has a garden that is 7 feet wide and 4 feet long she has 30 tomato plants to put in the garden each plant needs one square

ankoles [38]

A = L * W
A = 7 * 4
A = 28 sq ft....this is the area of her garden

and if each plant uses 1 sq ft, then there will be (30 - 28) = 2 plants leftover

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3 years ago

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Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

so the left side is

$2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n$

Also recall that

$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

so that the right side is

$b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)$

Solve for $c$ .

$2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}$

Now, the numerator increases more slowly than the denominator, since

$\dfrac{d}{dn}(2n(n-1)) = 4n - 2$

$\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2$

and for $n\ge5$ ,

$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

This means we only need to check if the claim is true for any $n\in\{1,2,3,4\}$ .

$n=1$ doesn't work, since that makes $c=0$ .

If $n=2$ , then

$c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0$

If $n=3$ , then

$c = \dfrac{12}{2^3 - 6 - 1} = 12$

If $n=4$ , then

$c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N$

There is only one value for which the claim is true, $c=12$ .

3 0

2 years ago

What type of polygon would a slice of a hexahedron at a vertex create? Explain. What type of polygon would a slice of an icosahe

harina [27]

Answer:

hexahedron: triangle or quadrilateral or pentagon
icosahedron: quadrilateral or pentagon

Step-by-step explanation:

Hexahedron

A hexahedron has 6 faces. A regular hexahedron is a cube. 3 square faces meet at each vertex.

If the hexahedron is not regular, depending on how those faces are arranged, a slice near a vertex may intersect 3, 4, or 5 faces. The first attachment shows 3- and 4-edges meeting at a vertex. If those two vertices were merged, then there would be 5 edges meeting at the vertex of the resulting pentagonal pyramid.

A slice near a vertex may create a triangle, quadrilateral, or pentagon.

Icosahedron

An icosahedron has 20 faces. The faces of a regular icosahedron are all equilateral triangles. 5 triangles meet at each vertex.

If the icosahedron is not regular, depending on how the faces are arranged, a slice near the vertex may intersect from 3 to 19 faces.

A slice near a vertex may create a polygon of 3 to 19 sides..

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3 years ago

True or false, there are two solutions to the equation |x-21|= -7

notsponge [240]

Answer:

false

Step-by-step explanation:

4 0

3 years ago

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