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Luda [366]
3 years ago
7

What's the answer for this multiplication problem????

Mathematics
1 answer:
borishaifa [10]3 years ago
6 0

Answer:

  see below

Step-by-step explanation:

You know the leading term will be the product of leading terms, so is ...

  (x^2)(3x^2) = 3x^4 . . . . . matches choices A, B, C

The x^3 term of the product will be the sum of the products of x and x^2 terms, so is ...

  (x^2)(2x) +(3x^2)(-5x) = 2x^3 -15x^3 = -13x^3 . . . . . matches choice A only

With very little work, we have identified the only viable answer choice:

  3x^4 -13x^3 -x^2 -11x +6

_____

You can work out the product using the distributive property 4 times: multiply each term of one polynomial by all terms of the other. Then collect terms.

A reasonable alternative is to identify the partial products that will make up any given term of the answer. Above we have shown how to find the x^4 and x^3 terms. The x^2 term will be the sum of products (x^2)(constant) +(x)(x), for a total of 3 contributors to that. Similar to the x^3 term, the x term of the product will be the sum of products (x)(constant). Of course, the final constant term in the result is only the product of the constants in each factor.

If you go about this systematically, then errors will not creep in, regardless of which method you use.

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3 years ago
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In ABC, D∈AB so that AD;DB=1 : 3. If ACDB = 12ft2 Find AACD and AABC
Sergeeva-Olga [200]

Answer:

Area of \triangleACD = 4 ft^2

Area of \triangleABC = 16 ft^2

Step-by-step explanation:

Given that:

D is a point on AB.

and ABC is a triangle.

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Area of \triangleCDB = 12 ft^2

Kindly refer to the attached image as per the given dimensions and values.

To find:

Area of \triangleACD and Area of \triangleABC = ?

Solution:

Formula for area of a triangle = \frac{1}{2}\times Base \times Height

The altitudes of triangles \triangleCDB and \triangleACD are equal in dimensions.

Therefore the area of triangles \triangleCDB and \triangleACD will be equal to the ratio of their bases.

Area of \triangleACD : Area of \triangleCDB = AD: DB = 1 : 3

\Rightarrow Area of \triangleACD = \frac{12}{3} = \bold{4 ft^2}

Area of \triangleABC = Area of \triangleACD + Area of \triangleCDB = 12 + 4 = <em>16</em> ft^2

Therefore, the answer is:

Area of \triangleACD = 4 ft^2

Area of \triangleABC = 16 ft^2

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