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Mathematics

1 answer:

marishachu [46]2 years ago

6 0

1) is A, parellel lines are lines that will never touch

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Classifying quadrilaterals

zmey [24]

Answer:

A quadrilateral is a polygon with four sides.

Practicly anything with 4 sides like square, Rectangle ,and rhombas

Im sorry if i didn't answer your question but i did not get what you were saying but here is an image that can maybe help you

7 0

2 years ago

What are the domain and range of the function f(x)= x^4 -6x^3 -4x^2 54x -45?

Agata [3.3K]

Domain is the numbers you can use
we can use all real numbers for this one

rangge is the numbers we get from inputting the domain

well, this is a 4th degree, so we need to find the minimum,because the leading coefient is positive so it opens up, and as x approaches negative and positive infinity, then f(x) approaches infnity

find minimum
take derivitive
f'(x)=4x^3-18x^2-8x+54
the zeroes are at about -1.652 and 1.9381 and 4.2145
we use a sign chart
the minimum occurs at hwere the derivitive changes from negative to positive
that is at -1.652 and 4.2145
evaluate f(-1.652) and f(4.2145)
f(-1.652)=-110.626
f(4.2145)=-22.124
the least value is -110.626
that is the minimum

so the domain is all real numbers
range is from -110.626 to infinity

8 0

3 years ago

What are the solutions of the equation 9x4 – 2x2 – 7 = 0? Use u substitution to solve. tions of the equation 9

skelet666 [1.2K]

Answer:

Step-by-step explanation:

Let $u^2=x^4\\u = x^2$

Subbing in:

$9u^2-2u-7=0$

a = 9, b = -2, c = -7

The product of a and c is the aboslute value of -63, so a*c = 63. We need 2 factors of 63 that will add to give us -2. The factors of 63 are {1, 63}, (3, 21}, {7, 9}. It looks like the combination of -9 and +7 will work because -9 + 7 = -2. Plug in accordingly:

$9u^2-9u+7u-7=0$

Group together in groups of 2:

$(9u^2-9u)+(7u-7)=0$

Now factor out what's common within each set of parenthesis:

$9u(u-1)+7(u-1)=0$

We know this combination "works" because the terms inside the parenthesis are identical. We can now factor those out and what's left goes together in another set of parenthesis:

$(u-1)(9u+7)=0$

Remember that $u=x^2$

so we sub back in and continue to factor. This was originally a fourth degree polynomial; that means we have 4 solutions.

$(x^2-1)(9x^2+7)=0$

The first two solutions are found withing the first set of parenthesis and the second two are found in other set of parenthesis. Factoring $(x^2-1)$ gives us that x = 1 and -1. The other set is a bit more tricky. If