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tino4ka555 [31]
3 years ago
10

Divide. Show your work. (6x^4-5x^3+6x^2)/2x^2

Mathematics
1 answer:
serg [7]3 years ago
6 0
We can do like we did before, distribute the denominator,

\bf \cfrac{6x^4-5x^3+6x^2}{2x^2}\implies \cfrac{6x^4}{2x^2}-\cfrac{5x^3}{2x^2}+\cfrac{6x^2}{2x^2}\implies \cfrac{6}{2}\cdot \cfrac{x^4}{x^2}-\cfrac{5}{2}\cdot \cfrac{x^3}{x^2}+\cfrac{6}{2}\cdot \cfrac{x^2}{x^2}
\\\\\\
3 x^4x^{-2}-\cfrac{5}{2}x^3x^{-2}+3x^2x^{-2}\implies 3x^{4-2}-\cfrac{5}{2}x^{3-2}+3x^{2-2}
\\\\\\
3x^2-\cfrac{5}{2}x+3x^0\implies 3x^2-\cfrac{5}{2}x+3
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Yo please help
Alex Ar [27]

Answer:

Options 1, 3 and 4 are true

Step-by-step explanation:

The legs EF and DF of the right triangle DEF have lengths of 24 units and 7 units, respectively. By the Pythagorean theorem,

DE^2=EF^2+DF^2\\ \\DE^2=24^2+7^2\\ \\DE^2=576+49\\ \\DE^2=625\\ \\DE=25\ units

Find trigonometric functions:

\sin \angle D=\dfrac{\text{Opposite leg}}{\text{Hypotenuse}}=\dfrac{EF}{DE}=\dfrac{24}{25}

\cos \angle E=\dfrac{\text{Adjacent leg}}{\text{Hypotenuse}}=\dfrac{EF}{DE}=\dfrac{24}{25}

\tan \angle D=\dfrac{\text{Opposite leg}}{\text{Adjacent leg}}=\dfrac{EF}{DF}=\dfrac{24}{7}

\sin \angle E=\dfrac{\text{Opposite leg}}{\text{Hypotenuse}}=\dfrac{DF}{DE}=\dfrac{7}{25}

Thus, options 1, 3 and 4 are true

5 0
3 years ago
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Factor: 3.x2 + 7x + 2​
iren [92.7K]

Answer:

x= -1/3 and -2

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5 0
3 years ago
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mojhsa [17]

Answer:

The scalar factor is 4/3

Step-by-step explanation:

The bottom of triangle B is 16 units, the bottom of triangle A is 12 units. So to find the scalar factor you have to divide B by A:

6/12=4/3

8 0
3 years ago
Which of these triangels apears not to be congruent to any others shown here check all that apply
masha68 [24]

Triangles are congruent when they have exactly the same three sides and exactly the same three angles.

The diagram shows two pairs of congruent triangle:

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Triangles A and B apears not to be congruent to any others triangles.

6 0
3 years ago
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