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Alenkasestr [34]
3 years ago
6

Step by step how to solve ? 16x^2-10.3x+10=

Mathematics
1 answer:
Stella [2.4K]3 years ago
8 0

Answer:

it is not possible result is not given so it is not possible

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Help please this is due soon! I need you to show the work as well <3 pic below
Dafna1 [17]

Answer:

perimeter: 4x+2 area: x^2+x

Step-by-step explanation:

for perimeter, you add x+x+x+x+1+1, which is 4x+2

for area, you multiply x*x+1, which would be x^2+x

7 0
3 years ago
Put Them in The box They Go in
pav-90 [236]

Answer:

\frac{1}{3}  \times r - 4 \\ 3 \times r - 4 \\  \frac{1}{3}  \times r + 4 \\ 3(r + 4)

5 0
3 years ago
Help me Please..
Fittoniya [83]
Just general definitons:

a TRInomial has 3 terms (tri means three)
a BInomial has 2 terms (bi means two)
a MONOmial has 1 term (mono means one)

the degree is the highest exponent found in the algebraic expression

so they should be pretty easy to solve with that information, but just in case:

1. trinomial, degree of 4
2. binomial, degree of 3
3. monomial, degree of 2

for the final question, all you have to do is plug in 2 for x, so

(2)^2 - 2(2) + 1

4 - 4 + 1

so the answer is 1
7 0
3 years ago
Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator
SpyIntel [72]
\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}
\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)

When x=3, you're left with

147=49a_1\implies a_1=\dfrac{147}{49}=3

When x=\sqrt2 or x=-\sqrt2, you're left with

\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}

Adding the two equations together gives -10=2a_5, or a_5=-5. Subtracting them gives 2\sqrt2=2\sqrt2a_4, a_4=1.

Now, you have

5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)
5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)
2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that a_2x^4=2x^4 and a_3(-3)(-2)=6a_3=-6. These alone tell you that you must have a_2=2 and a_3=-1.

So the partial fraction decomposition is

\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}
7 0
3 years ago
Vito has a fair 20-sided die with the faces
Hitman42 [59]

Answer:

15, C

Step-by-step explanation:

Any number is equally likely to appear. Therefore, we have 3/20 * 100 = 15. This is C.

Hope this helped!

~clouddragon

5 0
2 years ago
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