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Valentin [98]
3 years ago
6

Y=2x+5 y=3x−1 solve by subsitution

Mathematics
1 answer:
loris [4]3 years ago
5 0

Answer:

(6,17)

Step-by-step explanation:

2x+5=3x-1

-3x        -3x

-x+5=-1

-5      -5

-x=-6

^divided by -1 on both sides

x=6

y=2x+5

y=(2)(6)+5

y=17

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A cell reproduces by doubling every hour. If you start with a single cell, how many will you have in 24 hours?
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Answer: 4 cells.

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The figures below are similar. What are a) the ratio of the perimeters and b) the ratio of the areas of the larger figure to the
iren [92.7K]
A) ratio of the perimeters is the same as ratio of corresponding sides, so:

\dfrac{P_1}{P_2}=k=\dfrac{22}{14}=\dfrac{11}{7}\approx1,57

b) ratio of the areas is equal to k² so:

\dfrac{A_1}{A_2}=k^2=\left(\dfrac{11}{7}\right)^2=\dfrac{121}{49}\approx2,47
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( 2 − k ) + 3 ( − 4 k + 2 )<br> Find the Sum of the Linear Expressions:
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Step-by-step explanation:

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HELP ASAP!!!
Alenkasestr [34]
The answer would be 5
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3 years ago
Find an n-degree polynomial function with real coefficients satisfying the given condition.
kolezko [41]

In all cases, if f has real coefficients, then any complex roots occur in conjugate pairs, so if a+bi is a root, then so is a-bi. Also, by the fundamental theorem of algebra, if r_1,\ldots,r_n are roots to f, then for some constant a\in\mathbb R,

f(x)=a(x-r_1)\cdots(x-r_n)

1. If n=3 and f(3)=f(2i)=0, then

f(x)=a(x-3)(x-2i)(x+2i)=ax^3-3ax^2+4ax-12a

Given that f(-1)=50, we have

f(-1)=a(-1-3)(-1-2i)(-1+2i)=-20a=50\implies a=-\dfrac52

\implies\boxed{f(x)=-\dfrac52x^3+\dfrac{15}2x^2-10x+30}

2.

f(x)=a(x-4)(x-(-5+2i))(x-(-5-2i))=a x^3 + 6 a x^2 - 11 a x - 116 a

With f(2)=-636, we have

f(2)=a(2-4)(2+5-2i)(2+5+2i)=-106a=-636\implies a=6

\implies\boxed{f(x)=6x^3+36x^2-66x-696}

The rest are done in the same exact way.

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4 years ago
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