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Fudgin [204]
2 years ago
8

You start at (5, -1). You move up 5 units. Where do you end? On a coordinate plane

Mathematics
2 answers:
aliya0001 [1]2 years ago
4 0

Answer:

0,-1

Step-by-step explanation:

STatiana [176]2 years ago
4 0

Then I will end at (5,4)

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Dude i have 15 mins to finish this so pls help me
kirill [66]
A) 11 (32 divided by 2 = 16 | 16 minus 5 is 11)

b) 5 + x = y | 32 minus y = number of stuffed animals

c)
Veronica has 11 stuffed animals
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hope this helped gl on your test!! :”)
8 0
2 years ago
How do I solve this question?
asambeis [7]
This is a perfect square trnomial
(a+b)²=a²+2ab+b²
we see that a=5x and b=2

(5x)²+2(5x)(2)+2²=0
factor
(5x+2)²=0
set equal to zero
5x+2=0
5x=-2
x=-2/5
5 0
3 years ago
4/5 of a number is 16. What is the number?
Zepler [3.9K]

Answer:

20

Step-by-step explanation:

5 0
3 years ago
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The volume of water in a plastic bottle is 12.3 oz. Determine whether the data value is from a discrete or continuous data set.
Elena L [17]

Answer:

d

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Find a polynomial P(x) with real coefficients having a degree 4, leading coefficient 4, and zeros 3-i and 4i.
Alisiya [41]

now, this polynomial has roots of 3-i and 4i, namely 3 - i and 0 + 4i.

let's bear in mind that a complex root never comes all by her lonesome, her sibling is always with her, the conjugate, so if 3 - i is there, 3 + i is also coming along, likewise if 0 + 4i is there, her sibling 0 - 4i is also there.

\bf \begin{cases} x=3-i\implies &x-3+i=0\\ x=3+i\implies &x-3-i=0\\ x=4i\implies &x-4i=0\\ x=-4i\implies &x+4i=0 \end{cases}\\\\[-0.35em] ~\dotfill\\\\ (x-3+i)(x-3-i)(x-4i)(x+4i)=\stackrel{y}{0} \\[2em] \underset{\textit{difference of squares}}{[(x-3)+i][(x-3)-i]}\underset{\textit{difference of squares}}{[x-4i][x+4i]}=0

\bf [(x-3)^2-i^2][x^2-(4i)^2]=y\implies [(x-3)^2-(-1)][x^2-(4^2i^2)]=0 \\[2em] [(x-3)^2-(-1)][x^2-(16(-1))]=0\implies [(x-3)^2+1][x^2+16]=0 \\[2em] [(x^2-6x+9)+1][x^2+16]=y\implies (x^2-6x+10)(x^2+16)=0 \\\\\\ x^4-6x^3+10x^2+16x^2-96x+160=0 \\\\\\ x^4-6x^3+26x^2-96x+160=0 \\\\\\ \stackrel{\textit{multiplying both sides by 4}}{4(x^4-6x^3+26x^2-96x+160)=4(0)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 4x^4-24x^3+104x^2-384x+640=y~\hfill

3 0
3 years ago
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