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2 years ago

Please i need someone help.

Mathematics

1 answer:

natali 33 [55]2 years ago

3 0

Answer: I can help :)

Step-by-step explanation:

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If you were 2 years old in 2011 how old would you be in 2020?

Roman55 [17]

Answer:

11 years

Step-by-step explanation:

2 years that you were in 2011

2020-2011= 9

2+9=11

also meaning they were born in 2009

2020-2009 also equals 11

this prolly doesnt make sense but hope it helps :) <3

4 0

3 years ago

Read 2 more answers

Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?

likoan [24]

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=15,\:\quad a_n=2$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5$

$=\left(x+3\right)\left(2x^2-11x+5\right)$

$Factor: 2x^2-11x+5$

$2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)$

$=x\left(2x-1\right)-5\left(2x-1\right)$

$2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)$

$\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:}$

thus zeros of f(x) is

$x=-3,\:x=5,\:x=\frac{1}{2}$

5 0

3 years ago

Read 2 more answers

What’s the correct answer for this?

VMariaS [17]

Answer:

Second option is the correct answer

$(x+11)^2+(y+6)^2 = 324$

Step-by-step explanation:

$x^{2} +12y+22x+y^2-167=0\\x^{2}+22x +y^2 +12y-167=0\\(x^{2}+22x +121)-121 +(y^2 +12y+36)-36-167=0\\(x+11)^2+(y+6)^2-324 = 0\\\huge\red{\boxed{(x+11)^2+(y+6)^2 = 324}}\\$

5 0

3 years ago

Read 2 more answers

Please help!! Will give brainlist is answer right

Drupady [299]

Answer:

Step-by-step explanation:

8 0

2 years ago

pls help will give brainliest Rena used the steps below to evaluate the expression (StartFraction (x Superscript negative 3 Base

kaheart [24]

Answer:

First "order of operations" mistake: step 2

First arithmetic mistake: step 4

Step-by-step explanation:

As we understand Rena's work, she wants to simplify ...

$\left(\dfrac{x^{-3}y^{-2}}{2x^4y^{-4}}\right)^{-3}$

for x = -1 and y = 2.

Her work seems to be ...

Step 1

$\text{Substitute $x=-1$ and $y=2$ into the expression}\\\\\left(\dfrac{(-1)^{-3}2^{-2}}{2(-1)^42^{-4}}\right)^{-3}\qquad\text{no error}$

Step 2

$\text{Simplify the parentheses}\\\\\left(\dfrac{2^4}{2(-1)^4(-1)^32^2}\right)^{-3}=\left(\dfrac{2^2}{2(-1)^7}\right)^{-3}\qquad\text{order of operations error}$

Step 3

$\text{Evaluate the power to a power}\\\\\dfrac{2^{-6}}{2^{-3}(-1)^{21}}\qquad\text{no error}$

Step 4

$\text{Use reciprocals and find the value}\\\\\dfrac{1}{2^32^6(-1)^{21}}=\dfrac{1}{8\cdot 64\cdot (-1)}=\dfrac{-1}{512}\qquad\text{error: $2^3$ is used instead of $2^{-3}$}$

_____

So, the first arithmetic error is in Step 4. However, the order of operations requires exponents be evaluated first. Doing that makes step 2 look like ...

$\left(\dfrac{-\dfrac{1}{4}}{2(1)\dfrac{1}{16}}\right)^{-3}=(-2)^{-3}\qquad\text{proper Step 2}$

We expect your answer is supposed to be Step 4.

7 0

3 years ago