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3 years ago

√2+1 b) Rationalize the denominator: √2+1

Mathematics

1 answer:

Misha Larkins [42]3 years ago

4 0

Answer:

there is not a complete question here ...

$\sqrt{2} +1$ is not in a denominator to rationalize

if it were you would multiply the rational expression (the fraction)

by $\frac{\sqrt{2} -1}{\sqrt{2} -1}$ the result (in the denominator would end up being "2 -1"

which is 1

Step-by-step explanation:

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The ratio of blue to yellow is 4:5 and there are 15 yellow how many blue is there?

KATRIN_1 [288]

Answer: 12 blue marbles

Step-by-step explanation: Set up a proportion.

4/5 = x/15

Solve for x, the number of blue marbles. Cross multiply. You will get 60 = 5x

Divide by 5 on each side.

x = 12

There are 12 blue marbles.

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3 years ago

Nancy leaves the school and drives toward her town at 27 mph. greg leaves the same school 2 hours later and drives the same rout

Paul [167]

Nancy's rate=27 mph
Greg's rate=45 mph
t=Greg's time
t+2=Nancy's time
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27(t+2)=45t
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3 years ago

Greenville County, South Carolina, has 461,299 adult residents, of which 59,969 are 65 years or older. A survey wants to contact

Salsk061 [2.6K]

The proportion would be the number of the sample (x) over the population size (n)

Here we have x = 59,969 and n= 461,299

P = x/n = 59969/461299

P = 0.13

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3 years ago

Find a possible middle term to make this polynomial factorable:

mafiozo [28]

A. the polynomial can then be factored to (x+10)(x+2).

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3 years ago

The quadratic function g(x) = a.ca + bx+c has the

Mumz [18]

<em>The value of b is 14 and the value of c is 65</em>

<h2>Explanation:</h2>

The quadratic function is a function of the form:

$f(x)=ax^2+bx+c$

Here we know that the leading coefficient $a=1$ so we reduce our equation to:

$g(x)=x^2+bx+c$

The roots are those values at which $g(x)=0$

So:

$x^2+bx+c=0 \\ \\ First \ root: \\ \\ (-7+4i)^2+b(-7+4i)+c=0 \\ \\ (-7)^2-2(7)(4i)+(4i)^2-7b+4bi+c=0 \\ \\ 49-56i+16i^2-7b+4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49-56i+16(-1)-7b+4bi+c=0 \\ \\ 49-56i-16-7b+4bi+c=0 \\ \\ 33-56i-7b+4bi+c=0 \\ \\ \\$

$Second \ root: \\ \\ (-7-4i)^2+b(-7-4i)+c=0 \\ \\ (-1)^2(7+4i)^2+b(-7-4i)+c=0 \\ \\ (7)^2+2(7)(4i)+(4i)^2-7b-4bi+c=0 \\ \\ 49+56i+16i^2-7b-4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49+56i+16(-1)-7b-4bi+c=0 \\ \\ 49+56i-16-7b-4bi+c=0 \\ \\ 33+56i-7b-4bi+c=0$

So we have:

$(1) \ 33-56i-7b+4bi+c=0 \\ \\ (2) \ 33+56i-7b-4bi+c=0 \\ \\ \\ Subtract \ 2 \ from: \\ \\ 33-56i-7b+4bi+c-(33+56i-7b-4bi+c)=0 \\ \\ 33-56i-7b+4bi+c-33-56i+7b+4bi-c=0 \\ \\ \\ Combine \ like \ terms: \\ \\ 33-33-56i-56i-7b+7b+4bi+4bi+c-c=0 \\ \\ -112i+8bi=0 \\ \\ Isolating \ b: \\ \\ b=\frac{112i}{8i} \\ \\ \boxed{b=14}$

Finding c from (1):

$33-56i-7b+4bi+c=0 \\ \\ \\ Substituting \ b: \\ \\ 33-56i-7(14)+4(14)i+c=0 \\ \\ 33-56i-98+56i+c=0 \\ \\ -65+c=0 \\ \\ \boxed{c=65}$

<h2>Learn more:</h2>

Complex conjugate: brainly.com/question/2137496

#LearnWithBrainly

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3 years ago