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Norma-Jean [14]
3 years ago
9

What us number seven to length to the nearest centimeter

Mathematics
1 answer:
svet-max [94.6K]3 years ago
6 0
I can't understand. Be more specific and I could maybe help. :P
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What are the zeros of the quadratic function f(x) equals 6X squared +12 X -7
tamaranim1 [39]

The zeroes of the function will be \rm -1 \pm \dfrac{\sqrt{78}}{6}}

<h3>What is a Quadratic Function ?</h3>

A quadratic function is given by ax² +bx+c = f(x)

The given quadratic function is

f(x) = 6x² +12x - 7

The zeroes of the function will be given by

f(0) = 0

6x² +12x - 7 = 0

The zeroes are given by

\rm \dfrac {-b  \pm \sqrt{b^2 -4ac}}{2a}

here

a = 6

b = 12

c = -7

Therefore the zeroes of the function will be

\rm \dfrac {-12  \pm \sqrt{12^2 -4*6*(-7)}}{2 *6}

\rm \dfrac {-12  \pm \sqrt{312}}{12}

\rm -1 \pm \dfrac{\sqrt{78}}{6}}

Therefore these are the two zeroes of the function.

To know more about Quadratic Function

brainly.com/question/27918223

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7 0
2 years ago
What number is ten less than 658?
Karo-lina-s [1.5K]

Answer:

658-10=648

Step-by-step explanation:

subtract 10 from 658.

6 0
3 years ago
The graph shows the amount of money earned by two different workers.
Masteriza [31]

Answer:

The asnwer is B

Step-by-step explanation:

HOPE YOU DO WELL

3 0
3 years ago
Read 2 more answers
What Ln(z) is answer????!!
dimaraw [331]

Write <em>z</em> in exponential form:

z=1-i=\sqrt2 e^{-i\frac\pi4}

Then taking the logarithm, we get

\mathrm{Ln}(z)=\ln(\sqrt2) + \ln e^{-i\frac\pi4} = \boxed{\ln(\sqrt2)-\dfrac\pi4i}

so a is the correct answer.

3 0
3 years ago
I need help with part B plz help me asap
Sloan [31]

Answer:

see below    PLEASE GIVE BRAINLIEST!!

Step-by-step explanation:

A)  2.40A ≦ 9.60

B) You can choose to purchase between 0 and 4 avacados.  4 is the maximum you can buy because 2.40 x 4 = 9.60 and that is all the money you have and more than 4 would make the equation incorrect.  0 is the least you can buy since you can't purchase a negative amount of something.

So, A ≦ 4

6 0
3 years ago
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