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guapka [62]
3 years ago
6

How many terms are in the algebraic expression 3x^2 + 4y - 1

Mathematics
2 answers:
netineya [11]3 years ago
8 0

Answer:

<h2><u>3</u></h2>

Step-by-step explanation:

The terms of this given expression are 3x^2, 4y and 1. Thus, there are <u>three </u>terms.

ElenaW [278]3 years ago
4 0

Answer:

3

Step-by-step explanation:  There are 3 terms there

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Answer:

5:17pm

Step-by-step explanation:

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Step-by-step explanation:

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of Shares  Share                             (At 6% of total)        

100           $16.25          $1625         1625\times \frac{6}{100}=97.5      1625+97.5 = $1722.5      

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100           $13.22           $1322        1322\times \frac{6}{100}=79.32   1322+79.32=$1401.32

5 0
3 years ago
Assume that the terminal side of thetaθ passes through the point (negative 12 comma 5 )(−12,5) and find the values of trigonomet
zmey [24]

Answer:

\sin \theta = \dfrac{5}{13} and \sec \theta = -\dfrac{13}{12}

Step-by-step explanation:

Assume that the terminal side of thetaθ passes through the point (−12,5).

In ordered pair (-12,5), x-intercept is negative and y-intercept is positive. It means the point lies in 2nd quadrant.

Using Pythagoras theorem:

hypotenuse^2=perpendicular^2+base^2

hypotenuse^2=(5)^2+(12)^2

hypotenuse^2=25+144

hypotenuse^2=169

Taking square root on both sides.

hypotenuse=13

In a right angled triangle

\sin \theta = \dfrac{opposite}{hypotenuse}

\sin \theta = \dfrac{5}{13}

\sec \theta = \dfrac{hypotenuse}{adjacent}

\sec \theta = \dfrac{13}{12}

In second quadrant only sine and cosecant are positive.

\sin \theta = \dfrac{5}{13} and \sec \theta = -\dfrac{13}{12}

6 0
4 years ago
Which is equivalent to the expression below? 6x + 2x + y + 3y​
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Answer:

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6x + 2x + y + 3y

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What are the solutions to the quadratic equation below in factored form?
saw5 [17]

Answer:

\cfrac{-4}{3} and \cfrac{1}{2}.

Step-by-step explanation:

Equation:

\rm \:  \: (2x - 1)(3x + 4) = 0

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Set factors equal to zero,that is:

(2x - 1) = 0 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: ...(1)

(3x + 4) = 0 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: ...(2)

Solving for equation 1:

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Solving for equation 2:

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Hence,the solutions to the quadratic equation in factored form is \cfrac{-4}{3} and \cfrac{1}{2}.

\rule{225pt}{2pt}

Good luck on your assignment!

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