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Karo-lina-s [1.5K]
3 years ago
10

Identify the Terms, Constants, Coefficients and Variables in the following eqaution:

Mathematics
1 answer:
irinina [24]3 years ago
5 0

Answer:

<h2>13x - 1. Term</h2><h2>5y - 1. Term</h2><h2>13 - 3. Coefficient</h2><h2>2 - 2. Constant</h2><h2>5 - 3. Coefficient</h2><h2>y - 4. Variable</h2><h2>3 - 2. Constant</h2><h2>x - 4. Variable</h2>

Step-by-step explanation:

<h3>13x + 5y - 2 = 3</h3><h3 /><h3><u>Term</u> has a <em>number and also a variable</em>. </h3><h3>Ex.: 5y</h3><h3 /><h3><u>Constant</u> is any <em>number without a variable</em>.</h3><h3>Ex.: 3</h3><h3 /><h3><u>Coefficient</u> is the <em>number in a term</em>.</h3><h3>Ex.: 13</h3><h3 /><h3><u>Variable</u> is any <em>letter</em> in the equation.</h3><h3>Ex.: x</h3><h3 />

\tt{ \green{P} \orange{s} \red{y} \blue{x} \pink{c} \purple{h} \green{i} e}

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frosja888 [35]
You’re gonna have to use the quadratic formula here and a bit of thinking. We know y is height, and once it touches the ground, it will be 0.

(-b +- sqrt b^2 - 4ac)/2a
2 +- sqrt(4 + 48(400))/-32
(2 +- 138.58)/-32
We don’t want a negative time because it’ll make no sense. So do subtraction

(2-138.58)/-32 = 4.27 seconds aka 4.3 seconds
4 0
3 years ago
A rectangle is to be inscribed in an isosceles right triangle in such a way that one vertex of the rectangle is the intersection
svet-max [94.6K]

Answer:

x  =  2  cm

y  = 2  cm

A(max) =  4 cm²

Step-by-step explanation: See Annex

The right isosceles triangle has two 45° angles and the right angle.

tan 45°  =  1  =  x / 4 - y        or     x  =  4  -  y     y  =  4  -  x

A(r)  =  x* y

Area of the rectangle as a function of x

A(x)  =  x  *  (  4  -  x )       A(x)  =  4*x  -  x²

Tacking derivatives on both sides of the equation:

A´(x)  =  4 - 2*x             A´(x)  =  0            4   -  2*x  =  0

2*x  =  4

x  =  2  cm

And  y  =  4  - 2  =  2  cm

The rectangle of maximum area result to be a square of side 2 cm

A(max)  = 2*2  =  4 cm²

To find out if A(x) has a maximum in the point  x  =  2

We get the second derivative

A´´(x)  =  -2           A´´(x)  <  0   then A(x) has a maximum at  x = 2

5 0
3 years ago
Graph the solution to this system of inequalities in the coordinate plane.
Vsevolod [243]

By using Geogebra online graphing calculator, the solution to this system of inequalities are plotted in the image attached below.

<h3>What is an inequality?</h3>

An inequality is a mathematical relation that compares two (2) or more integers and variables in an equation based on any of the following:

  • Less than (<).
  • Greater than (>).
  • Less than or equal to (≤).
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Given the following system of inequalities:

3y > 2x + 12  ⇒ 3y - 2x > 12

2x + y ≤ -5

By using Geogebra online graphing calculator, the solution to this system of inequalities are plotted in the image attached below.

Read more on inequalities here: brainly.com/question/24372553

#SPJ1

3 0
2 years ago
What are the intercepts of the line?
Ann [662]
D is not the answer bc it is a 


8 0
3 years ago
Read 2 more answers
A leprechaun places a magic penny under a girls pillow. The next night there are 2 magic pennies under her pillow. The following
kiruha [24]

Answer:

<em>After </em><em>47</em><em> days she will have more than 90 trillion pennies.</em>

Step-by-step explanation:

At the beginning there was 1 penny. At the second day the amount of pennies under the pillow became 2.

The amount of pennies doubled each day. So the series is,

1,2,4,8,16,32,.....

This series is in geometric progression.

As the pennies from each of the previous days are not being stored away until more pennies magically appear so the sum of series will be,

S_n=\dfrac{a(r^n-1)}{r-1}

where,

a = initial term = 1,

r = common ratio = 2,

As we have find the number of days that would elapse before she has a total of more than 90 trillion, so

\Rightarrow 90\times 10^{12}\le \dfrac{1(2^n-1)}{2-1}

\Rightarrow 90\times 10^{12}\le \dfrac{2^n-1}{1}

\Rightarrow 90\times 10^{12}\le 2^n-1

\Rightarrow 2^n\ge 90\times 10^{12}+1

\Rightarrow \log 2^n\ge \log (90\times 10^{12}+1)

\Rightarrow n\times \log 2\ge \log (90\times 10^{12}+1)

\Rightarrow n \ge \dfrac{\log (90\times 10^{12}+1)}{\log 2}

\Rightarrow n \ge 46.4

\Rightarrow n\approx 47


8 0
3 years ago
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