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anygoal [31]
3 years ago
5

Which is the approximate solution to the system y = 0.5x + 3.5 and y = − 2/3 x + 1/3 shown on the graph? (–2.7, 2.1) (–2.1, 2.7)

(2.1, 2.7) (2.7, 2.1)
Mathematics
2 answers:
Volgvan3 years ago
6 0

Answer:

(–2.7, 2.1)

Step-by-step explanation:

edg. 2020 unit test

AlekseyPX3 years ago
5 0

Answer:

The approximate solution to the system is (-2.7, 2.1).

Step-by-step explanation:

To solve the system of equations \begin{bmatrix}y=0.5x+3.5\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix} you must:

\mathrm{Rationalize\:equations}\\\\\begin{bmatrix}y=\left(\frac{1}{2}\right)x+\left(\frac{7}{2}\right)\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}

\mathrm{Subsititute\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\begin{bmatrix}-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\end{bmatrix}

\mathrm{Isolate}\:x\:\mathrm{for}\:-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\\\\-\frac{2}{3}x=\frac{19}{6}+\frac{1}{2}x\\\\-\frac{7}{6}x=\frac{19}{6}\\\\6\left(-\frac{7}{6}x\right)=\frac{19\cdot \:6}{6}\\\\-7x=19\\\\x=-\frac{19}{7}\approx-2.7

\mathrm{For\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\mathrm{Subsititute\:}x=-\frac{19}{7}\\\\y=-\frac{2}{3}\left(-\frac{19}{7}\right)+\frac{1}{3}\\\\y=\frac{15}{7}\approx 2.1

The approximate solutions to the system of equations are:

x=-2.7 ,\:y=2.1

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3 years ago
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Question
maria [59]

Answer:

He needs 16 ounces of raisins and 8 ounces of nuts

Step-by-step explanation:

Let

x -----> the number of ounces of raisins

y -----> the number of ounces of nuts

we know that

x+y=24 -----> equation A

x=2y -----> equation B

Solve the system by substitution

Substitute equation B in equation A and solve for y

(2y)+y=24

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y=8 ounces of nuts

Find the value of x

x=2(8)=16 ounces of raisins

therefore

He needs 16 ounces of raisins and 8 ounces of nuts

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Find the volume of the solid formed by rotating the region bounded by the given curves about the indicated axis. y = x3/4, x = 0
vovangra [49]

Answer:

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Step-by-step explanation:

The volume formed is V = ∫πx²dy

Now, since y = x³/4, x = ∛(4y). Also if x = 0, y = 0³/4 = 0 and the curve intersects the line y = 1. So the limits of integration are y = 0 to y = 1

So, V = ∫₀¹πx²dy

= ∫₀¹π[ ∛(4y)]²dy

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= π(∛4)²∫₀¹y^³/₂dy

= π(∛4)²[y^⁵/₂]₀¹

= π(∛4)²[1^⁵/₂ - 0^⁵/₂]

= π(∛4)²[1 - 0]

= π(∛4)²

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