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Masteriza [31]
3 years ago
14

What is the solution to the system of equations below? Y=-1/4x+2 and 3y=-3/4x-6

Mathematics
1 answer:
fgiga [73]3 years ago
4 0

<u>Answer:</u>

There are infinite solutions to the system of equations given

<u>Solution: </u>

The given equations given are,

y=-\frac{x}{4}+2------ (i)

3 y=-\frac{3 x}{4}-6---- (ii)

Now putting the value of (i) in (ii) we get,

3 \times\left(-\frac{x}{4}+2\right)=-\frac{3 x}{4}-6

\Rightarrow-\frac{3 x}{4}+6=-\frac{3 x}{4}-6

\Rightarrow\frac{3 x}{4}-\frac{3 x}{4}=-12

\Rightarrow0 \times x=-12

\Rightarrowx=(\frac{-12}{0})=\infty

So, there are <em>infinite solutions</em> to the system.

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Solve the equation x4 – 26x2 + 25 = 0 in the real number system.
gogolik [260]

Answer:

x = -1, 1, -5 , 5.

Step-by-step explanation:

x4 – 26x2 + 25 = 0

Factoring:

(x^2 - 25)(x^2 - 1) = 0

So x^2 - 25 = 0 gives x^2 = 25 and x = +/- 5.

x^2 - 1  gives  x^2 = 1 and x = +/- 1.

4 0
4 years ago
The doubling period of a bacteria population is 10 minutes. At time t = 110 minutes, the bacterial population was 800.
goldenfox [79]

The initial population at time t = 0 was 0.39 while the size of the bacteria population after 4 hours was 6553600

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

An equation is an expression that shows the relationship between two or more numbers and variables.

Let y represent the bacteria population at time t. a represent the initial population at t = 0. Since the doubling period of a bacteria population is 10 minutes, hence:

y=a(2)^\frac{t}{10}

At time t = 110 minutes, the bacterial population was 800. Hence:

800=a(2)^\frac{110}{10} \\\\a = 0.39

At 4 hours (240 minutes):

y=0.39(2)^\frac{240}{10} =6553600

The initial population at time t = 0 was 0.39 while the size of the bacteria population after 4 hours was 6553600

Find out more on equation at: brainly.com/question/2972832

#SPJ1

4 0
2 years ago
A road has a 10% grade, meaning increasing 1 unit of rise to every 10 units of run.
fenix001 [56]

The grade is the ratio of rise to run, i.e. the slope aka the tangent.

\tan \theta = \dfrac{1}{10}

\theta = \arctan 0.1 \approx 5.711^\circ

Answer: (a) 6 degrees

For part b, the road is the hypotenuse c of a right triangle whose tangent of the small angle is 1/10.    The height h or rise is the side opposite the small angle.

\sin\theta = \dfrac h c

h = c \sin \theta

We could just take the sine of the angle we got but let's get it from the tangent exactly.

\cos^2 \theta + \sin ^2 \theta = 1

Dividing by squared cosine

1 + \tan ^2 \theta = 1/\cos^2 \theta = 1/(1- \sin^2 \theta)

(1- \sin^2 \theta) = 1/(1 + \tan^2 \theta)

\sin^2 \theta = 1 - 1/(1 + \tan^2 \theta)

\sin^2 \theta = 1 - 1/(1 + (1/10)^2) = 1-1/(101/100) = 1/101

h = c \sin \theta = 2 \sqrt{1/101} \approx 0.199

Answer: (b) Rise of 0.199 km

6 0
4 years ago
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