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Molodets [167]
3 years ago
5

Solve the following equations.x + 3y =0 and 5x + y = 15

Mathematics
1 answer:
Andrei [34K]3 years ago
3 0

Answer:

The required values are:-

x= 45/14 and y= -15/14

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I need help! (It's easy)
kati45 [8]
A.) it's a 90° angle 
     90=6x+4x+10
     90=10x+10
    -10         -10
     80=10x
     divide both sides by 10, x=8
B.) it's a 180° angle
     180=5x+13+3x+7
     180=8x+20
     -20        -20
     160=8x
     divide both sides by 8, x=20
C.) it's a 180° angle
     180=3x+5+2x+18+2x+17
     180=7x+40
     -40        -40
     140=7x
     divide both sides by 7, x=20
D.) it's a 180° angle
     180=90+30+x
     180=120+x
     -120  -120
     60=x
7 0
3 years ago
Amanufacturer of potato chips would like to know whether its bag filling machine works correctly at the 433 gram setting. It is
ankoles [38]

Answer:

There is not enough evidence to support the claim that the bags are under filled.

Step-by-step explanation:

Given :

Population mean, μ = 433

Sample size, n = 26

xbar = 427

Variance, s² = 324 ; Standard deviation, s = √324 = 18

The hypothesis :

H0 : μ = 433

H0 : μ < 433

The test statistic :

(xbar - μ) ÷ (s/√(n))

(427 - 433) / (18 / √26)

-6 / 3.5300904

T = -1.70

The Pvalue :

df = 26-1 = 25 ; α = 0.05

Pvalue = 0.0508

Since Pvalue > α ; WE fail to reject the Null and conclude that there is not enough evidence to support the claim that the bags are underfilled

3 0
3 years ago
PLEASE HELP! WILL MARK BRAINLIEST!
AfilCa [17]
Number 3 is the correct answer 
5 0
4 years ago
A student claims that 2i is the only imaginary root of a polynomial equation that has real coefficients. Explain the student's m
____ [38]

Answer:

The Fundamental Theorem of Algebra assures that any polynomial  f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i

Step-by-step explanation:

1) This claim is mistaken.

2) The Fundamental Theorem of Algebra assures that any polynomial  f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i with real coefficients.

a_{0}x^{n}+a_{1}x^{2}+....a_{1}x+a_{0}

For example:

3) Every time a polynomial equation, like a quadratic equation which is an univariate polynomial one, has its discriminant following this rule:

\Delta < 0\\b^{2}-4*a*c

We'll have <em>n </em>different complex roots, not necessarily 2i.

For example:

Taking 3 polynomial equations with real coefficients, with

\Delta < 0

-4x^2-x-2=0 \Rightarrow S=\left \{ x'=-\frac{1}{8}-i\frac{\sqrt{31}}{8},\:x''=-\frac{1}{8}+i\frac{\sqrt{31}}{8} \right \}\\-x^2-x-8=0 \Rightarrow S=\left\{\quad x'=-\frac{1}{2}-i\frac{\sqrt{31}}{2},\:x''=-\frac{1}{2}+i\frac{\sqrt{31}}{2} \right \}\\x^2-x+30=0\Rightarrow S=\left \{ x'=\frac{1}{2}+i\frac{\sqrt{119}}{2},\:x''=\frac{1}{2}-i\frac{\sqrt{119}}{2} \right \}\\(...)

2.2) For other Polynomial equations with real coefficients we can see other complex roots ≠ 2i. In this one we have also -2i

x^5\:-\:x^4\:+\:x^3\:-\:x^2\:-\:12x\:+\:12=0 \Rightarrow S=\left \{ x_{1}=1,\:x_{2}=-\sqrt{3},\:x_{3}=\sqrt{3},\:x_{4}=2i,\:x_{5}=-2i \right \}\\

4 0
3 years ago
8 3/8 -3 1/4 equals what
aleksandrvk [35]

Answer:

Exact Form: 41/8

Decimal form: 5.125

Mixed number form: 5 1/8

4 0
3 years ago
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