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mel-nik [20]
3 years ago
13

Can someone help me?

Mathematics
1 answer:
OverLord2011 [107]3 years ago
8 0

Answer:

52

Step-by-step explanation:

180-90-38=52

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3
ser-zykov [4K]

Answer: D. Subtracting 9 from each side

Step-by-step explanation:

Always start with numbers without variables first before doing anything with x. And make sure one side must not equal 0 unless you're doing polynomials.

5 0
3 years ago
What’s the correct answer for this question
HACTEHA [7]

All of them are true except AD = GH.

6 0
3 years ago
Solve the system of linear equations using either substitution or the elimination method.
Lelechka [254]

Answer:

x=2;y=2

Step-by-step explanation:

2x+y=6..........1

3x-y=4............2

adding equation1 with equation 2

2x+y+3x-y=6+4

5x=10

x=10/5=2

substituting vslue of y in equation 1

2×2+y=6

y=6-4

y=2

4 0
2 years ago
How to convert 37 \ 4 to a mixed number
Iteru [2.4K]
How many times does 4 go into 37?
Well, 4*9 is 36.
37 and 36 have a difference of 1. You take that one and put it over the 4.
Therefore, 37/4 into a mixed number is 9 and 1/4
8 0
2 years ago
find the zeros of following quadratic polynomial and verify the relationship between the zeros and the coefficient of the polyno
Cloud [144]

Answer:

\textsf{Zeros}: \quad x=\dfrac {\sqrt{3}}{2}, \:\:x=-\dfrac {4\sqrt{3}}{3}

Step-by-step explanation:

Rewrite the given polynomial in the form ax² + bx + c:

f(x)=2 \sqrt{3}x^2+5x-4 \sqrt{3}

To find the zeros, set the function to zero and solve for x using the quadratic formula.

\implies 2 \sqrt{3}x^2+5x-4 \sqrt{3}=0

<u>Quadratic formula</u>:

x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0

Therefore,

  • a = 2√3
  • b = 5
  • c = - 4√3

Substituting the values into the quadratic formula:

\implies x=\dfrac{-5 \pm \sqrt{5^2-4(2\sqrt{3})(-4\sqrt{3})} }{2(2\sqrt{3})}

\implies x=\dfrac {-5 \pm \sqrt {121}}{4\sqrt{3}}

\implies x=\dfrac {-5 \pm 11}{4\sqrt{3}}

\implies x=\dfrac {6}{4\sqrt{3}}, \:\:x=\dfrac {-16}{4\sqrt{3}}

\implies x=\dfrac {3}{2\sqrt{3}}, \:\:x=-\dfrac {4}{\sqrt{3}}

\implies x=\dfrac {\sqrt{3}}{2}, \:\:x=-\dfrac {4\sqrt{3}}{3}

The sum of the roots of a polynomial is -b/a:

\implies -\dfrac{b}{a}=-\dfrac{5}{2 \sqrt{3}}=-\dfrac{5\sqrt{3}}{6}

The sum of the found roots is:

\implies \left(\dfrac {\sqrt{3}}{2}\right)+\left(-\dfrac {4\sqrt{3}}{3}\right)=-\dfrac{5\sqrt{3}}{6}

Hence proving the sum of the roots is -b/a

The product of the roots of a polynomial is:  c/a

\implies \dfrac{c}{a}=\dfrac{-4\sqrt{3}}{2\sqrt{3}}=-2

The product of the found roots is:

\implies \left(\dfrac {\sqrt{3}}{2}\right)\left(-\dfrac {4\sqrt{3}}{3}\right)=-\dfrac{12}{6}=-2

Hence proving the product of the roots is c/a

Therefore, the relationship between the roots and the coefficients is verified.

8 0
1 year ago
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