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3 years ago

-6+3(2-4(-3)+6)+(-5)

Mathematics

1 answer:

FromTheMoon [43]3 years ago

3 0

Answer:

Step-by-step explanation:

Remove parenthesis.

−6+3(2−4⋅−3+6)−5

Multiply −4 by−3.

−6+3(2+12+6)−5

Add 2 and 12.

−6+3(14+6)−5

Add 14 and 6.

-6+3(20)-5

Multiply 3 by 20.

-6+60-5

Add -6+60.

54-5

Subtract 54 and 5.

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Solve 5x^2 − 3x + 17 = 9.

Vedmedyk [2.9K]

Answer:

The roots of the equation are x = $\frac{3+\sqrt{151}i}{10}$ and x = $\frac{3-\sqrt{151}i}{10}$

and there are no real roots of the equation given above

Step-by-step explanation:

To solve:

5x² − 3x + 17 = 9

⇒ 5x² − 3x + 17 - 9 = 0

⇒ 5x² − 3x + 8 = 0

Now,

the roots of the equation in the form ax² + bx + c = 0 is given as:

x = $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

in the above given equation

a = 5

b = -3

c = 8

therefore,

x = $\frac{-(-3)\pm\sqrt{(-3)^2-4\times5\times8}}{2\times5}$

x = $\frac{3\pm\sqrt{9-160}}{10}$

x = $\frac{3+\sqrt{-151}}{10}$ and x = $\frac{3-\sqrt{-151}}{10}$

x = $\frac{3+\sqrt{151}i}{10}$ and x = $\frac{3-\sqrt{151}i}{10}$

here i = √(-1)

Hence,

The roots of the equation are x = $\frac{3+\sqrt{151}i}{10}$ and x = $\frac{3-\sqrt{151}i}{10}$

and there are no real roots of the equation given above

7 0

2 years ago

Please somebody help me

Montano1993 [528]

60=4/5x
4/5=0.8
60/0.8=75
X=75

5 0

3 years ago

Read 2 more answers

The most accurate estimation 5+4

MariettaO [177]

The most accurate "estimation" for 5+4 is um... 9

4 0

3 years ago

Write these numbers as fractions. Simplify the answers. 0.5, 12.3, 0.07, 0.625

motikmotik

0.5 => 5/10 => 1/2

12.3 => 123/10

0.07 => 7/100

0.625 => 625/1000 => 25/40 => 5/8

4 0

3 years ago

Read 2 more answers

K^2+5k-6=0 quadratic formula

s2008m [1.1K]

I hope this helps you

1.k^2+5k-6=0

a=1

b=5

c= -6

disctirminant =b^2-4ac

disctirminant =5^2-4.1. (-6)

disctirminant =49

x1= -(5)+ square root of 49/2.1

x1=-5+7/2

x1=1

x2= -5- square root of 49/2.1

x2= -5-7/2

x2= -6

7 0

3 years ago