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lesantik [10]
2 years ago
14

Help pls :P im just bad at math

Mathematics
2 answers:
WINSTONCH [101]2 years ago
5 0
3.3g = 330cg
3.3g x 100cg/1g=3.3x100cg= 330cg
mojhsa [17]2 years ago
4 0

Answer:

sorry i don't under stand

Step-by-step explanation:

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Please help with the question in the picture!!
Greeley [361]

Answer:

Top: -5

Middle: -40

Bottom: -24

7 0
2 years ago
Help plz:))) I’ll mark u brainliest
dybincka [34]

Answer:

a

Step-by-step explanation:

8 0
2 years ago
Rotational symmetry is the quality a design has if it maintains all characteristics when it is rotated about an axis lying in it
mrs_skeptik [129]
Answer – TrueA design is said to have Rotational symmetry if all its characteristics remain the same after it has been rotated about an axis lying in its plane. In other words, rotational symmetry is the quality possessed by a shape has when it looks the same after it has undergone some rotation by a partial turn about an axis lying in its plane. Rotational symmetry is also referred to as radial symmetry.

5 0
3 years ago
Read 2 more answers
Use the quadratic formula to solve for the roots in the following equation. 4x 2 + 5x + 2 = 2x 2 + 7x – 1
Natali5045456 [20]

Answer:

So, The roots are x= \frac{1+\sqrt{5}i}{2} \,\, and \,\,x= \frac{1-\sqrt{5}i}{2}

Step-by-step explanation:

4x^2 + 5x + 2 = 2x^2 + 7x - 1

We need to solve the equation to find the roots using quadratic formula.

The quadratic formula is:

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

Rearranging the above equation:

4x^2 -2x^2+ 5x-7x + 2+1 =0

2x^2 -2x + 3 =0

Where a =2 , b=-2 and c =3 Putting values in quadratic equation and solving:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-2)\pm\sqrt{(-2)^2-4(2)(3)}}{2(2)}\\x=\frac{2\pm\sqrt{4-24}}{4}\\x=\frac{2\pm\sqrt{-20}}{4}\\\sqrt{-20} \,\,can\,\,be\,\, written\,\, as\,\, 2\sqrt{-5}\\ x=\frac{2\pm2\sqrt{-5}}{4}\\x=\frac{2+2\sqrt{-5}}{4} \,\, and \,\, x=\frac{2-2\sqrt{-5}}{4}\\x=\frac{2(1+\sqrt{-5})}{4} \,\, and \,\, x=\frac{2(1-\sqrt{-5})}{4}\\ x=\frac{1+\sqrt{-5}}{2} \,\, and \,\, x=\frac{1-\sqrt{-5}}{2}\\As \,\,we\,\, know\,\, \sqrt{-1} = i \\

x= \frac{1+\sqrt{5}i}{2} \,\, and \,\,x= \frac{1-\sqrt{5}i}{2}

So, The roots are x= \frac{1+\sqrt{5}i}{2} \,\, and \,\,x= \frac{1-\sqrt{5}i}{2}

6 0
3 years ago
Which set represents the domain of the function shown? (5 points)
Vikentia [17]

Answer:

To make it simple, when finding the domain all you have to do is look for the x within the points.

Looking at the first set you can tell that the domain is -3, 0, 4, and 11

The y or the f(x) would be 6, 2, 7, and 15

(Just saying y and f(x) are the same thing.)

For the next set the domain is 6, 2, 7, and 15

For the f(x) it is -3, 0, 4, and 11

Remeber x and Domain is the same thing so all you have to do is look for the x within the points of the plot, as for looking for the range, it is the same thing as y or f(x).

God Bless

Hope that this helps!!!!!!!!!!!!!!

7 0
2 years ago
Read 2 more answers
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