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3 years ago

Is 2/10 a simplest form

Mathematics

2 answers:

gayaneshka [121]3 years ago

6 0

You can divide the numerator and the denominator by 2 to get 1/5.

BARSIC [14]3 years ago

5 0

Nope, 1/5 is the simplest form of 2/10, since 2 and 10 are both be divisible ( can be divided) by 2 to get a new number of 1/5.

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Evaluate Functions Given f(x) = 2x + 6 and g(x) = 5x - 1 Evaluate f(-2)

ZanzabumX [31]

Answer:

f(-2) = -4x + 6

Step-by-step explanation:

f(x) = 2x + 6

f(-2) = -2(2x) + 6

f(-2) = -4x + 6

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2 years ago

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How to subtract radians?

Dafna1 [17]

You first need to make sure the denominators are equal, then when they are, you can add/subtract the numerators. Then you can place the result over the common denominator.

I hope this helps! :)

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3 years ago

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Each student at some college has a mathematics requirement M (to take at least one mathematics course) and a science requirement

Furkat [3]

all students = 150

M = 60

S = 45

M and S = 25

(a) At least one of the two requirements:

M or S = M + S - (M and S) = 60 + 45 - 25 = 80

(b) Exactly one of the two requirements:

(M or S) - (M and S) = 80 - 25 = 55

(all students) - (M or S) = 150 - 80 = 70

8 0

3 years ago

Write the standard form of the equation for the circle that passes through the points (2,31),(-15,14),(33,0)

stepladder [879]

Answer:

Step-by-step explanation:

Begin with the standard form of a circle as a conic:

$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$

For a circle, A and C will be the exact same, and B will equal 0. If B is non-zero, the equation represents a rotation of a conic, which is reserved for college-level courses. Shortening this, then:

$x^2+y^2+Dx+Ey+F=0$ is good enough for us for this. Start with the first point on the circle, (2, 31) and fill in the equation above with x and y:

$2^2+31^2+2D+31E+F=0$ which simplifies down to:

$(1):2D+31E+F=-965$

Do the same with the next point on the circle, (-15, 14):

$-15^2+14^2-15D+14E+F=0$ which simplifies down to:

$(2):-15D+14E+F=0$

Do the same with the last point, (33, 0):

$33^2+0^2+33D+0E+F=0$ which simplifies down to:

$(3):33D+F=-1089$

Now we will add (1) and (2) to get (4):

2D + 31E + F = -965

-15D + 14E + F = -421

Multiply the top equatio by -1 to get rid of the F terms:

-2D - 31E - F = 965

-15D + 14E + F = -421

which simplifies to

(4): -17D - 17E = 544

Now add (2) and (3) to get (5):

-15D + 14E + F = -421

33D + F = -1089

Multiply the bottom equation by -1 to get rid of the F terms:

-15D + 14E + F = -421

-33D - F = 1089

which simplifies to

(5): -48D + 14E = 668

Now add (4) and (5) together and eliminate the E terms:

-17D - 17E = 544

-48D + 14E = 668

In order to eliminate the E terms, multiply the top equation by 14 and the bottom equation by 17 to solve for D:

-238D - 238E = 7616

-816D + 238E = 11356

Which gives you that

D = -18

Now plug the value for D into (4) to find E:

-17(-18) - 17E = 544 and

306 - 17E = 544 and

-17E = 238 so

E = -14

Now plug the values for both D and E into (1) to find F:

2(-18) + 31(-14) + F = -965 and

-36 - 434 + F = -965 and

-470 + F = -965 so

F = -495

Now we can fill in the standard form of the conic:

$x^2+y^2-18x-14y=495$

but we're not done til we complete the square on both the x terms and the y terms (and I am assuming you know how to complete the square):

$(x^2-18x+81)+(y^2-14y+49)=495+81+49$ which simplifies to

$(x-9)^2+(y-7)^2=625$

The second choice down matches that equation, but the center they have there is not correct. The center of that circle is (9, 7) and they have it as being (7, 10) with a radius of 5. The radius is 25. The center is wrong in the equation that represents the circle as is the radius. Maybe let someone know that...

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3 years ago

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Pls help me Here is triangle ABC: No links pls

SOVA2 [1]

Answer:

$Given: AC=1$