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Anvisha [2.4K]
3 years ago
14

Equivalent fractions using the LCD 7/10 and 2/5

Mathematics
1 answer:
hammer [34]3 years ago
5 0
7/2,14/4,21/6,28/8,35/10,42/12,49/14,56/16,63/18,70/20
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Solve for h 1/2h-3=3/2-3h
Sergio039 [100]

Answer:

its 3

Step-by-step explanation:

i know that bc u keep mince

3 0
3 years ago
15h - m = 13h + q pleas helppp
tensa zangetsu [6.8K]
I don’t know what I’m solving for so I solved for h and with that it should be h=q/2+m/2

Explanation:

15h-m=13h+q
-13h -13h
———————-
2h-m=q
+m +m
———————-
2h=q+m
/2 /2 /2
———————-
H=q/2+m/2

8 0
3 years ago
Read 2 more answers
Help ME PLSSS 13 points
Greeley [361]

Answer:

6

Step-by-step explanation:

8 0
3 years ago
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Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in s
Jobisdone [24]

Answer:

f(x)=x^4-9x^2-50x-150

Step-by-step explanation:

Let f(x) be the polynomial function of minimum degree with real coefficients whose zeros are 5, -3, and -1 + 3i be f(x).

By the complex conjugate property of polynomials, -1-3i is also a root of this polynomial.

Therefore the polynomial in factored form is f(x)=(x-5)(x+3)(x-(-1+3i))(x-(-1+3i))

We expand to get:f(x)=(x^2-2x-15)(x^2+2x+10)

We expand further to get:\

f(x)=x^4-9x^2-50x-150

3 0
3 years ago
4.
marta [7]

Answer:

a) (x-2)(x-6) + (y-5)(y-11) =0\Rightarrow (x-4)^{2}+(y+3)^{2}=68\\C(4,-3) \:r=\sqrt{68} b) (a,b) (c, d) As long as (a,b) and (c,d) are the endpoints of of the diameter.

Step-by-step explanation:

a)

1) The reduced formula of the Circumference is given by:

(x-a)^{2}}+(y-b)^{2}=r^{2}

2) Let's expand the factored one into one closer to the pattern above:

x^{2}-8x+12+y^{2}+6y-55=0\Rightarrow x^{2}-8x+y^{2}+6y=43

3) Completing the square for both trinomials:

(x-4)^{2}+(y+3)^{2}=43+16+9\Rightarrow (x-4)^{2}+(y+3)^{2}=68

4) In the Reduced Formula, (x-a)^{2}}+(y-b)^{2}=r^{2},

C(a,b) \Rightarrow C(4,-3) \:the\:radius\:is\:r=\sqrt{68} \Rightarrow r=2\sqrt{17}

b) Using the previous example to show this:

When we  factor this way

(x-a)(x -c) + (y -b)(y-d) = 0

We are indeed, naming "a" and "b", the coordinates of (a, b) of the first endpoint and "b" and "d" the second endpoint as well.Id est, D (2, 5) and B (6,-11).

The radius, is \sqrt{68} \cong 8.25

So yes, the equation of the circle can be written as

(x-a)(x -c) + (y -b)(y-d) = 0

As long as (a,b) and (c,d) are the endpoints of of the diameter.

d_{AC}=d_{BC}=R

3 0
3 years ago
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