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zubka84 [21]
3 years ago
8

Pls help i need 5he ans to this question

Mathematics
1 answer:
Karolina [17]3 years ago
5 0
-(4/5)*(3/7)*(15/16)*(-14/9)   [two negatives make one positive]
=(4*3*15*14)/(5*7*16*9)   [simplify 4/16]
=(3*15*14)/(5*7*4*9)
=(3*3*14)/(7*4*9)
=(14)/(7*4)
=(2)/(4)
=1/2
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Chester Boles reviewed his credit card's monthly statement. It shows a $2,376.10 previous balance and this month's new purchases
ss7ja [257]

Initial balance, I = $2376.10 .

Total amount of purchase made, A = $( 875.22+65.75+45.22+21.23 ) = $1007.42 .

Total amount credit, c = $875.22 .

Fine, f = $45.30 .

Another purchase, a=\dfrac{2376.10\times 2.5}{100}=\$ 59.4025 .

So, balance left is :

B = I - A - f - a + c

B = 2376.10 - 1007.42 - 45.30 - 59.4025 + 875.22

B = $2139.1975

Hence, this is the required solution.

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3 years ago
Dwayne starts at the surface of the water and swims to -25 feet. From there he swims +10 feet to see the fish and then back down
d1i1m1o1n [39]
(-25) + 10 + (-5) = -20 feet is his current position. To reach zero, he should add 20 feet to his altitude.
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If the correlation between variable a and variable b is .20, then variable a explains _____ of the variance in variable
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3 years ago
The endpoints of a diameter of a circle are (6,2) and (−2,5) . What is the standard form of the equation of this circle?
Paul [167]

Answer:

The standard form of the given circle is

(x-2)^2+(y-\frac{7}{2})^2=73

Step-by-step explanation:

Given that the end points of a diameter of a circle are (6,2) and (-2,5);

Now to find the standard form of the equation of this circle:

The center is (h,k) of the circle is the midpoint of the given diameter

midpoint formula is M=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})

Let (x_{1},y_{1}) and (x_{2},y_{2}) be the given points (6,2) and (-2,5) respectively.

M=(\frac{6-2}{2},\frac{2+5}{2})

M=(\frac{4}{2},\frac{7}{2})

M=(2,\frac{7}{2})

Therefore the center (h,k) is (2,\frac{7}{2})

now to find the radius:

The diameter is the distance between the given points (6,2) and (-2,5)

d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

=\sqrt{(-2-6)^2+(5-2)^2}

=\sqrt{(-8)^2+(3)^2}

=\sqrt{64+9}

=\sqrt{73}

Therefore the radius is \sqrt{73}

i.e., r=\sqrt{73}

Therefore the standard form of the circle is

(x-h)^2+(y-k)^2=r^2

Now substituting the center  and radiuswe get

(x-2)^2+(y-\frac{7}{2})^2=(\sqrt{73})^2

(x-2)^2+(y-\frac{7}{2})^2=73

Therefore the standard form of the given circle is

(x-2)^2+(y-\frac{7}{2})^2=73

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3 years ago
Whats bigger 7/12 or 3/4
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3/4 will be the bigger value of the two
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