Complete the two- column proof

1 answer:

7 0

The correct answers are a) given; b) substitution property; c) subtraction property of equality; d) division property of equality; and e) reflexive property of equality.

Explanation:
The first line of your proof just restates what was given to us; therefore the reason is "given."

Between the first step and the second step, 11x was changed to 88; this is because x=8, and we substitute that in for x. That would give us 11*8=88, with the justification being substitution.

Between the second line and third line, 88 is cancelled from the left side and the -1 becomes -89; this happens because in order to cancel a positive 88, you must subtract it from both sides. This is the subtraction property of equality.

Between the third line and fourth line, -6 is cancelled from the left side and on the right side, -89 is changed to 89/6. This happens because in order to cancel the -6 that is beside the y (which means multiplication), we must divide both sides; this is the division property of equality.

The last step is to rewrite the equation with y on the left and the answer on the right; the reflexive property allows us to do that.

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Estimate the sum or difference 6/10 + 7/12

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6/10 + 7/12 = 142/120 = 71/60 = 1.18 (One whole number, and 11/60)

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Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)

Mnenie [13.5K]

One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
$(x-h)^2+(y-k)^2=r^2$
the center is (h,k) and the radius is r

and the distance formula is
distance between $(x_1,y_1)$ and $(x_2,y_2)$ is
$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle

so
$(x+2)^2+(y-3)^2=18$
$(x-(-2))^2+(y-3)^2=(\sqrt{18})^2$
$(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2$

the radius is $3\sqrt{2}$
center is (-2,3)

find distance between (8,4) and (-2,3)

$D=\sqrt{(8-(-2))^2+(4-3)^2}$
$D=\sqrt{(8+2)^2+(1)^2}$
$D=\sqrt{10^2+1}$
$D=\sqrt{100+1}$
$D=\sqrt{101}$

$r=3\sqrt{2}$ ≈4.2
$D=\sqrt{101}$ ≈10.04

do r<D

(8,4) is outside the circle

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2 years ago

Solve the following proportion. х 42 35 10

anzhelika [568]

Answer:

Step-by-step explanation:

The proportion is equal to $\frac{x}{10}=\frac{42}{35}$