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iogann1982 [59]

3 years ago

11

Barry's deck of cards contains 40 blue cards and 60 red cards max's deck of cards contains the same number of blue cards but the

ratio blue cards to red cards 5:9 how many total cards does max's deck of cards contain?
NEED THIS ASAP

1 answer:

Rudiy273 years ago

5 0

Max's deck also has 40 blue cards, and for every 5 blue cards in his deck there are 9 red cards. His blue cards can be divided exactly into 8 groups (8 times 5 is 40), so he should also be able to divide his red cards into exactly 8 groups of 9 cards each. So there are 72 red cards in his deck (8 times 9 is 72).

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The graph of y=1/x is shown below. Draw the graph of its inverse.

lord [1]

Y=1/x solve for x

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4 0

3 years ago

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Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle

Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

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

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x+3)^{2} +(y-4)^{2}=r^{2}$

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

$r=\sqrt{(2-4)^{2}+(-6+3)^{2}}$

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

$r=\sqrt{13}\ units$

The equation of the circle is equal to

$(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}$

$(x+3)^{2} +(y-4)^{2}=13$

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

$(10+3)^{2} +(4-4)^{2}=13$

$(13)^{2} +(0)^{2}=13$

$169=13$ -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x-1)^{2} +(y-3)^{2}=r^{2}$

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

$r=\sqrt{(6-3)^{2}+(2-1)^{2}}$

$r=\sqrt{(3)^{2}+(1)^{2}}$

$r=\sqrt{10}\ units$

The equation of the circle is equal to

$(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}$

$(x-1)^{2} +(y-3)^{2}=10$

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

$(11-1)^{2} +(5-3)^{2}=10$

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7 0

3 years ago

How do you solve the problem?

sleet_krkn [62]

3x-17+x+40+2x-5=180
6x=162
x=27
(From left to right) 67,64, 49

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3 years ago

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Artyom0805 [142]

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Then recalling that $2^3=8$ , you have

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3 years ago

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goblinko [34]

Answer

hello :

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5 0

3 years ago

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