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Simora [160]
2 years ago
9

Last year the girls’ basketball team had 8 fifth-grade students and 7 sixth-grade students. What was the ratio of sixth-grade st

udents to fifth-grade students on the team?
Mathematics
2 answers:
attashe74 [19]2 years ago
6 0

7:8 it could also be written as 7 to 8 or 7/8

Since we need to find out SIXTH to FIFTH grades we have to put the sixth graders first. Hope this helps and have a great day!

timofeeve [1]2 years ago
4 0

Answer:

the ratio of the sixth-grade students to fifth-grade students on the team is 7 : 8

Step-by-step explanation:

Given Parameters

Fifth grade students of the basket team = 8

Sixth grade students of the basket team = 7

Required;

Ratio of the sixth-grade students to fifth-grade students on the team

Let F represent the fifth graders

F = 8

Let S represent the sixth graders

S = 7

The ratio of S to F is represented mathematically by S : F

By substituton

S : F becomes

7 : 8

Hence, the ratio of the sixth-grade students to fifth-grade students on the team is 7 : 8

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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

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(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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104=10 -----> is not true

therefore

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So, Heat (Q) = m×s×ΔT

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or,           Q = 3024 × 10³ joule

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