0.5ab-0.3bc +2.4cd; 1.2bc + 0.2cd -3.3de; 8.6ab-1.4cd +17de

I need help or my report card is gonna have a f its coming out tommorow

OverLord2011 [107]

Answer:

F.

A.

Step-by-step explanation:

hope this helps you

8 0

3 years ago

What is the value of x in the equation −6 + x = −2? (5 points) a 8 b 4 c −4 d −8

Ymorist [56]

The correct answer is B)4

5 0

3 years ago

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

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

$(10)^{2} +(2)^{2}=10$

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

therefore

The point is not on the circle

The statement is false

7 0

3 years ago

Diego says that 108 cubes with an edge length of 1/3 inch are needed to fill a rectangle prism that is 3 inches by 1 inch by 1 1

RUDIKE [14]

Answer:

Step-by-step explanation:

Option C) 4 cubic inches is the volume of the rectangular prism. Step-by-step explanation: It is given that, 108 cubes with an edge length of 1/3 inch are needed to fill a rectangular prism. Therefore, the volume of the rectangular prism is equal to the volume of the 108 cubes needed to fill that rectangular prism.

3 0

3 years ago

Read 2 more answers

4x + 7 PLEASEE HELPPPPPPPPPPP

nasty-shy [4]

Answer:

thats the simplest form it can be unless you want it expressed as a graph ....

7 0

3 years ago

Read 2 more answers