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3 years ago

What is the standard equation for a circle with the center (2,1) and passing through (0, 1)

2 answers:

Volgvan3 years ago

4 0

The radius is the distance from the center to any point. The distance from (2,1) to (0,1) is of course 2. The general formula for a circle with radius r and center (a,b) is

$(x-a)^2 + (y-b)^2 = r^2$

In our case we get

$(x-2)^2 + (y-1)^2 = 4$

BigorU [14]3 years ago

3 0

So this is what I got (x-2)^2+(y-1)^2=4

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The gardening club at school is growing vegetables. The club has 300 square feet of planting beds. Cucumber plants require 6 squ

nadya68 [22]

The question doesn't ask anything in particular, I will show the set of inequalities defined in the problem.

Answer:

System of inequalities:

$6c+4t\leq 300$

$c>0\\t > 0\\c+t\geq 60$

Step-by-step explanation:

Inequalities

The express relations between expressions with a sign other than the equal sign. Common relationals are 'less than', 'greater than', 'not equal to', and many others.

The gardening club at school has 300 square feet of planting beds to plant cucumber and tomato. Each cucumber plant requires 6 square feet of growing space and each tomato plant requires 4 square feet of growing space. We know the total area cannot exceed 300 square feet, so

$6c+4t\leq 300$

Being c and t the number of cucumber and tomato plants respectively.

We also know the students want to plant some of each type of plant and have at least 60 plants. This lead us to more conditions

$c>0\\t > 0\\c+t\geq 60$

Note: The set of inequalities shown is not enough to uniquely solve the problem. We need something to maximize or minimize to optimize c and t

8 0

3 years ago

All the fraction that are greater than 1\2

never [62]

Hi There!

Change all the fractions to decimal you do that by dividing numerator by the denominator. Remember $\frac{1}{2}$ as a decimal is 0.5

A= $\frac{3}{7}$ = 0.43

B= $\frac{5}{9}$ = 0.60

C= $\frac{8}{10}$ = 0.8

D= $\frac{5}{12}$ = 0.42

E= $\frac{3}{5}$ = 0.6

Based, on the data above B,C,E are greater than $\frac{1}{2}$ because they all have decimals greater than 0.5, B= 0.6, C= 0.8, E=0.6

So, $\frac{5}{9} ,\frac{8}{10} ,\frac{3}{5}$ are all greater than $\frac{1}{2}$

Hope This Helps :)

6 0

3 years ago

I'M GIVING BRAINLIEST

Karolina [17]

I don’t get want u need help with.

4 0

3 years ago

Can someone please do this for me, It would be amazing! sorry that's all the points I have :(

Serggg [28]

If i was you i would look up every box in the word bank then find examples of each box and do the best I could.

-hope this helps!

7 0

3 years ago

Write an equation of the perpendicular bisector of the segment with endpoints G 9,8       and H 3,2      .

Norma-Jean [14]

Answer:

y = -x + 11

Step-by-step explanation:

The equation of a straight line is is given by:

y = mx + b; where m is the slope and b is the y intercept

The equation of the line joining G(9, 8) and H(3, 2) is given as:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\y-8=\frac{2-8}{3-9}(x-9)\\\\y-8=x-9\\\\y=x-1$

The perpendicular bisector of the line joining G(9, 8) and H(3, 2) is perpendicular to the line joining G(9, 8) and H(3, 2) and passes through the midpoint of line joining G(9, 8) and H(3, 2).

Let (x, y) be the midpoint of the line joining G(9, 8) and H(3, 2). Hence:

x = (9 + 3)/2 = 6

y = (8 + 2)/2 = 5

The midpoint = (6, 5)

Two lines are perpendicular if the product of their slopes is -1.

The line joining G(9, 8) and H(3, 2) has a slope of 1, hence, the slope of the perpendicular bisector would be -1.

This means that the perpendicular bisector has a slope of -1 and passes through (6, 5). Using:

$y-y_1=m(x-x_1)\\\\y-5=-1(x-6)\\\\y-5=-x+6\\\\y=-x+11$

The equation of the perpendicular bisector is y = -x + 11

8 0

3 years ago