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2 years ago

Can anyone identify the center & radius of these

Mathematics

1 answer:

kobusy [5.1K]2 years ago

6 0

Answer:

43 is the answer to your question

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Please I need help. Thnx

Mkey [24]

Answer:

this is 37

Step-by-step explanation:

because 9+9+2+2+11+4=37

3 0

3 years ago

Read 2 more answers

Can you help or tell answer for numbers 1 2 3 4

Masteriza [31]

Here you go, learn from this child.

6 0

3 years ago

PEOPLE WHO ARE GOOD AT MATH CAN YOU PLEASE HELP ME!!!

OlgaM077 [116]

The cost of using 19 HCF of water is $32.49

Given in the question:

The monthly cost (in dollars) of the water use (in dollars) is a linear function of the amount of water used (in hundreds of cubic feet, HCF)

The cost for using 17 HCF of water is using $32.13

and, the cost of using 35 HCF is $61.83.

To find the cost of using 19 HCF of water.

Now, According to the question:

The cost for using 17 HCF of water is $32.13

and, the cost of using 35 HCF is $61.83.

To find the slope:

(17, 32.13) and (35, 61.83)

Slope = (61.83 - 32.13)/ (35 - 17) = 1.65

We know that:

Formula of slope :

y = mx + b

32.13 = 1.65 x 17 + b

b = 1.14

The equation will be :

C(x) = 1.65x + 1.14

Now, To find the cost of using 19 HCF of water.

C(19) = 1.65 × 19 + 1.14

C(19) = $32.49

Hence, the cost of using 19 HCF of water is $32.49.

Learn more about Slopes at:

brainly.com/question/3605446

#SPJ1

6 0

1 year ago

Y = 0.95X 7.95<br> Write and solve a linear equation to find how many songs he downloaded

Dmitry [639]

0.95 x 7.95 = 7.5525

5 0

3 years ago

Solve these linear equations by Cramer's Rules Xj=det Bj / det A:

timurjin [86]

Answer:

(a) $x_1=-2,x_2=1$

(b) $x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}$

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix $A$ and the matrix $B_j$ for each variable. The matrix $A$ is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get $A$ more easily.

$2x_1+5x_2=1\\x_1+4x_2=2$

$\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]$

To get $B_1$ , replace in the matrix A the 1st column with the results of the equations:

$B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]$

To get $B_2$ , replace in the matrix A the 2nd column with the results of the equations:

$B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]$

Apply the rule to solve $x_1$ :

$x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2$

In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable $x_1$ in one of the equations and solve for $x_2$ :

$2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1$

(b) In this system, follow the same steps,ust remember $B_3$ is formed by replacing the 3rd column of A with the results of the equations:

$2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0$

$\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]$

$B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]$

$B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]$

$B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]$

$x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}$

$x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}$

$x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}$

6 0

3 years ago