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3 years ago

4p3 - 3px=ypls give me answer fast

1 answer:

5 0

Answer:idk if it right but what i got was
y=12p-3px

explanation:
4•3=12 then just add the P
then since 3 is by itself it would remain -3px

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28.

Mkey [24]

D.
x-intercept = 5
y-intercept = 15

4 0

3 years ago

PLEASE ANSWER IF YOU CAN Cameron did 4 jumping jacks on Wednesday, 9 jumping jacks on Thursday, 19 jumping jacks on Friday, and

Blizzard [7]

Answer:

My answer is 64

Step-by-step explanation:

How I got my answer is that you write the numbers and the difference of the each 2 numbers. When you see the pattern is 5, 10, 20, and it might as well be 25 because you are going to start all over with the pattern

6 0

3 years ago

Read 2 more answers

Solve for x. what is 1900=1300(1+.0625)^x?

Ksenya-84 [330]

First rearrange as:
$\frac{1900}{1300}=1.0625^{x}$
Taking logs of both sides, we get:
$log\ \frac{1900}{1300}=x\ log1.0625$
$x=\frac{log\frac{1900}{1300}}{log\ 1.0625}=6.26$
x = 6.26.

7 0

3 years ago

Read 2 more answers

What is the measure of Arc BA? An explanation would be much appreciated.

yKpoI14uk [10]

Answer:

m<BA = 180degrees

Step-by-step explanation:

First you must know that measure of <BA is 180 degrees and m<CD = m<BA

Since <CD = 16z - 12, hence;

180 = 16z - 12

16z = 180++12

16z = 192

z = 192/16

z = 12

Get <BA

m<BA = 16z - 12

m<BA = 16(12) - 12

m<BA = 192 - 12

m<BA = 180degrees

8 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

4 years ago

4p3 - 3px=ypls give me answer fast ​

4p3 - 3px=ypls give me answer fast