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4 years ago

Perpendicular Lines please help me guys I'm lostt

Mathematics

1 answer:

mamaluj [8]4 years ago

4 0

9 = 2(6) - b
9 = 12 - b
-3 = b

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Solve for x. 2x + 5 = x - 8

Gemiola [76]

2x+5=x-8
-x on both sides
x+5=8
-5 on both sides
x=-13

7 0

4 years ago

(a.) what is the GCF of 33 and 42 using prime factors?

Gennadij [26K]

$(a)\\33|\fbox3\\11|11\\.\ 1|\\---------\\42|2\\21|\fbox3\\.\ 7|7\\.\ 1|\\----------\\GCF(33;\ 42)=3$

$(b)\\2\underline{|72\ \ \ \ \ 30}\\.3\underline{|36\ \ \ \ \ 15}\\...\underline{|12\ \ \ \ \ \ 5}\\\\GCF(72;\ 30)=2\times3=6$

$(c)\\factors\ of\ 27:\ \fbox1;\ 3;\ 9;\ 27\\\\factors\ of\ 76:\ \fbox1;\ 2;\ 4;\ 19;\ 38;\ 76\\\\factors\ of\ 34:\ \fbox1;\ 2;\ 17;\ 34\\\\GCF(27;\ 76;\ 34)=1\\\\The\ numbers\ 27;\ 76\ and\ 34\ are\ coprime\ numbers.$

6 0

4 years ago

Use elimination to solve the following system of equations:5x + 5y = -30, 2x - y = 15

elixir [45]

Add the equations in order to solve for the first variable. Plug this value into the other equations in order to solve for the remaining variables.

(3,-9)

5 0

3 years ago

Read 2 more answers

Find the value of x .<br><br> 12.5,-10,-7.5,x ; The mean is 11.5

Tems11 [23]

Answer:

Step-by-step explanation:

The mean is obtained by adding up the values in the data set and then divide by the number of values that you added:

$\bar X=\sum{X_i}\\$

Given our mean as 11.5 and n=4:

$\bar X=\frac{1}{n}\sum{X_i}\\\\\\\\\therefore\sum{X_i}\\=11.5\times 4=46$

#Knowing our sum as 65, we can obtain x by subtracting sum of known values from 65:

$x=46-(12.5+-10+-7.5)\\=51$

Hence, the value of x is 51

8 0

4 years ago

A male student of the author has a measured pulse rate of 52 beats per

Liono4ka [1.6K]

Answer:

$z=-1.49$

Step-by-step explanation:

$\text{Standard Score, z} =\dfrac{X-\mu}{\sigma} $ where:\\\\Mean Pulse rate, \mu =67.3$ beats per minute\\Standard Deviation, \sigma = 10.3$ beats per minute.\\$

For a male student who has a measured pulse rate of 52 beats per minute.

Raw Score, X =52 beats per minute.

Therefore:

$\text{Standard Score, z} =\dfrac{52-67.3}{10.3}\\z=-1.49$

Since the usual pulse rates are within 2 standard deviations of the mean, a z-score of -1.49 tells us that the selected student's pulse rate is within the usual pulse rates.

8 0

4 years ago