Hm good question. Maybe you can measure the temperature in a place and measure another temperature in a other place and add it together.
1. 35x + 0.15y
=3500x + 15y [multiplying the equation by 100]
= 700x + 3y [dividing the equation by 5]
2. Let x be an even integer
x + (x+2) + (x+2+2)
x+x+2+x+4
3x+6
3. 1/5*(r+7) - 9s
(r+7)/5 - 9s
4. x=cows
y= goats + sheep
x= (y+140)/2
Hello.
First, draw a picture of two parallel lines and a transversal.
Find the two same side angles, either the same side interior or the same side exterior.
We do not know the measurement of the angle, so we'll label it x.
The second angle is 20° smaller than the other one so we'll label is x-20
Same side angles are supplementary. Meaning that they add up to 180°
Next, make an equation.
1st angle + 2nd angle = 180°
Now substitute the first and second angle to x and x-20
The substituted equation should now look like this.
x+x-20=180
solve it and you'll get...
x=100
But we're not done.
The angle labeled x is 100°
The other one is x-20
And since x=100, the second angle is 100-20=80°
Now use linear pairs, more same side angles, vertical angles, alternate angles, and linear pairs, to label the rest. Remember that each angle is either 100° or 80°.
Hope this helped!
-Emily
Answer:
KL = 50
Step-by-step explanation:
∆JML is similar to ∆JNL. it follows that:
[tex] \frac{JM}{JN} = \frac{JL}{JK} [\tex]
JM = 4 + 20 = 24
JN = 4
JL = 10 + KL
JK = 10
Plug in the values
[tex] \frac{24}{4} = \frac{10 + KL}{10} [\tex]
[tex] 6 = \frac{10 + KL}{10} [\tex]
Multiply both sides by 10
[tex] 6*10 = \frac{10 + KL}{10}*10 [\tex]
[tex] 60 = 10 + KL [\tex]
Subtract 10 from each side
[tex] 60 - 10 = KL [\tex]
[tex] 50 = KL [\tex]
KL = 50
Answer:
Step-by-step explanation:
<u>Expression on the board :</u>
![12.2x + 50.6 y + 3(1.4x - 2.6y)\\\\= 12.2x + 50.6y + 4.2x - 7.8y\\\\= 16.4x + 42.8y](https://tex.z-dn.net/?f=12.2x%20%2B%2050.6%20y%20%2B%203%281.4x%20-%202.6y%29%5C%5C%5C%5C%3D%2012.2x%20%2B%2050.6y%20%2B%204.2x%20-%207.8y%5C%5C%5C%5C%3D%2016.4x%20%2B%2042.8y)
<u>Expression by the student </u>:
<em><u> Expression 1 </u></em>:
![4 ( 4.1x + 10.7 y ) \\\\= ( 4 \times 4.1x ) + ( 4 \times 10.7 y)\\\\= 16.4x + 42.8y\\\\](https://tex.z-dn.net/?f=4%20%28%204.1x%20%2B%2010.7%20y%20%29%20%5C%5C%5C%5C%3D%20%20%28%204%20%5Ctimes%204.1x%20%29%20%2B%20%28%204%20%5Ctimes%2010.7%20y%29%5C%5C%5C%5C%3D%2016.4x%20%2B%2042.8y%5C%5C%5C%5C)
The expression 1 is equivalent to the expression on the board. Because the coefficients of x and y are same on both.
<em><u>Expression 2 :</u></em>
<em><u /></em>
<em><u /></em>
The expression 2 is also equivalent to the expression on the board. Because the coefficients of x and y are the same.