The graph will cross at the coordinates (-2, 9)
<h3>How to solve equations?</h3>
y = 3x + 15
y = 3 - 3x
y = 3x + 15
Hence,
when x = 2
y = 3(2) + 15 = 21
when x = 0
y = 3(0) + 15 = 15
y = 3 - 3x
when x = 2
y = 3 - 3(2)
y = 3 - 6
y = -3
when x = 0
y = 3 - 3(0)
y = 3
Therefore, let's check if the equation will cross.
y = 3x + 15
y = 3 - 3x
using substitution,
3 - 3x = 3x + 15
3 - 15 = 3x + 3x
- 12 = 6x
x = -12 / 6
x = -2
y = 3 - 3(-2)
y = 3 + 6
y = 9
Therefore, the graph will cross at the coordinates (-2, 9)
learn more on equations here: brainly.com/question/19297665
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Answer:
I don't really know but one triangle is 180 degrees so two added together is 360 degrees...
but ohh I remembered
a=c+d (since a+b=180 and c+d+b=180)
e=g+h (since e+f=180 and g+h+f= 180)
and the fact is that c+d+g+h+b+f=the angle sum of quadrilateral PQRS
a+e+b+f= the angle sum of quadrilateral PQRS (by substitutions)
a+e+b+f=360 (four angles are from one origin so it's 360 degrees )
Hi!
These are not <span>equivalent fractions because 8/10 reduces to 4/5, not 3/5, therefore, they are not </span>equivalent.
Answer:
Add 4 to both sides
x<7
Step-by-step explanation:
- Square <em>:</em><em>4</em><em> </em><em>sides</em><em>,</em><em>each</em><em> </em><em>side</em><em> </em><em>9</em><em>0</em><em> </em><em>degrees</em><em> </em><em>.</em><em>.</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>:each angle is 90 degrees
: 4 diagonals
- Rectangle: 4 sides..4 angles..2 diagonals..
- Rhombus : <u>4</u><u> </u><u>sides</u><u>.</u><u>.</u><u>2</u><u> </u><u>angles</u><u>.</u><u>.</u><u>.</u><u>2</u><u> </u><u>diagonals</u><u>.</u><u>.</u>
- Parallelogram : 4 sides..4 angles..2 diagonals..
- Quadrilateral: 4 sides...4 angles...4 diagonals..
<em>If</em><em> </em><em>this</em><em> answer</em><em> helps</em><em><u> you</u></em><em><u> plz</u></em><em><u> mark</u></em><em><u> as</u></em><em><u> brainlist</u></em><em><u>.</u></em><em><u>.</u></em>