Answer:
Irregular
Step-by-step explanation:
C.pentagon
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THANK YOU


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<h3>I'm solving it using substitution method:-</h3>
<h3>7x+2y=3 {given}</h3>
<h3>=>7x=3-2y</h3>
<h3>=>x=(3-2y)/7-------(1)</h3>
<h3>x-3y=30 {given}</h3>
<h3>=>x=30+3y</h3>
<h3>=>(3-2y)/7=30+3y {putting the value of x from eqn 1}</h3>
<h3>=>3-2y=210+21y</h3>
<h3>=>3-210=21y+2y</h3>
<h3>=>-207=23y</h3>
<h3>=>y= -207/23= -9</h3>
<h3>putting the value of y on eqn(1):-</h3>
<h3>x=(3-2y)/7</h3>
<h3>x=>(3-2(-9))/7=(3+18)/7=21/7=3</h3>
<h2>Hence, x=3, y= -9</h2>
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Answer:
r = 14
Step-by-step explanation:
Use distributive property:
4r + 2r - 16 = 12 + 4r
Combine like terms:
6r - 16 = 12 + 4r
Isolate the variable:
6r = 28 + 4r
2r = 28
r = 14
Answer:
Step-by-step explanation:
x has to be equal to or less than a -5 to get (4 + x) to be a negative number.
x has to be equal to or greater than -10 for (11 + x) to be a positive number.
Therefore, any number between -10 and -5 will work.