<u><em>Answer:</em></u>
<u><em>1. x=4 or x=10</em></u>
<u><em>2. x=-10 or x=10</em></u>
<em>No equivalent</em>
<em />
Step-by-step explanation:
1. |x-3|=7 and x-3=7
<em>Absoulte rules.</em>
<em>x-3=-7 and x-3=7</em>
<em>Add 3 both sides.</em>
<em>x-3+3=-7+3</em>
<em>Simplify.</em>
<em>-7+3=-4</em>
<em>x=-4</em>
<em>x-3=7</em>
<em>Add by 3 both sides.</em>
<em>x-3+3=7+3=10</em>
<em>x=10</em>
<em>X=-4, and X=10 is the correct answer.</em>
<em>________________________________</em>
<em>2. |x|-3=7</em>
<em>Add by 3 both sides.</em>
<em>|x|-3+3=7+3</em>
<em>Simplify.</em>
<em>7+3=10</em>
<em>X=10</em>
<em>X=-10 and X=10 is the correct answer.</em>
<em>______________________________</em>
Answer:
13 sq cm.
Step-by-step explanation:
To find the area of this irregular figure, we will need to find the areas of a rectangle and triangle within this figure.
To find the area of the rectangle, we can simply do:
1 × 4= 4 sq cm.
Now, this is where it gets a little trickier. To find the area of the triangle, we will first need to determine the base and height.
We can see from the diagram that to find the base, we need to subtract:
7-4= 3 cm. This is the base of ONLY the triangle.
To find the height, we will need to do:
5+1= 6 cm. This is the total height of the entirety of the triangle.
We can now use the Triangle Area formula: 
(bh), to calculate the area of the triangle.
(6*3)= 9 sq cm.
Finally, add the areas of the rectangle and triangle together, giving you:
9+4= 13 sq cm.
(Y+22) + 3 = Y + 22 + 3 = Y + 25
Answer:
what i am supposed to answer
Step-by-step explanation:
Answer:
It would be "a sentence that includes two independent clauses connected by a conjunction" which is Choice B.
Choice A wouldn't work as there needs to be two or more independent clauses for it to be a Compound Sentence.
<u>Choice B would work, as it follows the definition of a Compound Sentence.</u>
Choice C wouldn't work, that would be a Complex Sentence
Choice D wouldn't, that would be a Simple Sentence.
Also, note that this question is something for the Language Arts section.
