An <em>algebraic expression</em> is one that consists of both <u>number(s)</u> and an <u>alphabet(s)</u>. The <em>required</em> answers are:
i. Distance from Chenoa's <u>house</u> to the <em>coffee shop</em> = 6.0 miles
ii. D<u>istance</u> from <em>coffee shop</em> to Chenoa's <u>school</u> = 1.5 miles
iii. <em>Distance</em> from Chenoa's <u>house</u> to her <u>school</u> = 7.5 miles
An <em>algebraic expression</em> is one that consists of both <u>number(s)</u> and an <u>alphabet(s)</u>. The <em>alphabet</em> is referred to as the <u>unknown</u> whose <u>value</u> has to be <em>determined</em>.
In the given question, let the <u>distance</u> from the <em>coffee shop</em> to Chenoa's <u>school</u> be represented by y.
So that;
The <u>distance</u> from Chenoa's house to the <em>coffee shop</em> = (2y + 3) miles.
The <em>total distance</em> from Chenoa's <u>house</u> to her <u>school </u>= 5y.
This implies that:
(2y + 3) + y = 5y
3y + 3 = 5y
3 = 5y - 3y
2y = 3
y = 
= 1.5
The <em>distance</em> from the <em>coffee shop</em> to Chenoa's <u>school</u> is 1.5 miles.
Thus;
(2y + 3) = ( 2(1.5) + 3)
= 6
The <u>distance</u> from Chenoa's <u>house</u> to the <em>coffee shop</em> is 6 miles.
And,
5y = 5(1.5)
= 7.5
The <em>total distance</em> from Chenoa's <u>house</u> to her <u>school</u> is 7.5 miles.
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Answer:
The number line is shown below.
Step-by-step explanation:
We need to represent number that is less than 3 on a number line.
Let x be numbers.
So,
.
Now, we have a number which represents x<3. Here 3 is not included in the solution set, so there is an open circle on 3 and left side of 3 are in the solution set.
The number line is shown below.
6x - 4 < 8
Add 4 to both sides:
6x < 12
Divide both sides by 6:
x < 2
Answer: x < 2
Answer:
object is 59.30m away from the base of the cliff
Step-by-step explanation:
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