Answer:
10
Step-by-step explanation:
The equation of the line that is parallel to the line whose equation is 3x-2y=7 would be y = 3/2x + b, in which b can be any real number.
How are parallel straight lines related?
Parallel lines have the same slope since the slope is like a measure of steepness and since parallel lines are of the same steepness, thus, are of the same slope.
We have been given a parallel line with has equation
3x-2y=7
In order to solve this, the slope of the original line.
3x - 2y = 7
-2y = -3x + 7
y = 3/2x - 7/2
thus its slope is 3/2.
thus, the slope of the needed line is 3/2 too.
we know that any line that is parallel to that would have this slope.
So anything is written in the form:
y = 3/2x + b
The equation of the line that is parallel to the line whose equation is 3x-2y=7 would be y = 3/2x + b, in which b can be any real number.
Learn more about parallel lines here:
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<h2>
<u>Answer:</u></h2>
⟶ 2³ × 3² is the prime factorization for which one of these choices?
Let's check,
1) 6 = 3 × 2 [So, obviously not this choice]
2) 25 = 5 × 5 = 5² [Not this either]
3) 36 = 3 × 2 × 2 × 3 = 3² × 2² [Doesn't match with 2³ × 3²]
4) 72 = 2 × 2 × 2 × 3 × 3 = <u>2</u><u>³</u><u> </u><u>×</u><u> </u><u>3</u><u>²</u><u> </u>[Matches]
⟶ The answer is, choice <u>7</u><u>2</u><u>.</u>

The first evening she received x calls.
The second evening she received 7 fewer calls than the first evening, so she received x-7 calls.
The third evening she received 3 times as many calls as the first evening, so she received 3x calls.
Over the three evenings she received 73 calls altogether.

The first evening she received 16 calls, the second evening she received 9 calls, the third evening she received 48 calls.
Complete question :
a ladder 5 meters long is leaning against a wall. the base of the ladder is sliding away from the wall at a rate of 1 meter per second. how fast is the top of the ladder sliding down the wall at the instant when the base is 4 meters from the wall?
Answer:
-4/3 m/s
Step-by-step explanation:
Using Pythagoras :
a² = x² + c²
5² = x² + y² - - - (1)
The horizontal distance x with respect to time t = 1m/sec
dx/dt = 1m/sec
To obtain the vertical height 'y' with respect to time t when x = 4
5² = x² + y²
Wen x = 4
5² = 4² + y²
25 = 16 + y²
y =√(25 - 16)
y = 3
Differentiate (1) with respect to t
x² + y² = 25 - ---- - (1)
2x * dx /dt + 2y dy/dt = 0
dx/dt = 1
x = 4.
y = 3
2(4)(1) + 2(3) * dy/dt = 0
8 + 6dy/dt = 0
6 dy/dt = - 8
dy/dt = - 8/6
dy/dt = - 4/3