Find the point of intersection of the lines x-3 = 0 and y-5 = 0.

Mathematics

1 answer:

Vikki [24]3 years ago

3 0

Answer:

x = 3 and y = 5

Step-by-step explanation:

Given equations are :

x-3 = 0 ....(1)

y-5 = 0 ....(2)

We need to find the point of intersection of the lines.

Add 3 both sides in equation (1).

x-3+3=3

x = 3

Now, add 5 to both sides of the equation.

y-5+5=0+5

y = 5

Hence, the required point of intersection of the lines is x = 3 and y = 5.

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A biker travels the 120 mi from New York , NY to Wilmington, DE in 8 hours. A bus, a truck, and a rider on a horse take the same

pickupchik [31]

Answer:

First, remember the relation:

Distance = Speed*Time.

We know that the distance is 120mi.

A) For the biker we know that:

Time = 8hours.

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B) For the bus we know that:

Time = 2 hours.

Speed = Distance/Time = 120mi/2h = 60 mi/h.

C) For the truck we know that:

Time = 2.5 hours.

Speed = Distance/Time = 120mi/2.5h = 48 mi/h.

D) For the rider on a horse we know that:

Time = 18.75 hours.

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E) The equation for the speed is:

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4 0

3 years ago

Please help me figure out this problem

tiny-mole [99]

Answer:

$f(g(x))=f\left(-\dfrac{3x+21}{5}\right)=x\\\\g(f(x))=g\left(-\dfrac{5}{3}x-7\right)=x$

Step-by-step explanation:

Fill in the given expressions and simplify.

$f(g(x))=f\left(-\dfrac{3x+21}{5}\right)=-\dfrac{5}{3}\left(-\dfrac{3x+21}{5}\right)-7\\\\=\left(-\dfrac{5}{3}\right)\left(-\dfrac{3x}{5}\right)-\left(\dfrac{5}{3}\right)\left(-\dfrac{21}{5}\right)-7\\\\=\dfrac{15x}{15}+\dfrac{5\cdot 21}{3\cdot 5}-7=x+7-7=x$