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Sloan [31]
3 years ago
11

Find and x and y PLEASE HELP

Mathematics
1 answer:
Bogdan [553]3 years ago
6 0

Answer:

Step-by-step explanation:

x is 79 and y is 47

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Train X is traveling at a constant speed of 30 miles per hour and Train Y is traveling at a constant speed of 40 miles per hour.
topjm [15]

Answer:

3.5 hours

Step-by-step explanation:

Speed of train X=30 mph

Speed of train Y=40  mph

Relative speed When the two trains travelling in same direction

Relative speed=40-30=10 mph

Total distance =25+10=35 miles

We have to find the time when train Y is 10 miles ahead of train X.

We know that

Time=\frac{Distance}{Relative\;speed}

Using the formula

Then, we get

Time=\frac{35}{10}=3.5 hours

Hence, it will be 3.5 hours until train Y is 10 miles ahead of train X.

7 0
3 years ago
Read 2 more answers
I'm dying of boredom. who else wanna talk ^-^​
sertanlavr [38]

Answer:

me

Step-by-step explanation:

8 0
3 years ago
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Which equation has the solutions x=1+or-square root of 5?
stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

<u>case a)</u> x^{2}+2x+4=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

x^{2}+2x=-4

Complete the square. Remember to balance the equation by adding the same constants to each side.

x^{2}+2x+1=-4+1

x^{2}+2x+1=-3

Rewrite as perfect squares

(x+1)^{2}=-3

(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i

therefore

case a) is not the solution of the problem

<u>case b)</u> x^{2}-2x+4=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

x^{2}-2x=-4

Complete the square. Remember to balance the equation by adding the same constants to each side.

x^{2}-2x+1=-4+1

x^{2}-2x+1=-3

Rewrite as perfect squares

(x-1)^{2}=-3

(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i

therefore

case b) is not the solution of the problem

<u>case c)</u> x^{2}+2x-4=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

x^{2}+2x=4

Complete the square. Remember to balance the equation by adding the same constants to each side.

x^{2}+2x+1=4+1

x^{2}+2x+1=5

Rewrite as perfect squares

(x+1)^{2}=5

(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}

therefore

case c) is not the solution of the problem

<u>case d)</u> x^{2}-2x-4=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

x^{2}-2x=4

Complete the square. Remember to balance the equation by adding the same constants to each side.

x^{2}-2x+1=4+1

x^{2}-2x+1=5

Rewrite as perfect squares

(x-1)^{2}=5

(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}

therefore

case d) is the solution of the problem

therefore

<u>the answer is</u>

x^{2}-2x-4=0

5 0
3 years ago
Read 2 more answers
Answer ASAP, please..
Phantasy [73]

Answer:

g(x) = (x-3)^{2} -4

Step-by-step explanation:

The graph is shifted 3 units to the right which is represented by -3 (in the parentheses) and is shifted four units down.

6 0
3 years ago
Multiply and simplify
emmasim [6.3K]

Answer:

B

Step-by-step explanation:

It's easy:

\frac{4}{16}* \frac{y^5}{y^2}

\frac{y^3}{4}

So, the answer is B.

8 0
3 years ago
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