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3 years ago

11

Find and x and y PLEASE HELP

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

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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