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2 years ago

Please answer this asap

Mathematics

1 answer:

Ymorist [56]2 years ago

8 0

30yd it is ur welcome have a nice nice day bteee

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Find an odd natural numbers x such that LCM (x, 40) = 1400

vesna_86 [32]

Answer:

Let x be odd such that LCM {x,40} = 1400 .

Since 1400 = 23×52×7 , then

x ∈ {5m×7n∣(m,n) ∈ {0,1,2}×{0,1}} .

By testing these values, we find that x = 175 .

5 0

3 years ago

Kaylee put an empty cup underneath a leaking faucet. After 1 1/2 hours, Kaylee collected 1/4 cup of water. What is the rate in c

Tju [1.3M]

Answer:

I think it is 6/1

Step-by-step explanation:

You are wrong because 0.167 into a fraction is 167/1000 and that isn't one of the options but when I did it 1 1/2=3/2=1.5 and 1/4=0.25 so I did 1.5 divide by 0.25 and it gave me 6 so that why I think that my answer is correct and yours is not.

4 0

3 years ago

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.

case a) $x^{2}+2x+4=0$

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

case b) $x^{2}-2x+4=0$

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

case c) $x^{2}+2x-4=0$

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

case d) $x^{2}-2x-4=0$

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

the answer is

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

5 0

2 years ago

Read 2 more answers

What is the system equation of y=1/4x : x+4y=8?

Masja [62]

Since x+4y = 8 the thing is y = 1/4*x but if you get 1/4 of x then multiply it by 4 when you put it back in the problem, which means x=4 y=1

3 0

2 years ago

Plz Help fast WILL MARK BRAINLIEST

Annette [7]

I hope this help a=

6 0

2 years ago