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4 years ago

10

Two trains are moving towards each other. One train moves at a speed of 50 mph, and the other train travels at 70 mph. How soon

will they meet, if originally the distance between them was 390 miles?

2 answers:

nikitadnepr [17]4 years ago

8 0

Answer:

3 hours and 15 minuets

Step-by-step explanation:

First you add the speeds together

120 mph

Second you divide 390 by 120

390/120=3.25

maks197457 [2]4 years ago

7 0

Answer:

in 270 miles

Step-by-step explanation:

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Answer:

13.5m

Step-by-step explanation:

The boy is 1.5m tall so you subtract that by 15m and you get 13.5m

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3 years ago

M³×n=189 Find m and n

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Answer:

m=3, n = 7

Step-by-step explanation:

Assuming m and n are integers, the following are the possible values of m and n

Let m = 1, then m³ = 1 and n= 189

If m = 2 we get 8n = 189 and 189/8 is not an integer so m ≠ 2

If m = 3, we get 27n = 189 and n = 189/27 = 7

Any higher integer values of m will not result in n being an integer and I am assuming m > 1 so the correct answer is m=3, n = 7

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2 years ago

The difference of two trinomials is x2 − 10x + 2. If one of the trinomials is 3x2 − 11x − 4, then which expression could be the

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Answer:

$4x^2-21x-2$

Step-by-step explanation:

Given that:

Difference of two trinomials is $x^2 - 10x + 2$

One of the two trinomials is $3x^2 - 11x - 4$

To find:

The other trinomial = ?

Four options are:

$2x2 - x - 2 \\2x2 + x + 6 \\4x2 + 21x + 6\\ 4x2 - 21x - 2$

Solution:

Let the two trinomials be A and B.

Given A - B = $x^2 - 10x + 2$

B = $3x^2 - 11x - 4$

We have to find the other trinomial A.

A - B = $x^2 - 10x + 2$

A - ( $3x^2 - 11x - 4$ ) = $x^2 - 10x + 2$

$\Rightarrow$ A = $x^2 - 10x + 2$ + ( $3x^2 - 11x - 4$ )

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3 years ago

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Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

$x^2+(2x^2-3)^2=2$

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$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

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$x^2-1=0$

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$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

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