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4 years ago

Murad ran in a track race on Tuesday he finished the race in 420 seconds how many minutes did it take to run the race?

2 answers:

max2010maxim [7]4 years ago

7 0

The key information here is the time he took to run the race - 420 seconds

We know that there are 60 seconds in 1 minute so to find the number of minutes we divide 420 by 60 to get 7.

Murad took 7 minutes to run the race

Dovator [93]4 years ago

7 0

Answer:

the first one is 1906.8 and the second one is 4.767

Step-by-step explanation:

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stealth61 [152]

Step-by-step explanation:

(x+4)3=2

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3x=-12+2

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divide both sides by 3

x=-10/3

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3 years ago

F(x) = - x ^ 2 + 8x - 2

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3 years ago

Easy question, Easy points LIMTED TIME. LOTS OF POINTS

Rudik [331]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Suppose u = (1,2,3), v = (2, -3, -4), and w = (1,0,-2). (a) Find u-2v + 3w. (b) Write the vector (3,4,5) as a linear combination

marishachu [46]

<h2>Answer with explanation:</h2>

We are given vectors u,v and w as follows:

$u=(1,2,3)\,\ v=(2,-3,-4),\ and\ w=(1,0,-2)$

(a)

Now we are asked to find the value of:

$u-2v+3w$

The value is calculated as follows:

$u-2v+3w=(1,2,3)-2(2,-3,-4)+3(1,0,-2)\\\\i.e.\\\\u-2v+3w=(1,2,3)+(-2\times 2,-2\times -3,-2\times -4)+(3\times 1,3\times 0,3\times -2)\\\\i.e.\\\\u-2v+3w=(1,2,3)+(-4,6,8)+(3,0,-6)\\\\i.e.\\\\u-2v+3w=(1-4+3,2+6+0,3+8-6)\\\\i.e.\\\\u-2v+3w=(0,8,5)$

Hence,

$u-2v+3w=(0,8,5)$

(b)

We are asked to represent (3,4,5) as a linear combination of u, v, and w.

Let:

$(3,4,5)=au+bv+cw$

$(3,4,5)=a(1,2,3)+b(2,-3,-4)+c(1,0,-2)$

i.e.

$(3,4,5)=(a,2a,3a)+(2b,-3b,-4b)+(c,0,-2c)\\\\i.e.\\\\(3,4,5)=(a+2b+c,2a-3b,3a-4b-2c)\\\\i.e.\\\\a+2b+c=3----------(1)$

$2a-3b=4-------------(2)$

$3a-4b-2c=5------------(3)$

on multiplying equation(1) by 2 an then adding equation (1) to equation (3)

$5a=11\\\\i.e.\\\\a=\dfrac{11}{5}$

on putting the value of a in equation (2) we get:

$2\times \dfrac{11}{5}-3b=4\\\\i.e.\\\\\dfrac{22}{5}-3b=4\\\\i.e.\\\\3b=\dfrac{22}{5}-4\\\\i.e.\\\\3b=\dfrac{2}{5}\\\\i.e.\\\\b=\dfrac{2}{15}$

Now on putting the value of a and b in equation (1)

$\dfrac{11}{5}+2\times \dfrac{2}{15}+c=3\\\\i.e.\\\\\dfrac{11}{5}+\dfrac{4}{15}+c=3\\\\i.e.\\\\c=3-\dfrac{11}{5}-\dfrac{4}{15}\\\\i.e.\\\\c=\dfrac{3\times 15-11\times 3-4}{15}\\\\i.e.\\\\c=\dfrac{45-33-4}{15}\\\\i.e.\\\\c=\dfrac{8}{15}$

Hence, we have:

$(3,4,5)=\dfrac{11}{5}(1,2,3)+\dfrac{2}{15}(2,-3,-4)+\dfrac{8}{15}(1,0,-2)\\\\i.e.\\\\(3,4,5)=\dfrac{11}{5}u+\dfrac{2}{15}v+\dfrac{8}{15}w$

5 0

4 years ago

You made two deposits to your bank accounts this month. One deposit was $17.92, and the second deposit was $15.33. Your balance

WITCHER [35]

$72.31 is the total amount AFTER ADDING TWO DEPOSITS

So, if we want to know the amount before deposits, we just need to take the total of previous two deposits

The expressions that show your balance would be

$72.31 – $17.92 – $15.33; $39.06

6 0

3 years ago