Test quest and I’m stuck plz help

A team can row 40 km in 2 hours when rowing with the current, but only

dexar [7]

<h3>Answer: 14 km per hour</h3>

=====================================================

Work Shown:

x = rowing speed when there is no current

c = speed of the current

speeds are in km per hour

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Going downstream, i.e. with the current:

x+c = faster speed since the current is pushing you

distance = rate*time

d = r*t

40 = (x+c)*2

x+c = 40/2

x+c = 20

c = 20-x

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Going upstream, i.e. against the current:

x-c = slower speed since the current is working against you

d = r*t

16 = (x-c)*2

16 = 2x-2c

2x-2c = 16

2(x-c) = 16

x-c = 16/2

x-c = 8

x-(20-x) = 8 ... plug in c = 20-x

x-20+x = 8

2x-20 = 8

2x = 8+20

2x = 28

x = 28/2

x = 14 km per hour is the rowing speed when there is no current

Side note: the speed of the current is c = 20-x = 20-14 = 6 km per hour.

3 0

2 years ago

If F (x) = x + 4 what is f (5)

Scrat [10]

Answer:

9

Step-by-step explanation:

f=x=f=5+4=9

5 0

2 years ago

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How would I solve 16x^4-41x^2+25=0 ???

sveticcg [70]

$16{ x }^{ 4 }-41{ x }^{ 2 }+25=0$

${ x }^{ 4 }={ ({ x }^{ 2 }) }^{ 2 }\\ \\ 16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0$

First of all to make our equation simpler, we'll equal $x^{2}$ to a variable like 'a'.

So,

${ x }^{ 2 }=a$

Now let's plug $x^{2}$ 's value (a) into the equation.

$16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0\\ \\ { x }^{ 2 }=a\\ \\ 16{ (a) }^{ 2 }-41{ a }+25=0$

Now we turned our equation into a quadratic equation.

(The variable 'a' will have a solution set of two solutions, but 'x' , which is the variable of our first equation will have a solution set of four solutions since it is a quartic equation (<span>fourth-degree <span>equation) )

Let's solve for a.

The formula used to solve quadratic equations ;

$\frac { -b\pm \sqrt { { b }^{ 2 }-4\cdot t\cdot c } }{ 2\cdot t }$

The formula is used in an equation formed like this :
</span></span>
$t{ x }^{ 2 }+bx+c=0$

In our equation,

t=16 , b=-41 and c=25

Let's plug the values in the formula to solve.

$t=16\quad b=-41\quad c=25\\ \\ \frac { -(-41)\pm \sqrt { -(41)^{ 2 }-4\cdot 16\cdot 25 } }{ 2\cdot 16 } \\ \\ \frac { 41\pm \sqrt { 1681-1600 } }{ 32 } \\ \\ \frac { 41\pm \sqrt { 81 } }{ 32 } \\ \\ \frac { 41\pm 9 }{ 32 }$

So the solution set :

$\frac { 41+9 }{ 32 } =\frac { 50 }{ 32 } \\ \\ \frac { 41-9 }{ 32 } =\frac { 32 }{ 32 } =1\\ \\ a\quad =\quad \left\{ \frac { 50 }{ 32 } ,\quad 1 \right\}$

We found a's value.

Remember,

${ x }^{ 2 }=a$

So after we found a's solution set, that means.

${ x }^{ 2 }=\frac { 50 }{ 32 }$

and

${ x }^{ 2 }=1$

We'll also solve this equations to find x's solution set :)

${ x }^{ 2 }=\frac { 50 }{ 32 } \\ \\ \frac { 50 }{ 32 } =\frac { 25 }{ 16 } \\ \\ { x }^{ 2 }=\frac { 25 }{ 16 } \\ \\ \sqrt { { x }^{ 2 } } =\sqrt { \frac { 25 }{ 16 } } \\ \\ x=\quad \pm \frac { 5 }{ 4 }$

${ x }^{ 2 }=1\\ \\ \sqrt { { x }^{ 2 } } =\sqrt { 1 } \\ \\ x=\quad \pm 1$

So the values x has are :

$\frac { 5 }{ 4 }$ , $-\frac { 5 }{ 4 }$ , $1$ and $-1$

Solution set :

$x=\quad \left\{ \frac { 5 }{ 4 } \quad ,\quad -\frac { 5 }{ 4 } \quad ,\quad 1\quad ,\quad -1 \right\}$

I hope this was clear enough. If not please ask :)

3 0

3 years ago

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Only 1/3 of the students who run in each event win medals at the track meet if nine students are running the mile how many of th

taurus [48]

1/3 of 9 is 3. 3 will win the mile

5 0

3 years ago

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Estimate then record the product $473x9

ladessa [460]

So because 9 and 10 are close to each other, and the fact that 10 is really easy to multiply by, you do 473 x 10 which is 4730, because you just add a 0 to the end.
This means that an estimate is 4730 or if you want to be less specific, <4730, because 9 is less than 10.
The actual product though is 4257.

7 0

4 years ago

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