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4 years ago

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Pratap puri rowed 10 miles down a river in 2 hours, but the return trip took him 2 and one half hours. Find the rate pratap can

row in still water and find the rate of the current.

1 answer:

N76 [4]4 years ago

6 0

Answer:

<em>Rate of Pratap in still water is 4.5 miles/hour and rate of current is 0.5 miles/hour.</em>

Step-by-step explanation:

Pratap Puri rowed 10 miles down a river in 2 hours, but the return trip took him 2.5 hours.

We know that, $Speed = \frac{Distance}{Time}$

So, the <u>speed of Pratap with the current</u> will be: $(\frac{10}{2})miles/hour = 5$ miles/hour

and the <u>speed of Pratap against the current</u> will be: $(\frac{10}{2.5})miles/hour = 4$ miles/hour.

Suppose, the rate of Pratap in still water is $x$ and the rate of current is $y$ .

So, the equations will be........

$x+y= 5 .............................. (1)\\ \\ x-y=4 .............................. (2)$

Adding equation (1) and (2) , we will get......

$2x=9\\ \\ x=\frac{9}{2}= 4.5$

Now, plugging this $x=4.5$ into equation (1), we will get.....

$4.5+y=5\\ \\ y=5-4.5 =0.5$

Thus, Pratap can row at 4.5 miles per hour in still water and the rate of the current is 0.5 miles/hour.

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Which products result in a perfect square trinomial? Select three options.

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Answer:

Second option: $(xy+x)(xy+x)$

Third option: $(2x-3)(-3+2x)$

Fifth option: $(4y^2+25)(25+4y^2)$

Step-by-step explanation:

By definition, a perfect square trinomial can be obtained by squaring binomials.

Then:

$(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2\\\\(a-b)^2=(a-b)(a-b)=a^2-2ab+b^2$

Knowing this, to obtain a perfect square trinomial, the binomials that you multiply must be equals.

Therefore, the products result in a perfect square trinomial are:

$(xy+x)(xy+x)=(xy+x)^2=(xy)^2+2(xy)(x)+x^2=x^2y^2+2x^2y+x^2$

$(2x-3)(-3+2x)=(2x-3)^2=(2x)^2-2(2x)(3)+3^2=4x^2-12x+9$

$(4y^2+25)(25+4y^2)=(4y^2+25)^2=(4y^2)^2+2(4y^2)(25)+25^2=16y^4+200y^2+625$

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4 years ago

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1/4(x-2) = 10 NEED DONE ASAP

kari74 [83]

Answer:

x = 42

Step-by-step explanation:

Step 1 :

1

Simplify —

4

Equation at the end of step 1 :

1

(— • (x - 2)) - 10 = 0

4

Step 2 :

Equation at the end of step 2 :

(x - 2)

——————— - 10 = 0

4

Step 3 :

Rewriting the whole as an Equivalent Fraction :

3.1 Subtracting a whole from a fraction

Rewrite the whole as a fraction using 4 as the denominator :

10 10 • 4

10 = —— = ——————

1 4

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

3.2 Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

(x-2) - (10 • 4) x - 42

———————————————— = ——————

4 4

Equation at the end of step 3 :

x - 42

—————— = 0

4

Step 4 :

When a fraction equals zero :

4.1 When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, multiplys both sides of the equation by the denominator.

Here's how:

x-42

———— • 4 = 0 • 4

4

Now, on the left hand side, the 4 cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :

x-42 = 0

Solving a Single Variable Equation :

4.2 Solve : x-42 = 0

Add 42 to both sides of the equation :

x = 42

One solution was found :

x = 42

Processing ends successfully

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3 years ago

Given the vectors w = <-5, 3> and z = <1, 4>, find the results of the vector subtractions.

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Answer:

Given two linearly independent vectors w and z

We want -w-z

Hence apply a negative gradient to w

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So <5,-3>-<1,4> = (<5-1>,<-3-4>)

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The quotient of

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suter [353]

Answer:

$\theta = 30^\circ, 330^\circ$

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Step-by-step explanation:

Given [Missing from the question]

Equation:

$cos\theta = \frac{\sqrt 3}{2}$

Interval:

$0 \le \theta \le 360$

$0 \le \theta \le 2\pi$

Required

Determine the values of $\theta$

The given expression:

$cos\theta = \frac{\sqrt 3}{2}$

... shows that the value of $\theta$ is positive

The cosine of an angle has positive values in the first and the fourth quadrants.

So, we have:

$cos\theta = \frac{\sqrt 3}{2}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{\sqrt 3}{2})$

$\theta = 30$ --- In the first quadrant

In the fourth quadrant, the value is:

$\theta = 360 -30$

$\theta = 330$

So, the values of $\theta$ in degrees are:

$\theta = 30^\circ, 330^\circ$

Convert to radians (Multiply both angles by $\pi/180$ )

So, we have:

$\theta = \frac{30 * \pi}{180}, \frac{330 * \pi}{180}$

$\theta = \frac{\pi}{6}, \frac{33 * \pi}{18}$

$\theta = \frac{\pi}{6}, \frac{11 * \pi}{6}$

$\theta = \frac{\pi}{6}, \frac{11\pi}{6}$

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3 years ago