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3 years ago

The length of a rectangle is the sum of the width and 4. The area of the rectangle is 32

Mathematics

1 answer:

Sauron [17]3 years ago

4 0

Answer:

The width of rectangle w = 4 units

Step-by-step explanation:

We are given:

Width of rectangle = w

Length of rectangle = w+ 4

Area of rectangle = 32 units

We need to find width of the rectangle

The formula used is: $Area\:of\:rectangle=Length\times Width$

Putting values and finding width of rectangle

$Area\:of\:rectangle=Length\times Width\\32=(w+4)\times w\\32=w^2+4w\\w^2+4w-32=0$

We need to solve this quadratic equation to find width (w)

We will use factorisation and break the middle term.

$w^2+4w-32=0\\w^2+8w-4w-32=0\\w(w+8)-4(w+8)=0\\(w-4)(w+8)=0\\w-4=0\:or\:w+8=0\\w=4\:or\:w=-8$

Since width of rectangle can't be negative, so rejecting w=-8

Therefore, the width of rectangle w = 4 units

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What is the solution to the equation 41+5 -3- 7+5?

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Answer:

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Step-by-step explanation:

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What is the solution to StartFraction 5 over 6 EndFraction x minus one-third greater-than 1 and one-third?

Vikentia [17]

Answer:

The correct answer is x > 2.

Step-by-step explanation:

$\dfrac{5}{6} x - \dfrac{1}{3} > 1\dfrac{1}{3}$

An inequality compares two quantities unlike an equality. An inequality is written with either a greater than ( > ) or lower than ( < ) or greater than equal to ( $\geq$ ) or less than equal to ( $\leq$ ) signs. We solve the above given inequality to find the solutions of the unknown x.

$\dfrac{5}{6} x - \dfrac{1}{3} > 1\dfrac{1}{3} \\\dfrac{5}{6} x - \dfrac{1}{3} > \dfrac{4}{3} \\\dfrac{5}{6} x > \dfrac{1}{3} + \dfrac{4}{3} \\\dfrac{5}{6} x > \dfrac{5}{3}\\x > 2$

Firstly we change the right hand side quantity to fraction.

We then transfer the - $\frac{1}{3}$ to the right hand side and add them. The inequality sign does not change as we are simply adding or subtracting terms from both the ends.

Finally we divide both sides with $\frac{6}{5}$ to get the required solution. The inequality sign does not change as we are multiplying both the ends with a positive quantity.

This gives us the answer as x > 2.

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3 years ago

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Given that sec (x) = 2 and cosec (x) is negative,

weeeeeb [17]

Answer:

i) sin(2x) = $-\frac{\sqrt{3}}{2}$

ii) cot(x+360) = $-\frac{\sqrt{3}}{3}$

iii) sin(x-180) = $\frac{\sqrt{3}}{2}$

Step-by-step explanation:

sec(x) = 2

Since cos(x) is reciprocal of sec(x), this means:

cos(x) = $\frac{1}{2}$

cosec(x) is negative , this means sin(x) is also negative. The only quadrant where cos(x), sec(x) are positive and sin(x), cosec(x) are negative is the 4th quadrant. Hence the terminal arm of the angle x is in 4th quadrant.

Part i)

sin(2x) can be simplified as:

sin(2x) = 2 sin(x) cos(x)

First we need to find the value of sin(x). According to Pythagorean identity:

$sin^{2}(x)=1-cos^{2}(x)\\\\ sin(x)=\pm \sqrt{1-cos^{2}(x)}$

Since, angle is in 4th quadrant, sin(x) will be negative. So considering the negative value of sin(x) and substituting the value of cos(x), we get:

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

So,

$sin(2x)=2 \times -\frac{\sqrt{3} }{2} \times \frac{1}{2}\\\\ sin(2x)=-\frac{\sqrt{3}}{2}$

Part ii)

We have to find cot(x + 360)

An addition of 360 degrees to the angle brings it back to the same terminal point. So the trigonometric ratios of the original angle and new angle after adding 360 or any multiple of 360 stay the same. i.e.

cot(x + 360) = cot(x)

$cot(x) = \frac{cos(x)}{sin(x)}\\$

Using the values, we get:

$cot(x)=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2} }\\\\ cot(x)=-\frac{\sqrt{3}}{3}$

Part iii)

We need to find the value of sin(x - 180)

sin(x - 180) = - sin(x)

Addition or subtraction of 180 degrees changes the angle by 2 quadrants. The sign of sin(x) becomes opposite if the angle jumps by 2 quadrants. For example, sin(x) is positive in 1st quadrant and negative in 3rd quadrant.

So,

sin(x - 180) = $-(-\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{2}$