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3 years ago

A room has an area of 121 ft2 but carpeting is only sold in m2. How much carpeting is needed to carpet the room?

Mathematics

2 answers:

stepladder [879]3 years ago

8 0

The correct answer is D. 11.24 m2

kykrilka [37]3 years ago

7 0

The correct answer for the question that is being presented above is this one: "D.)11.24 m2."
First, we need to know the conversion between feet to meters.
1 meter = 3.28 feet
1 meter^2 = 10.7584 ft^2

Given the value of 121 ft^2,
= 121 ft^2 * (1 m^2 / 10.7584 ft^2)
= 11.2470 m^2

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Solve the equation for all real values of x.<br> cosxtanx - 2 cos x=-1

IceJOKER [234]

Solution to equation $cosxtanx - 2 cos^2 x=-1$ for all real values of x is $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$ .

<u>Step-by-step explanation:</u>

Here we have , $cosxtanx - 2 cos^2 x=-1$ . Let's solve :

⇒ $cosxtanx - 2 cos^2 x=-1$

⇒ $cosx(\frac{sinx}{cosx}) - 2 cos^2 x=-1$

⇒ $sinx = 2 cos^2 x-1$

⇒ $sinx = 2 (1-sin^2x)-1$

⇒ $sinx = 1-2sin^2x$

⇒ $2sin^2x+sinx-1=0$

By quadratic formula :

⇒ $sinx = \frac{-b \pm \sqrt{b^2-4ac} }{2a}$

⇒ $sinx = \frac{-1 \pm \sqrt{1^2-4(2)(-1)} }{2(2)}$

⇒ $sinx = \frac{-1 \pm3}{4}$

⇒ $sinx = \frac{1}{2} , sinx =-1$

⇒ $sinx = sin\frac{\pi}{6} , sinx = sin\frac{3\pi}{2}$

⇒ $x=\frac{\pi}{6} , x=\frac{3\pi}{2}$

But at $x=\frac{3\pi}{2}$ we have equation undefined as $cos\frac{3\pi}{2}=0$ . Hence only solution is :

⇒ $x=\frac{\pi}{6}$

Since , $sin(\pi -x)=sinx$

⇒ $x=\pi -\frac{\pi}{6} = \frac{5\pi}{6}$

Now , General Solution is given by :

⇒ $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$

Therefore , Solution to equation $cosxtanx - 2 cos^2 x=-1$ for all real values of x is $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$ .

3 0

3 years ago

The ratio of girls to boys in a class is 3:5 what fraction of the class are girls?

Wewaii [24]

Answer:

Hello!!

If the ratio of girls to boys in a class is 3:5, the fraction of the class that are girls is...

$\frac{3}{5}$

Hope this helps!!

8 0

3 years ago

This question has been deleted

Serhud [2]

What is the question

7 0

2 years ago

In Exercises 10 and 11, points B and D are points of tangency. Find the value(s) of x.<br>

Lemur [1.5K]

In both cases,

$AB^2=AD^2$

(as a consequence of the interesecting secant-tangent theorem)

So we have

10.

$(4x+7)^2=(6x-3)^2$

$16x^2+56x+49=36x^2-36x+9$

$20x^2-92x-40=0$

$5x^2-23x-10=0$

$(5x+2)(x-5)=0\implies\boxed{x=5}$

(omit the negative solution because that would make at least one of AB or AD have negative length)

11.

$(4x^2-18x-10)^2=(x^2+x+4)^2$

$16x^4-144x^3+244x^2+360x+100=x^4+2x^3+9x^2+8x+16$

$15x^4-146x^3+235x^2+352x+84=0$

$(x-7)(3x+2)(5x^2-17x-6)=0\implies\boxed{x=-\dfrac23\text{ or }x=7}$

(again, omit the solutions that would give a negative length for either AB or AD)

7 0

3 years ago

In quadrilateral ABCD, AD is congruent to BC, and AD is parallel to BC. Andre has written a

Dahasolnce [82]

Answer:

AD is Congruent to BC and it's given, (given just means that it was already said or stated, and you don't need to do the work to find it) Angle DAC would be congruent to angle BAC (if that doesn't work, rearrange them to look like angle CAB). AC would be congruent to DB. Triangle ADC would be congruent to triangle BCD by (since i don't know exactly which way the letters are arranged is would either be SAS or SSA) and that's because you know two of the sides are congruent to each other and one angle.

I tried hard to sketch out what the shape looked like based on the information given, and that's because I need a visual of what the shape looks like. Sorry, this took so long to answer.

5 0

7 months ago