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3 years ago

Two squares have sides 12cm and 4cm respectively. What is the ratio of their areas

Mathematics

1 answer:

zaharov [31]3 years ago

7 0

Answer:

9:1

Step-by-step explanation:

square with side 12cm area= 144cm

square with side 4cm area= 16 cm

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I WILL GIVE BRAINLIEST PLEASE HELP!!!

luda_lava [24]

Answer:

Step-by-step explanation:

interchange x and y

y= 1/(x - 2)

x = 1 / (y - 2) Multiply both sides by y - 2

x(y - 2) = 1 Remove the brackets

xy - 2x = 1 Add x to both sides

xy = 1 + 2x Divide by x

y = (1 + 2x)/x

The brackets have been destroyed. The answer is as I've given it.

y = (1/x) + 2

5 0

3 years ago

How to solve this trig

n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

$\displaystyle \large{ cos \theta = - \frac{1}{2} }$

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

$\displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }$

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$

Both values are apart of the interval. Hence,

$\displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }$

Second Question

Isolate sin(4 theta).

$\displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }$

Rationalize the denominator.

$\displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }$

The problem here is 4 beside theta. What we are going to do is to expand the interval.

$\displaystyle \large{0 \leqslant \theta < 2 \pi}$

Multiply whole by 4.

$\displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}$

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$

Find Q4

$\displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }$

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

$\displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}$

Hence:-

For Q3

$\displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }$

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

$\displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }$

Therefore:-

$\displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }$

Then we divide all these values by 4.

$\displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }$

Let me know if you have any questions!

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3 years ago

Explain the different ways you can name an angle

nydimaria [60]

Answer:

When you say that, I think directly to Triangles. SO i am going to state that.

There is acute: Less than 90 degrees

There is Obtuse: more than 90 degrees

There is Right: 90 degrees

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4 years ago

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lys-0071 [83]

96 oz is greater than 5 lb

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3 years ago

5. As two boats are coming to the end of a race, a top view of their positions is

Dafna1 [17]

The distance of the boat on the right to the finish line is 297.29 m and the distance of the boat on the left to the finish is 294.11 m, so the boat on the left will reach the finish line faster.

<h3>How do you know which boat will reach the finish line faster? </h3>

The image shows two triangles of which we know an angle and the value of the opposite leg. Therefore, we can perform a mathematical operation to know the value of the hypotenuse of these triangles, which would be the distance of the boats to the goal as shown below.

h = hypotenuse

Right Boat

Sen 22° = $\frac{110 m}{h}$

h = $\frac{110 m}{ Sin 22}$

h = $\frac{110 m}{0.37}$

h = 297,29 m

Left boat

Sen 20° = $\frac{100}{h}$

h = $\frac{|00 m}{ Sin 20}$

h = $\frac{100 m}{0.34}$

h = 294,11 m

According to the above, the distance of the boat on the left is less, so it can be inferred that this boat will reach the finish line faster.

Learn more about triangles in: brainly.com/question/2269348

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2 years ago

Two squares have sides 12cm and 4cm respectively. What is the ratio of their areas​

Two squares have sides 12cm and 4cm respectively. What is the ratio of their areas