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2 years ago

Sorry if it is not clear the question asks, Two squares of side x cm are removed from a square of side 3x cm as shown

Mathematics

1 answer:

shtirl [24]2 years ago

5 0

$(3x)^2-2\cdot x^2=9x^2-2x^2=7x^2\\\\
\frac{7x^2}{9x^2}=\boxed{\frac{7}{9}}$

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What is the side length of the eectagkes

dedylja [7]

1. 5 in and 1/3 in: Area = 5/3 in^2

2. 5 in and 4/3 in: Area= 20/3 in^2

3. 5/2 in and 4/3 in: Area=10/3 in^2

4. 7/6 in and 6/7 in: Area = 1 in^2

Step-by-step explanation:

1. 5 in and 1/3 in

Here,

$Let\\l=5\\w=\frac{1}{3}\\Area=l*w\\=5*\frac{1}{3}\\=\frac{5}{3}\ in^2$

2. 5 in and 4/3 in

Here,

$l= 5\ in\\w=\frac{4}{3}\ in\\Area=l*w\\=5*\frac{4}{3}\\=\frac{20}{3}\ in^2$

3. 5/2 in and 4/3 in

$let\\l=\frac{5}{2}\\w=\frac{4}{3}\\Area= \frac{5}{2}*\frac{4}{3}\\=\frac{20}{6}=\frac{10}{3}\ in^2$

4. 7/6 in and 6/7 in

Let

$l=\frac{6}{7}\\w=\frac{7}{6}\\Area=l*w\\=\frac{6}{7}*\frac{7}{6}\\=1\ in^2$

Hence,

1. 5 in and 1/3 in: Area = 5/3 in^2

2. 5 in and 4/3 in: Area= 20/3 in^2

3. 5/2 in and 4/3 in: Area=10/3 in^2

4. 7/6 in and 6/7 in: Area = 1 in^2

Keywords: Rectangle, Area

Learn more about rectangles at:

#LearnwithBrainly

8 0

2 years ago

Complete the equation. 5x + 5.2 10.0

skad [1K]

I do t think it’s right so far but x=1.04

8 0

2 years ago

What is half of 7 the whole number

Volgvan

Answer:

3 1/2

Step-by-step explanation:

You have to do 7 divided by 2 or 1/2 x 7

6 0

2 years ago

Read 2 more answers

33. Suppose you were on a planet where the

tatyana61 [14]

Answer:

a) 3.675 m

b) 3.67m

Explanation:

We are given acceleration due to gravity on earth = $9.8ms^-2$

And on planet given = $2.0ms^-2$

A) Since the maximum jump height is given by the formula

$\mathrm{H}=\frac{\left(\mathrm{v} 0^{2} \times \sin 2 \emptyset\right)}{2 \mathrm{g}}$

Where H = max jump height,

v0 = velocity of jump,

Ø = angle of jump and

g = acceleration due to gravity

Considering velocity and angle in both cases

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 2}{\mathrm{g} 1}$

Where H1 = jump height on given planet,

H2 = jump height on earth = 0.75m (given)

g1 = 2.0 $ms^-2$ and

g2 = 9.8 $ms^-2$

Substituting these values we get H1 = 3.675m which is the required answer

B) Formula to find height of ball thrown is given by

$\mathrm{h}=(\mathrm{v} 0 * \mathrm{t})+\frac{\mathrm{a} *\left(t^{2}\right)}{2}$

which is due to projectile motion of ball

Now h = max height,

v0 = initial velocity = 0,

t = time of motion,

a = acceleration = g = acceleration due to gravity

Considering t = same on both places we can write

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 1}{\mathrm{g} 2}$

where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations

substituting h2 = 18m, g1 = 2.0 $ms^-2$ and g2 = 9.8 $ms^-2$

We get h1 = 3.67m which is the required height

6 0

2 years ago

Someone please help me I suck @ simplifying thank you

ycow [4]

The anwer is 67/72 and can not be simpled any more

5 0

2 years ago