All categories

3 years ago

Select the two figures that are similar to the 5 by 10 figure that is shown.

Mathematics

1 answer:

Lapatulllka [165]3 years ago

6 0

Answer:

Figure (i) and (iv)

Step-by-step explanation:

Given:

Optional figure is given in attached file.

We need to find two figures that are similar to the 5 by 10 figure.

All the given figure are $m\times n$ form.

Where m represent the number of rows and n represent the number of columns.

Solution:

Observe that in the given figure 5 by 10, the number of rows is 5 and number of columns is 10, that is, the number of columns is double of that the number of rows.

So we need to find two such figures whose number of columns is double of the number of rows.

From the given figures, figure (i) the number of rows is 2 and number of columns is 4, which is double of number of rows. so it is similar to 5 by 10 figure.

Similarly in figure (iv), the number of rows is 4 and number of columns is 8. so the number of columns is double the number of rows, so it is similar to the figure 5 by 10.

Therefore, the two figures that are similar to 5 by 10 figure are given in attached file such as (i) and (iv).

You might be interested in

Write the expression without using exponents (-2y)^4

JulsSmile [24]

Answer:

16yyyy

Step-by-step explanation:

Getting rid of exponents:

(-2y)^4 =
(-2)^4*y^4 =
16yyyy

5 0

3 years ago

Read 2 more answers

2. How many factors are represented in the expression (x + 2)(x - 3)?

Valentin [98]

Step-by-step explanation:

(X+2)(x-3)4211334433434

8 0

3 years ago

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

Read 2 more answers

Joe tracked the height of his Dalmatian puppy from age 13 weeks to 25 weeks. He graphed the data and determined the line of best

nadya68 [22]

Answer:

22.5 weeks

Step-by-step explanation:

y = 1.12x + 28.78 represents the height of the puppy. Here that height is 54 cm:

Then 54 = y = 1.12x + 28.78

Combining the constants, we get 1.12x = 25.22. Dividing both sides by 1.12, x turns out to be 22.5 weeks.

3 0

3 years ago

Determine the measure of an angle (in radians) that cuts off an arc length of 6.4 inches in a circle with a 6.4 inch radius.

serg [7]

1 radian
because s= r × angle measurement in radian
we have s= arc , r = radian so we got the value

5 0

3 years ago