All categories

3 years ago

Anybody know the answer to A and B? Understandable if not!

Mathematics

1 answer:

Orlov [11]3 years ago

5 0

6 sides, 7 length for one side
6x7 = 42 for the first one

52 for one side, has 5 sides
52x5 = 260 for the second.

You might be interested in

Lee wants to buy two pairs of shoes. One pair costs $37.07 and the other pair costs $42.48. A pair of socks cost $8.50. If Lee b

daser333 [38]

Answer:

$96.55

Step-by-step explanation:

The computation of the money he would require is shown below:

Since lee wants to buy both pairs of shoes and the two pairs of socks

So the amount of money he needed is

= $37.07 + $42.48 + 2 × $8.50

= $37.07 + $42.48 + $17

= $96.55

5 0

3 years ago

there are 120 seats in the balcony of theatre .if this 1/5 of the total seats, what's the total number of seats in the theater

natka813 [3]

Answer:

600 seats

Step-by-step explanation:

We know there is 120 seats on a balcony and that is $\frac{1}{5}$ of the total seats, we want to find the total number of seats so,

120 seats = $\frac{1}{5}$

We know that a whole fraction sums to 1 so for example $\frac{1}{5} +\frac{4}{5} =\frac{5}{5} =1$

120 seats = $\frac{1}{5}$

Multiply both sides to make it a whole fraction

600 seats = $\frac{5}{5} =1$

So if there is 120 seats and that number is only $\frac{1}{5}$ of the total number of seats then the total number of seats is 600

3 0

3 years ago

Read 2 more answers

What’s the x axis of -4x+8y=-16

balu736 [363]

Answer:

Step-by-step explanation:

When need to find x axis y is =0

So now you can find x

The answer of x is equally to the xaxis.

If need to find y axis then please let x =0.

4 0

3 years ago

Would you say that pythagoras’ theorem holds true for hexagons as well? why or why not

kifflom [539]

No, the Pythgorean Theorem only applies to right triangles. To get two right triangles, you divide a rectangle diagonally. Dividing a hexagon into two pieces produces trapezoids or pentagons depending where it is divided.

7 0

4 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