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3 years ago

Darren drove his car 23 kilometers to work each day how many meters did he drive to work each day

Mathematics

1 answer:

SOVA2 [1]3 years ago

3 0

Answer:

Step-by-step explanation:

If this is worded right, then the answer to your question is in your question

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A room has an area of 121 ft squared for carpeting is only sold in meters squared how much carpeting is needed to carpet the roo

Naya [18.7K]

We know that

1 ft-----------> 0.3048 m
then
1 ft²---------> (0.3048)²------> 0.092903 m²
so
121 ft²----------> 121*0.092903----------> 11.24 m²

the answer is
11.24 m²

6 0

2 years ago

If a triangle has sides that are 21 and 6 what is the range for third side x?

VikaD [51]

Answer:

15<x<27

Step-by-step explanation:

Rule for the sides of a triangle:

The sum of the two smallest sides of a triangle must be greater than the biggest side.

In this question:

Sides of 6, 21 and x. We have to find the range for x.

If 21 is the largest side:

Two smallest are 6 and x.

x + 6 > 21

x > 21 - 6

x > 15

If x is the largest side:

Two smallest and 6 and 21. So

21 + 6 > x

27 > x

x < 27

Then

x has to be greater than 15 and smaller than 27. So the answer is:

15<x<27

8 0

2 years ago

Why is the answer to this integral's denominator have 1+pi^2

ss7ja [257]

It comes from integrating by parts twice. Let

$I = \displaystyle \int e^n \sin(\pi n) \, dn$

Recall the IBP formula,

$\displaystyle \int u \, dv = uv - \int v \, du$

Let

$u = \sin(\pi n) \implies du = \pi \cos(\pi n) \, dn$

$dv = e^n \, dn \implies v = e^n$

Then

$\displaystyle I = e^n \sin(\pi n) - \pi \int e^n \cos(\pi n) \, dn$

Apply IBP once more, with

$u = \cos(\pi n) \implies du = -\pi \sin(\pi n) \, dn$

$dv = e^n \, dn \implies v = e^n$

Notice that the ∫ v du term contains the original integral, so that

$\displaystyle I = e^n \sin(\pi n) - \pi \left(e^n \cos(\pi n) + \pi \int e^n \sin(\pi n) \, dn\right)$

$\displaystyle I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n - \pi^2 I$

$\displaystyle (1 + \pi^2) I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n$

$\implies \displaystyle I = \frac{\left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n}{1+\pi^2} + C$

6 0

2 years ago

Find the surface area of the right triangular prism (above) using its net (below).

lakkis [162]

Answer:

144 units²

Step-by-step explanation:

The net of the right triangular prism consists of 3 rectangles and 2 equal triangles

Let's solve for the area of each:

✔️Area of rectangle 1 = L*W

L = 11

W = 3

Area of rectangle 1 = 11*3 = 33 units²

✔️Area of rectangle 2 = L*W

L = 11

W = 4

Area of rectangle 2 = 11*4 = 44 units²

✔️Area of rectangle 3 = L*W

L = 11

W = 5

Area of rectangle 3 = 11*5 = 55 units²

✔️Area of the two triangles = 2(½*base*height)

base = 4

height = 3

Area of the two traingles = 2(½*4*3)

= 12 units²

✔️Surface area of the right triangle = area of rectangle 1 + area of rectangle 2 + area of rectangle 3 + area of the two triangles

= 33 + 44 + 55 + 12

= 144 units²

7 0

2 years ago

Y = 3 sin2x, y = 0, 0 ≤ x ≤ π; about the x−axis

Dominik [7]

I assume you're revolving the region with those bounds about the x-axis, and supposed to find the volume.

Via the disk method,

$\displaystyle\pi\int_0^\pi(3\sin2x)^2\,\mathrm dx=9\pi\int_0^\pi\sin^22x\,\mathrm dx$

Recall the half-angle identity for sine:

$\sin^2t=\dfrac{1-\cos2t}2$
$\implies\displaystyle\frac{9\pi}2\int_0^\pi(1-\cos4x)\,\mathrm dx$
$=\displaystyle\frac{9\pi}2\left(x-\frac14\sin4x\right)\bigg|_{x=0}^{x=\pi}$
$=\dfrac{9\pi^2}2$

8 0

3 years ago