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3 years ago

. A factory produces 32 tables in an 8-hour day. Calculate how many

Mathematics

1 answer:

Nutka1998 [239]3 years ago

8 0

Answer:

A) 4

B)280

C)480

Step-by-step explanation:

1. 32÷8 =4

multiply 4 by the times of hours in that day frame

and boom you got your answers

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What is the area of a sector with a central angle of 108 degrees and a diameter of 21.2 cm?

Rasek [7]

Answer:

$105.84\ cm^2$

Step-by-step explanation:

step 1

Find the area of the circle

The area of the circle is

$A=\pi r^{2}$

we have

$r=21.2/2=10.6\ cm$ ----> the radius is half the diameter

substitute

$A=\pi (10.6)^{2}$

$A=112.36\pi\ cm^2$

step 2

Find the area of a sector

Remember that

The area of the circle subtends a central angle of 360 degrees

using proportion

Find out the area of a sector with a central angle of 108 degrees

$\frac{112.36\pi}{360^o}=\frac{x}{108^o}\\\\x=112.36\pi(108)/360\\\\x=33.708\pi\ cm^2$

assume

$\pi=3.14$

substitute

$33.708(3.14)=105.84\ cm^2$

7 0

3 years ago

Colleen plans to start a small business by making and selling skirts. A sewing machine costs $250, and she estimates that she wi

Dmitrij [34]

Let C(x) be the function which calculates the total cost of making x skirts.

There is going to be a fixed constant, 250, to which we will add 15$ per x skirts.

So C(x)=250+15x

Answer: 250+15x

7 0

3 years ago

Read 2 more answers

Domain Range -3 6 5 3 -2. 1

uranmaximum [27]

Hello there!

Domain={-3,5,2}
Range={6,3,1}

Hope this helps

Have a great day/night

3 0

3 years ago

Given the following trigonometric ratio, enumerate the meaning ratio

Likurg_2 [28]

Answer:

The trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$

Step-by-step explanation:

From Trigonometry we know the following definitions for each trigonometric ratio:

Sine

$\sin \theta = \frac{y}{h}$ (1)

Cosine

$\cos \theta = \frac{x}{h}$ (2)

Tangent

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x}$ (3)

Cotangent

$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{y}$ (4)

Secant

$\sec \theta = \frac{1}{\cos \theta} = \frac{h}{x}$ (5)

Cosecant

$\csc \theta = \frac{1}{\sin \theta} = \frac{h}{y}$ (6)

Where:

$x$ - Adjacent leg.

$y$ - Opposite leg.

$h$ - Hypotenuse.

The length of the hypotenuse is determined by the Pythagorean Theorem:

$h = \sqrt{x^{2}+y^{2}}$

If $y = AC$ and $x = BC$ , then the trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$

6 0

3 years ago

Let w = x2 + y2 + z2, x = uv, y = u cos(v), z = u sin(v). use the chain rule to find ∂w ∂u when (u, v) = (9, 0).

jasenka [17]

By the chain rule,

$\dfrac{\partial w}{\partial u}=\dfrac{\partial w}{\partial x}\cdot\dfrac{\partial x}{\partial u}+\dfrac{\partial w}{\partial y}\cdot\dfrac{\partial y}{\partial u}+\dfrac{\partial w}{\partial z}\cdot\dfrac{\partial z}{\partial u}$
$\dfrac{\partial w}{\partial u}=2xv+2y\cos v+2z\sin v$

We also have $x(9,0)=0$ , $y(9,0)=9$ , and $z(9,0)=0$ , so at this point we get

$\dfrac{\partial w}{\partial u}(9,0)=2\cdot9\cdot\cos0=18$

7 0

3 years ago