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3 years ago

Consider the following pair of equations:

1 answer:

ivann1987 [24]3 years ago

3 0

Y=x+4 and y=-2x-2 When they are equal it follows that y=y at that point as well so we can say:

x+4=-2x-2 add 2x to both sides

3x+4=-2 subtract 4 from both sides

3x=-6 divide both sides by 3

x=-2, now we can use this value in either original equation to solve for y.
y=2

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0.004 of all math majors at a certain university double major in music. Express this decimal as a percent.

RUDIKE [14]

I think the answer is 4%

3 0

3 years ago

What is 300 is 10 times as much as

Over [174]

30 because 30x10=300

5 0

3 years ago

Collin is building a deck on the back of his house. He has enough lumber for the deck to be 144 square feet. The length should b

joja [24]

Width = x
length = x+10

Area of deck = 144 square feet

Area = length * width

144 = x(x+10)

Solving for x in a quadratic:

x²+10x = 144
x²+10x-144 = 0

Factor:
(x+18)(x-8) = 0

Solving for x:
x = 8, x = -18

Dimensions cannot be negative, therefore x = 8, only.

Length = x + 10
Length = 18

Width = 8

4 0

3 years ago

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago

Cube-shaped blocks are packed into a cube-shaped storage container. The edge of the storage container is 2 1/2 feet. The edge le

GarryVolchara [31]

Answer:

Volume of one cube shaped block is, 0.125 cubic feet

Step-by-step explanation:

Volume of a cube(V) is given by:

$V = a^3$ .....[1]

where a is the edge length of the cube.

As per the statement:

Cube-shaped blocks are packed into a cube-shaped storage container. The edge of the storage container is 2 1/2 feet.

⇒ $\text{Edge length of storage} = 2\frac{1}{2} = 2.5 ft$

It is also given that:

The edge length of each block is 1/5 the edge length of the storage container

⇒ $\text{Edge length of each block (a)} = \frac{1}{5} \cdot 2.5 = 0.5 ft$

Substitute this in [1] we have;

$V = (0.5)^3 = 0.125 ft^3$

Therefore, volume, in cubic feet, of one cube-shaped block is, 0.125

4 0

4 years ago