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3 years ago

Solve for j show work 8j-5+j=67

Mathematics

1 answer:

andrew11 [14]3 years ago

8 0

Answer:

Step-by-step explanation:

8j-5+j=67

8J+j

9J-5=67

+5 +5

9J=72

/9 /9

j=8

If this helps please put brainliest

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What is greater 3.26 or 3 5/8

Rudiy27

3 5/8 because 5/8 is 0.625 so 3.625 is greater than 3.26

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4 years ago

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Celia and Jake bought 4 pizzas that cost $7 each and bread sticks that cost $3 they spilt the cost between them. Which equations

Mumz [18]

Answer:

E = (3b+28)/2

T=3b+28

E= t/2

Step-by-step explanation:

The coat of the 4 pizzas would be $28, and an unknown amount of breadsticks that cost 3 dollars each.

Im going to use B as the amount of breadsticks bc i dont know what its supposed to be but that should be the correct answer.

T=3b+28

E= t/2

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3 years ago

A cube is packed with decorative pebbles. If the cube has a side length of 2 inches, and each pebble weighs on average 0.5 lb pe

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We can solve this problem by first solving for the total volume of the cube. The formula for volume of cube is given as:

V cube = s^3

Where s is the length of one side which is equivalent to 2 inches. Therefore:

V cube = (2 inches)^3

V cube = 8 cubic inch

Assuming that all the pebble fills the cube without any spaces, then the total weight of the pebbles in the cube would simply be:

Total weight = 0.5 lb per cubic inch * 8 cubic inch

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Therefore, the total weight of the pebbles in the cube is 4 lbs.

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3 years ago

Please help :c the first word is find

xeze [42]

Answer:

The values of $r_{2}$ and $\alpha_{2}$ are 2 and 150º.

Step-by-step explanation:

The complete statement is:

Find $\alpha_{2}$ and $r_{2}$ such that $\sin \theta - \sqrt{3}\cdot \cos \theta = r_{2}\cdot \cos (\theta - \alpha_{2})$ .

We proceed to use the following trigonometric identity:

$\cos (\theta - \alpha_{2}) = \cos \theta \cdot \cos \alpha_{2} +\sin \theta \cdot \sin \alpha_{2}$ (1)

$\sin \theta -\sqrt{3}\cdot \cos \theta = r_{2}\cdot \cos \theta \cdot \cos \alpha_{2}+r_{2}\cdot \sin \theta \cdot \sin \alpha_{2}$

By direct comparison we derive these expressions:

$r_{2}\cdot \sin \alpha_{2} = 1$ (2)

$r_{2}\cdot \cos \alpha_{2} = -\sqrt{3}$ (3)

By dividing (2) by (3), we have the following formula:

$\tan \alpha_{2} = -\frac{1}{\sqrt{3}}$

$\tan \alpha_{2} = -\frac{\sqrt{3}}{3}$

The tangent function is negative at second and fourth quadrants. That is:

$\alpha_{2} = \tan^{-1} \left(-\frac{\sqrt{3}}{3} \right)$

There are at least two solutions:

$\alpha_{2,1} = 150^{\circ}$ , $\alpha_{2,2} = 330^{\circ}$

And the value of $r_{2}$ :

$r_{2}^{2}\cdot \sin^{2}\alpha_{2} + r_{2}^{2}\cdot \cos^{2}\alpha_{2} = 4$

$r_{2}^{2} = 4$

$r_{2} = 2$

The values of $r_{2}$ and $\alpha_{2}$ are 2 and 150º.

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4 years ago

What is 8 by the power of -1

ElenaW [278]

$x^{-m}=\frac{1}{x^m}$

so
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3 years ago