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nexus9112 [7]
3 years ago
6

Perform the indicated operation. 6 1/3 + 8 3/4

Mathematics
1 answer:
belka [17]3 years ago
6 0

Answer:

15 1/12

Step-by-step explanation:

First step, add the whole numbers ie 6+8=14

Second, add fractions, 1/3+3/4

The lcm of denominators is 12 hence (4+9)/12=13/12=1 1/12

14+1 1/12=15 1/12

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Seven hotdogs and four hamburgers cost $13. Four hotdogs and seven hamburgers cost $14.50. How much does one hamburger cost and
viva [34]

Answer:

hotogs=$1.50  hamburgers=$1

Step-by-step explanation:

h=hotdog b=hamburger

7h+4b=13

4h+7b=14.5

(7h+4b=13)-4b

4h+7b=14.5

 

h=-0.57b+1.86

4h+7b=14.5

8 0
3 years ago
5 dollars for 2 cans of tuna
Lera25 [3.4K]

Answer:

5 times 2=10 well i guess this was a joke right?

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
I really need your help, but I don't get any.
STatiana [176]

Answer:

Step-by-step explanation:

x < -12, so x+12 < 0

| x-(-12) | = | x+12 | = -x-12

5 0
2 years ago
The perimeter of a rectangular garden is 108 meters. The length is 6 meters longer than twice the width. Find the dimensions of
faltersainse [42]

The answer is that the width is 16 m

7 0
2 years ago
Find all solutions of each equation on the interval 0 ≤ x &lt; 2π.
Korvikt [17]

Answer:

x = 0 or x = \pi.

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}.

\displaystyle \sec{x} = \frac{1}{\cos{x}}.

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, \sin^{2}{x} = 1 - \cos^{2}{x}. Therefore, for the square of tangents,

\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}.

This equation will thus become:

\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2.

To simplify the calculations, replace all \cos^{2}{x} with another variable. For example, let u = \cos^{2}{x}. Keep in mind that 0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1.

\displaystyle \frac{1 - u}{u^{2}} + \frac{2}{u} - \frac{1 - u}{u} = 2.

\displaystyle \frac{(1 - u) + u - u \cdot (1- u)}{u^{2}} = 2.

Solve this equation for u:

\displaystyle \frac{u^{2} + 1}{u^{2}} = 2.

u^{2} + 1 = 2 u^{2}.

u^{2} = 1.

Given that 0 \le u \le 1, u = 1 is the only possible solution.

\cos^{2}{x} = 1,

x = k \pi, where k\in \mathbb{Z} (i.e., k is an integer.)

Given that 0 \le x < 2\pi,

0 \le k.

k = 0 or k = 1. Accordingly,

x = 0 or x = \pi.

8 0
2 years ago
Read 2 more answers
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