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serg [7]
3 years ago
7

Which of the following trigonometric ratios are correct?

Mathematics
1 answer:
anastassius [24]3 years ago
4 0

Answer:

Step-by-step explanation:

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At a sale, coats were sold for $104 each. If the coats originally cost $160 each, what
Anvisha [2.4K]

Answer: I think it's 35%

With original price $160 and 35% discount,

Price after discount: $104.00

You saved: $56.00

The term discount can be used to refer to many forms of reduction in the price of a good or service. The two most common types of discounts are discounts in which you get a percent off, or a fixed amount off.

A percent off of a price typically refers to getting some percent, say 10%, off of the original price of the product or service. For example, if a good costs $45, with a 10% discount, the final price would be calculated by subtracting 10% of $45, from $45, or equivalently, calculating 90% of $45:

10% of $45 = 0.10 × 45 = $4.50

$45 – $4.50 = $40.50

or

90% of $45 = 0.90 × 45 = $40.50

In this example, you are saving 10%, or $4.50.

Step-by-step explanation:

hope this helps, sorry if I'm wrong.

4 0
1 year ago
The swimming team has 84 members. The number of boys on the team is six more than twice the number of girls. How many boys and g
Kamila [148]

Answer:

the number of boys and girls on a team is 58 and 26 respectively

Step-by-step explanation:

Given that

Total members in swimming team is 84

Let us assume the number of girls be x

so, the number of boys would be

y = 2x + 6

Now the equation is

x + y = 84

x + 2x + 6 = 84

3x = 78

x = 26

And, y = 2 × 26 + 6

= 52 + 6

= 58

Hence, the number of boys and girls on a team is 58 and 26 respectively

5 0
3 years ago
What is the measure of
EleoNora [17]

Answer:

129 degrees.

Step-by-step explanation:

As we can see, angle DCF is made up of 2 different angles, DCE and ECF, the sum of the measure of these two angles is equal to the measure of angle DCF.

75+54=129 degrees

Hope it helps!

4 0
3 years ago
In how many ways can we seat 3 pairs of siblings in a row of 7 chairs, so that nobody sits next to their sibling
monitta

Answer:

1,968

Step-by-step explanation:

Let x₁ and x₂, y₁ and y₂, and z₁ and z₂ represent the 3 pairs of siblings, and let;

Set X represent the set where the siblings x₁ and x₂ sit together

Set Y represent the set where the siblings y₁ and y₂ sit together

Set Z represent the set where the siblings z₁ and z₂ sit together

We have;

Where the three siblings don't sit together given as X^c∩Y^c∩Z^c

By set theory, we have;

\left | X^c \cap Y^c \cap Z^c  \right | = \left | X^c \cup Y^c \cup Z^c  \right | =  \left | U  \right | - \left | X \cup Y \cup Z  \right |

\left | U  \right | - \left | X \cup Y \cup Z  \right | = \left | U  \right | - \left (\left | X \right | +  \left | Y\right | +  \left | Z\right | -  \left | X \cap Y\right | -  \left | X \cap Z\right | -  \left | Y\cap Z\right | +  \left | X \cap Y \cap Z\right | \right)

Therefore;

\left | X^c \cap Y^c \cap Z^c  \right | = \left | U  \right | - \left (\left | X \right | +  \left | Y\right | +  \left | Z\right | -  \left | X \cap Y\right | -  \left | X \cap Z\right | -  \left | Y\cap Z\right | +  \left | X \cap Y \cap Z\right | \right)

Where;

\left | U\right | = The number of ways the 3 pairs of siblings can sit on the 7 chairs = 7!

\left | X\right | = The number of ways x₁ and x₂ can sit together on the 7 chairs = 2 × 6!

\left | Y\right | = The number of ways y₁ and y₂ can sit together on the 7 chairs = 2 × 6!

\left | Z\right | = The number of ways z₁ and z₂ can sit together on the 7 chairs = 2 × 6!

\left | X \cap Y\right | = The number of ways x₁ and x₂ and y₁ and y₂ can sit together on the 7 chairs = 2 × 2 × 5!

\left | X \cap Z\right | = The number of ways x₁ and x₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

\left | Y \cap Z\right | = The number of ways y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

\left | X \cap Y \cap Z\right | = The number of ways x₁ and x₂,  y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 2 × 4!

Therefore, we get;

\left | X^c \cap Y^c \cap Z^c  \right | = 7! - (2×6! + 2×6! + 2×6! - 2 × 2 × 5! - 2 × 2 × 5! - 2 × 2 × 5! + 2 × 2 × 2 × 4!)

\left | X^c \cap Y^c \cap Z^c  \right | = 5,040 - 3072 = 1,968

The number of ways where the three siblings don't sit together given as \left | X^c \cap Y^c \cap Z^c  \right |  = 1,968

5 0
3 years ago
Equivalent Fractions
Maru [420]

C is the correct answer


3/5 x 2 =6/10


6 0
3 years ago
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