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2 years ago

13

A shop owner offered a 20% discount off the regular price of a mirror. The amount of discount is $3 what is the regular price of

the mirror.

1 answer:

Alisiya [41]2 years ago

5 0

Answer: $15

Step-by-step explanation:

The normal price of the mirror is unknown we do not know that yet. So, let y represent the unknown number. The discount is 20 percent, we turn 20 into a decimal which would be 0.2. The amount of discount is $3. So, 0.2y equals to $3.

Then we divided.

y = 3 ÷ 0.2

And when you divide that is would be 15.

So, the regular price of the mirror is $15.

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Mathematics <br><br> Solving inequality

lions [1.4K]

Answer:

x _> 9 I dont have the symbol on my phone so I just put the greater than sign there

Step-by-step explanation:

1. Subtract 3 from both sides

5x > 48 - 3

2. simplofy 48 - 3 to 45

5x > 45

3. Divide both sides by 5

x > 45/5

4. Then simplify 45/5 to 9 which gives us the answer

And again I dont have the sign thats why I used the greater than sign

3 0

3 years ago

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

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Answer:

a. 86-degrees; angle-2 corresponds to the given angle.

Step-by-step explanation:

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