Ask question

Ask question

All categories

3 years ago

15

Mel bought a box of markers for b dollars, a shoulder bag that cost twice as much as the box of markers and a pen that cost $6 l

ess than the shoulder bag. Write the cost of the pen in terms of b,

1 answer:

vovikov84 [41]3 years ago

7 0

Since the box costed b dollars and the shoulder bag 2b (twice the amount of B), the shoulder bag costs 2b - 6.

Hopefully this helps! <3

You might be interested in

I need help please help

vesna_86 [32]

Answer:

C. (x-2)^2=7

Step-by-step explanation:

What you have to do first is to get rid of the 3 in the x^2 so you will divide 3 by the whole thing to get x^2-4x=3.

You will then divide -4 by 2 and then square the answer to get 4.

You will then add the 4 into the end of the equation and to the 3 to get x^2-4x+4=7.

You will then make into a perfect square (x-2)^2=7 and that is your answer.

5 0

3 years ago

Dding and Subtracting Decimals struction Active Finding an Account Balance Dewan's bank account balance is -$16.75. He deposits

inessss [21]

Answer: 24.34 is his balance.

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Please help!<br><br> Worth 10 points!

bonufazy [111]

10% of 140 is 14.
to solve you first must do 10/100 = 14/x
cross multiply and you get 10x = 1400
divide 1400 by 10 and you get 140

6 0

3 years ago

Read 2 more answers

Solve for c<br>16 — 20 - 40

givi [52]

Answer:

c=50

Step-by-step explanation:

6 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

8 0

3 years ago