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Mandarinka [93]
3 years ago
9

From January to June, a company spent $60.00 per month on office supplies. In July, the price of office supplies increased by 15

% and remained the same for the rest of the year. How much did the company spend on office supplies for the year?
Mathematics
2 answers:
pochemuha3 years ago
5 0
The company would spend $69.00 a month on office supplies for the rest of the year.
Sunny_sXe [5.5K]3 years ago
5 0
For august -through- december the company spent $345
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Brianna has 0.15 liter of a liquid. She will use 0.4 of this liquid. How many liters will she use?
alukav5142 [94]

Answer: 0.0375

Step-by-step explanation:

0.15 divided by 4 = 0.0375

4 0
3 years ago
Simplify (12a5−6a−10a3)−(10a−2a5−14a4) . Write the answer in standard form
almond37 [142]

Answer:

=14a^5+14a^4-10a^3-16a

Step-by-step explanation:

\left(12a^5-6a-10a^3\right)-\left(10a-2a^5-14a^4\right)\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=12a^5-6a-10a^3-\left(10a-2a^5-14a^4\right)\\-\left(10a-2a^5-14a^4\right):\quad -10a+2a^5+14a^4\\-\left(10a-2a^5-14a^4\right)\\\mathrm{Distribute\:parentheses}\\=-\left(10a\right)-\left(-2a^5\right)-\left(-14a^4\right)\\Apply\:minus-plus\:rules\\-\left(-a\right)=a,\:\:\:-\left(a\right)=-a\\=-10a+2a^5+14a^4\\=12a^5-6a-10a^3-10a+2a^5+14a^4

\mathrm{Simplify}\:12a^5-6a-10a^3-10a+2a^5+14a^4:\quad 14a^5+14a^4-10a^3-16a12a^5-6a-10a^3-10a+2a^5+14a^4\\Group\:like\:terms\\=12a^5+2a^5+14a^4-10a^3-6a-10a\\\mathrm{Add\:similar\:elements:}\:12a^5+2a^5=14a^5\\=14a^5+14a^4-10a^3-6a-10a\\\mathrm{Add\:similar\:elements:}\:-6a-10a=-16a\\=14a^5+14a^4-10a^3-16a

4 0
3 years ago
QUESTION IN THE ATTACHMENT
eimsori [14]

Answer:

A. The sum of the first 10th term is 100.

B. The sum of the nth term is n²

Step-by-step explanation:

Data obtained from the question include:

Sum of 20th term (S20) = 400

Sum of 40th term (S40) = 1600

Sum of 10th term (S10) =..?

Sum of nth term (Sn) =..?

Recall:

Sn = n/2[2a + (n – 1)d]

Sn is the sum of the nth term.

n is the number of term.

a is the first term.

d is the common difference

We'll begin by calculating the first term and the common difference. This is illustrated below:

Sn = n/2 [2a + (n – 1)d]

S20 = 20/2 [2a + (20 – 1)d]

S20= 10 [2a + 19d]

S20 = 20a + 190d

But:

S20 = 400

400 = 20a + 190d .......(1)

S40 = 40/2 [2a + (40 – 1)d]

S40 = 20 [2a + 39d]

S40 = 40a + 780d

But

S40 = 1600

1600 = 40a + 780d....... (2)

400 = 20a + 190d .......(1)

1600 = 40a + 780d....... (2)

Solve by elimination method

Multiply equation 1 by 40 and multiply equation 2 by 20 as shown below:

40 x equation 1:

40 x (400 = 20a + 190d)

16000 = 800a + 7600. ........ (3)

20 x equation 2:

20 x (1600 = 40a + 780d)

32000 = 800a + 15600d......... (4)

Subtract equation 3 from equation 4

Equation 4 – Equation 3

32000 = 800a + 15600d

– 16000 = 800a + 7600d

16000 = 8000d

Divide both side by 8000

d = 16000/8000

d = 2

Substituting the value of d into equation 1

400 = 20a + 190d

d = 2

400 = 20a + (190 x 2)

400 = 20a + 380

Collect like terms

400 – 380 = 20a

20 = 20a

Divide both side by 20

a = 20/20

a = 1

Therefore,

First term (a) = 1.

Common difference (d) = 2.

A. Determination of the sum of the 10th term.

First term (a) = 1.

Common difference (d) = 2

Number of term (n) = 10

Sum of 10th term (S10) =..?

Sn = n/2 [2a + (n – 1)d]

S10 = 10/2 [2x1 + (10 – 1)2]

S10 = 5 [2 + 9x2]

S10 = 5 [2 + 18]

S10 = 5 x 20

S10 = 100

Therefore, the sum of the first 10th term is 100.

B. Determination of the sum of the nth term.

First term (a) = 1.

Common difference (d) = 2

Sum of nth term (Sn) =..?

Sn = n/2 [2a + (n – 1)d]

Sn = n/2 [2x1 + (n – 1)2]

Sn = n/2 [2 + 2n – 2]

Sn = n/2 [2 – 2 + 2n ]

Sn = n/2 [ 2n ]

Sn = n²

Therefore, the sum of the nth term is n²

6 0
3 years ago
-7-5x+5x+7 simplified​
AysviL [449]

Answer:

0

Step-by-step explanation:

adding -7 to 7, since they are both constants (no variable) gives you 0. Adding -5x and 5x since they both have an x term gives you 0x also written as 0 (0*any number is 0). this gives you a final answer of 0

4 0
3 years ago
Read 2 more answers
To rent an instrument from the school it costs a $50 insurance fee and then $350 per day of use. If Sally rents her instrument f
Ne4ueva [31]

Answer:

$2,500

Step-by-step explanation:

$350 per day for 25 days

350*25=2450

Add the 50 for insurance and that makes it 2500

If its wrong im sorry

5 0
2 years ago
Read 2 more answers
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