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sergeinik [125]

4 years ago

10

Last week Malia spent $13,000 on advertising. This week, she plans to spend three times as much. Next week, she wants to spend 6

0% of what she spent the previous two weeks. How much should she plan to spend on advertising next week?

1 answer:

PSYCHO15rus [73]4 years ago

4 0

Answer:

31200 dollars

Step-by-step explanation:

you multiply 13 by 4, because its last week and this week, then find 60 percent of that. Proportion: x/52000=60/100. When solved, it is $31200

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Tatami mats are used as a floor covering in japan. One possible layout uses four identical rectangular mats and one square mat,

Maksim231197 [3]

a) Given

Total area = 81 feet squared.

Let the area of the square mat = x feet.

From question, area of the rectangular mat is twice that of square mat.

So, the area of the rectangular mat $= 2 x$ feet.

Now,

Total area = 81 feet squared

Then,

$4(2 x) + x = 81\\ \\ 9 x = 81\\ \\ x = \frac{81}{9}\\ \\ x = 9\\$

Hence the area of the one rectangular mat = 2*9 = 18 feet squared

b) Let the width of the rectangular mat = y feet.

Then length of the mat = 2 y feet.

Area of the rectangle $= 2 y* y = 2 y^{2}$

and area of rectangle = 18 feet squared.

So,

$2 y^{2} = 18\\ y^{2} = 9\\ y= \pm3\\$

Width can not be negative.

Hence, the width of the rectangular mat = 3 feet

and length of the rectangular mat = 3*2 = 6 feet

8 0

3 years ago

Product of (y^3 - 125) ÷ (y - 5)

Ludmilka [50]

Answer:

$y^{2}+5y+25$

Step-by-step explanation:

8 0

3 years ago

Help me asap! I will give you marks

Law Incorporation [45]

Recall the binomial theorem.

$(a+b)^n = \displaystyle \sum_{k=0}^n \binom nk a^{n-k} b^k$

1. The binomial expansion of $\left(1+\frac x3\right)^7$ is

$\left(1 + \dfrac x3\right)^7 = \displaystyle\sum_{k=0}^7 \binom 7k 1^{7-k} \left(\frac x3\right)^k = \sum_{k=0}^7 \binom 7k \frac{x^k}{3^k}$

Observe that

$k = 1 \implies \dbinom 71 \left(\dfrac x3\right)^1 = \dfrac73 x$

$k = 2 \implies \dbinom 72 \left(\dfrac x3\right)^2 = \dfrac73 x^2$

When we multiply these by $8-9x$ ,

• $8$ and $\frac73 x^2$ combine to make $\frac{56}3 x^2$

• $-9x$ and $\frac73 x$ combine to make $-\frac{63}3 x^2 = -21x^2$

and the sum of these terms is

$\dfrac{56}3 x^2 - 21x^2 = \boxed{-\dfrac73 x^2}$

2. The binomial expansion is

$\left(2a - \dfrac b2\right)^8 = \displaystyle \sum_{k=0}^8 \binom 8k (2a)^{8-k} \left(-\frac b2\right)^k = \sum_{k=0}^8 \binom 8k 2^{8-2k} a^{8-k} b^k$

We get the $a^6b^2$ term when $k=2$ :

$k=2 \implies \dbinom 82 2^{8-2\cdot2} a^{8-2} b^2 = 28 \cdot2^4 a^6 b^2 = \boxed{448} \, a^6b^2$

6 0

1 year ago

What are the quotient and remainder when 777 is divided by 21?

DIA [1.3K]

777 divided by 21 = 34 with a remainder of 3

5 0

3 years ago

A glass of skim milk supplies 0.1 mg of iron, 8.5 g of protein, and 1 g of carbohydrates. A quarter pound of lean red meat provi

Tems11 [23]

Answer:

glasses of milk=4

quatar pounds servings of meat =5

two slices of whole grains bread =8

Step-by-step explanation:

Let m= milk

r=meat

b=bread

0.1m+3.4r+2.2b=35 (1)

8.5m+22r+10b=224 (2)

m+20r+12b=200 (3)

Multiply (1) by 85 and (2) by 1

8.5m+289r+187b=2,975 (4)

8.5m+22r+10b=224 (5)

Subtract (5) from (4)

267r+177b=2,751 (6)

Recall,

8.5m+22r+10b=224 (2)

m+20r+12b=200 (3)

Multiply (2) by 1 and (3) by 8.5

8.5m+22r+10b=224 (7)

8.5m+170r+102b=1700 (8)

Subtract (7) from (8)

148r+92b=1,476 (9)

Recall,

267r+177b=2,751 (6) and

148r+92b=1,476 (9)

Multiply (6) by 148 and (9) by 267

26196b=407,148 (10)

24564b=394,092 (11)

Subtract (11) from (10)

1,632b=13,056

Divide both sides by 1,632

b=13,056/1,632

b=8

Substitute the value of b in (9)

148r+92b=1,476 (9)

148r+92(8)=1,476

148r+736=1,476

148r=1,476-736

148r=740

Divide both sides by 148

r=740/148

r= 5

Substitute the value of b and r in (1)

0.1m+3.4r+2.2b=35 (1)

0.1m+3.4(5)+2.2(8)=35

0.1m+17+17.6=35

0.1m+34.6=35

0.1m=35-34.6

0.1m=0.4

Divide both sides by 0.1

m=0.4/0.1

m=4

4 glasses of milk, 5 quatar pounds servings of meat and 8 two slices of whole grains bread will supply the required nutrients.

8 0

4 years ago