Question 2.1
Solve the following problem. Round to one decimal place if
necessary. If your answer is correct you will see an image
appear on your screen.

The total surface area of the doghouse is 1452 ft²

The area occupied by a three-dimensional object by its outer surface is called the surface area.

The dog house has many surfaces including the roofs . The total surface area is the sum of all the area of the surfaces.

area of the roof part = 2( 13×11) + 2( 12× 10)×1/2

= 286 + 720

= 1006 ft²

surface area of the building

= 2( lb + lh + bh) -bh

= 2( 10× 11 + 10×8 + 11×8) - 10×11

= 2( 110+80+88)-110

= 2( 278) -110

= 556 -110

= 446ft²

The total surface area = 1006 + 446

= 1452 ft²

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Step-by-step explanation:

\(\text{Let the original price of the phone be } P.\)

\(P(1+0.2\%)(1-10\%)=15000\)

\(P=\frac{15000}{1.2\cdot0.9}\)

\(P=13888.89\)

Answer:

(c) No; There is an exponent on a variable x that is not a whole number

Step-by-step explanation:

Given

\(f(x) = x^2 - \sqrt[5]{x}\)

Required

Determine if the function is a polynomial function of not

A polynomial function is represented as:

\(f(x) =ax^n + bx^{n-1} + cx^{n-2}+........+z\)

Where n is an integer and the least power is 0

So, we have:

\(f(x) = x^2 - \sqrt[5]{x}\)

The degree of x is \(\frac{1}{5}\) ----- i.e. not an integer

Hence, the function is not a polynomial

Answer:

D

Step-by-step explanation:

D because you have to multiply 8 and 2 for area for that side, then multiply it by two, since there are two of those sides. C is the only one that shows this.

The total amount the man had = 52,200 rupees. Out of this, he gave 21,750 rupees to his son, 13,050 rupees to his daughter, and 17,400 rupees to his wife , the total amount given away by the man = 21,750 + 13,050 + 17,400 = 52,200 rupees.

A man gave 5/12 of his money to his son, 3/7 of the remainder to his daughter, and the remaining to his wife. If his wife gets Rs. 8,700, what is the total amount?

The given problem can be solved using the concept of ratios and fractions. Let us solve the problem step-by-step.Assume the man had x rupees with him.The man gave 5/12 of his money to his son.

The remaining amount left with the man = x - 5x/12= (12x/12) - (5x/12) = (7x/12)The man gave 3/7 of the remainder to his daughter.'

Amount left with the man after giving it to his son = (7x/12)The amount given to the daughter = (3/7) x (7x/12)= (3x/4)The remaining amount left with the man = (7x/12) - (3x/4)= (7x/12) - (9x/12) = - (2x/12) = - (x/6) (As the man has given more money than what he had with him).

Therefore, the daughter's amount is (3x/4) and the remaining amount left with the man is (x/6).The man gave all the remaining amount to his wife.

Therefore, the amount given to the wife is (x/6) = 8700Let us find the value of x.x/6 = 8700 x = 6 x 8700 = 52,200

Therefore, the man had 52,200 rupees with him.He gave 5/12 of his money to his son. Therefore, the amount given to his son is (5/12) x 52,200 = 21,750 rupees.

The remaining amount left with the man = (7/12) x 52,200 = 30,450 rupees.He gave 3/7 of the remainder to his daughter. Therefore, the amount given to his daughter is (3/7) x 30,450 = 13,050 rupees.

The amount left with the man = (4/7) x 30,450 = 17,400 rupees.The man gave 17,400 rupees to his wife.
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Answer:

6x² + 5x + 1

Step-by-step explanation:

(4x² + x − 4) + (2x² + 4x + 5)

1. Expand the brackets to get
4x² + x − 4 + 2x² + 4x + 5

2. Group common terms (terms with same power as well as the constant)
4x² + 2x² + x + 4x - 4 + 5

3. Simplify
4x² + 2x² = 6x²
x + 4x = 5x
-4 + 5 = 1

4. Solution is
6x² + 5x + 1

Answer:

No.

Step-by-step explanation:

In similar triangles, all the angles are congruent.

b = 180 - 90 - 43 = 47

c = 180 - 90 - 48 = 42

1 set of angles is 90, 43, and 47

And the other is 90, 48, and 42.

Meaning the triangles are not similar.

In a case whereby Juan buys candy that costs $8 per pound and will spend at least $48 on candy  the possible numbers of pounds he will buy written in inequality is is p>5 (at least 5 pounds).

Inequality in math refers to a statement that one quantity is either greater than, less than, or not equal to another quantity. Inequalities are often represented using symbols such as "<" (less than), ">" (greater than), or "≤" (less than or equal to) and "≥" (greater than or equal to). For example, the inequality "x > 3" states that the variable "x" is greater than the value of 3. The inequality "y ≤ 5" means that the variable "y" is less than or equal to 5.

From the question, p = number of pounds Juan will buy.

candy=costs $8 per pound

$20/$4 = 5 pounds

p>5

at least 5 pounds

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Answer:

c

Step-by-step explanation:

assume base 10

-logy = x

\( \frac{1}{ log(y) } = x\)

log base x y = 5 turns into the format x^ 5 = y

implement that to get c

Answer:more

Step-by-step explanation:

A zero has the same meaning as the vertex.

The point where two lines meet forming an angle or set of angles is referred to as a vertex. This can also be called a point of origin especially when considering the axis of a Cartesian coordinate.

The vertex can be taken as zero since it is referred to as the point of origin. This is different to the axis of symmetry; as the axis of symmetry is one that splits a given plotted points i.e. a graph into two equal parts.

Thus, a zero has the same meaning as the vertex.

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Answer:

1. 2,772

2.1,071

3.4,641

4. 36,610

5. 4,096

6. 36,270

7. 2,403

8. 3,320

Answer:

28

Step-by-step explanation:

a^2+b^2=c^2

a^2+96^2=100^2

a^2+9216=10000

a^2=784

a^2= √784

a=28

Answer:

Step-by-step explanation:

Let,

The unknown leg be = x

According to Pythagoras Theorem,

\( {hypotenuse}^{2} - {base}^{2} = {height}^{2} \)

\( = > {100}^{2} - {96}^{2} = {x}^{2} \)

\( = > {x}^{2} = 10000 - 9216\)

\( = > {x}^{2} = 784\)

\( = > x = \sqrt{784} \)

=> x = 28

Hence,

the another leg is 28 in. (Ans)

Answer:

28 mph  

Step-by-step explanation:

I dont know if this helped but if it did then your welcome? I dont know if this helped sorry TwT

The term "100 points" usually indicates a high level of achievement or success in a particular area.

The term "100 points" can refer to a number of different things depending on the context. For example, in sports, it could refer to a team scoring 100 points in a game.

In wine tasting, it could refer to a perfect score given to a wine by a critic or judge. In school, it could refer to receiving a perfect score on an exam or assignment worth 100 points.

It's important to note that while a perfect score of 100 points is often seen as the ultimate goal, it's not always necessary or even possible in some situations. In many cases, simply doing your best and striving for improvement is enough to be considered successful.

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Answer:

Step-by-step explanation:

If Robert is 7 1/2 months old you need to multiply 7 1/2 by 12 because there are 12 months in a year.  

7x12=84  

1/2x12=6

6+84=90

We need to know about scientific notation to solve this problem. The number of seconds the playlist will play is 3.99 x 10^5.

Expressing a number in scientific notation means writing the number as a multiple of ten with ten raised to some power. We use scientific notation for numbers that are very small or very big. When writing numbers in scientific notation we always have one number other than zero before the decimal point. In this question we know that there are 1900 songs and each song has average time of 3.5 minutes, we need to convert the minutes into seconds and multiply the number of seconds of each song to 1900.

1 min=60 seconds

3.5 mins=3.5 x 60=210 seconds

210 x1900=399000 seconds

399000=3.99 x 10^5

Therefore the number of seconds that the playlist will play is 3.99x 10^5

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Answer:

P( red, red, red no replacement ) =  2/17

P( 3 cards with same rank, no replacement) =  1/425

Step-by-step explanation:

There are 52 cards, 26 are red

P( red) =  red/total =26/ 52 = 1/2

Now draw the second card without replacement

There are 51 cards, 25 are red

P( red) =  red/total =25/51

Now draw the third card without replacement

There are 50 cards, 24 are red

P( red) =  red/total =24/50 = 12/25

P( red, red, red no replacement ) = 1/2 * 25/51 * 12/25 = 2/17

P ( all have the same rank)

There are 52 cards,  4 of each rank

We don't care what the first rank is, now the second and third draw have to have the same rank as the first

P( rank) = 1/1

Now draw the second card without replacement

There are 51 cards, 3 are left with the same rank

P( same rank) = cards with same rank/total = 3/51

Now draw the third card without replacement

There are 50 cards, 2 are left with the same rank

P( same rank) = cards with same rank/total = 2/50 = 1/25

P( 3 cards with same rank, no replacement) = 1 * 3/51*1/25 = 1/425

As the hint suggests, convert to spherical coordinates using

x = p cos(u) sin(v)

y = p sin(u) sin(v)

z = p cos(v)

dV = dx dy dz = p² sin(v) dp du dv

Then U is the set

\(U = \left\{ (p,u,v) \mid 0\le p\le2 \text{ and } 0\le u\le \dfrac{\pi}2 \text{ and } 0\le v\le\dfrac{\pi}2\right\}\)

and the integral of y over U is

\(\displaystyle \iiint_U y \, dV = \iiint_U p\sin(u)\sin(v) \cdot p^2 \sin(v) \, dV\)

\(\displaystyle \iiint_U y \, dV = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \int_0^2 p^3 \sin(u) \sin^2(v) \, dp \, du \, dv \)

\( \displaystyle \iiint_U y \, dV = \frac{2^4-0^4}4 \int_0^{\frac\pi2} \int_0^{\frac\pi2} \sin(u) \sin^2(v) \, du \, dv \)

\(\displaystyle \iiint_U y \, dV = 4 \cdot \left(-\cos\left(\frac\pi2\right) + \cos(0)\right) \int_0^{\frac\pi2} \sin^2(v) \, dv \)

\(\displaystyle \iiint_U y \, dV = 4 \cdot \frac12 \int_0^{\frac\pi2} (1-\cos(2v)) \, dv \)

\(\displaystyle \iiint_U y \, dV = 2 \left(\left(\frac\pi2 - \frac12 \sin\left(2\cdot\frac\pi2\right)\right) - \left(0 - \frac12 \sin\left(2\cdot0\right)\right) \right)\)

\(\displaystyle \iiint_U y \, dV = \pi - \sin(\pi) = \boxed{\pi}\)

Let's start by defining some variables:

Let's call Jamal's age "J".

Let's call Alex's age "A".

From the problem statement, we know that:

J = A - 5 (Jamal is 5 years younger than Alex)

J * A = 300 (the product of their ages is 300)

We can use the first equation to substitute J in the second equation:

(A - 5) * A = 300

Expanding the left-hand side:

A^2 - 5A = 300

Moving all terms to the left-hand side:

A^2 - 5A - 300 = 0

We can solve for A using the quadratic formula:

A = (-(-5) ± sqrt((-5)^2 - 41(-300))) / (2*1)

A = (5 ± sqrt(25 + 1200)) / 2

A = (5 ± sqrt(1225)) / 2

Since A is a positive integer, we take the positive solution:

A = (5 + 35) / 2 = 20

Now we can use the first equation to find J:

J = A - 5 = 20 - 5 = 15

Therefore, Jamal is 15 years old.

Answer: 3

Step-by-step explanation:

Given: A sample of 100 production workers is obtained. The workers are classified by gender (male, female) and by age (under 20, 20−29, 30−39, and 40 or over).

Here,

Total classes as per gender: k = 2 (male female)

Total classes as per age : m = 4  (under 20, 20−29, 30−39, and 40 or over)

Now , the degree of freedom = (k-1) ( m-1)

=  (2-1) (4-1)

= (1)(3)

= 3

Hence, there are 3 degree of freedom.

1.Bricks and timber purchased for construction. (Debit: Bricks - Rs 60,000, Debit: Timber - Rs 35,000, Credit: Bank - Rs 95,000)

2.Transfer of Rs 13,000 to fixed deposit account. (Debit: Fixed Deposit - Rs 13,000, Credit: Current Account - Rs 13,000)

3.Appointment of Mr. S.N. Rao as Accountant. (Debit: Salary Expense - Rs 300, Debit: Security Deposit - Rs 1,000, Credit: Accountant - Rs 300)

4.Goods sold to Shruti with discounts. (Debit: Accounts Receivable - Shruti - Rs 80,000, Credit: Sales - Rs 80,000)

5.Goods purchased from Richa with discounts. (Debit: Purchases - Rs 60,000, Credit: Accounts Payable - Richa - Rs 60,000)

6.Goods sold to Shilpa with discounts and received payment. (Debit: Accounts Receivable - Shilpa - Rs 50,000, Credit: Sales - Rs 50,000)

7.Goods sold to Garima with discounts and received partial payment. (Debit: Accounts Receivable - Garima - Rs 1,00,000, Credit: Sales - Rs 1,00,000)

8.Purchase of land with additional charges. (Debit: Land - Rs 2,00,000, Debit: Brokerage Expense - Rs 2,000, Debit: Registration Charges - Rs 15,000, Credit: Bank - Rs 2,17,000)

9.Proprietor took goods and cash for personal use. (Debit: Proprietor's Drawings - Rs 65,000, Credit: Goods - Rs 25,000, Credit: Cash - Rs 40,000)

10.Goods sold to Charu with profit and discount. (Debit: Accounts Receivable - Charu - Rs 1,20,000, Credit: Sales - Rs 1,20,000)

11.Rent paid for the building. (Debit: Rent Expense - Rs 60,000, Credit: Bank - Rs 60,000)

12.Goods sold to Sunil with profit and discount. (Debit: Accounts Receivable - Sunil - Rs 24,000, Credit: Sales - Rs 24,000)

13.Purchased goods from Nanda and supplied to Helen. (Debit: Purchases - Rs 1,000, Debit: Accounts Payable - Nanda - Rs 1,000, Credit: Accounts Receivable - Helen - Rs 1,300, Credit: Sales - Rs 1,300)

14.Purchased goods from Rohit and Sons and supplied to Madan. (Debit: Purchases - Rs 2,700, Credit: Accounts Payable - Rohit and Sons - Rs 2,700, Debit: Accounts Receivable - Madan - Rs 3,000, Credit: Sales - Rs 3,000)

Here are the journal entries for the given transactions:

1. Bricks and timber purchased for construction:

Debit: Bricks (Asset) - Rs 60,000

Debit: Timber (Asset) - Rs 35,000

Credit: Bank (Liability) - Rs 95,000

2. Transfer to fixed deposit account:

Debit: Fixed Deposit (Asset) - Rs 13,000

Credit: Current Account (Asset) - Rs 13,000

3. Appointment of Mr. S.N. Rao as Accountant:

Debit: Salary Expense (Expense) - Rs 300

Debit: Security Deposit (Asset) - Rs 1,000

Credit: Accountant (Liability) - Rs 300

4. Goods sold to Shruti:

Debit: Accounts Receivable - Shruti (Asset) - Rs 80,000

Credit: Sales (Income) - Rs 80,000

5. Goods purchased from Richa:

Debit: Purchases (Expense) - Rs 60,000

Credit: Accounts Payable - Richa (Liability) - Rs 60,000

6. Goods sold to Shilpa:

Debit: Accounts Receivable - Shilpa (Asset) - Rs 50,000

Credit: Sales (Income) - Rs 50,000

7. Goods sold to Garima:

Debit: Accounts Receivable - Garima (Asset) - Rs 1,00,000

Credit: Sales (Income) - Rs 1,00,000

8.Purchase of land:

Debit: Land (Asset) - Rs 2,00,000

Debit: Brokerage Expense (Expense) - Rs 2,000

Debit: Registration Charges (Expense) - Rs 15,000

Credit: Bank (Liability) - Rs 2,17,000

9. Goods and cash taken away by proprietor:

Debit: Proprietor's Drawings (Equity) - Rs 65,000

Credit: Goods (Asset) - Rs 25,000

Credit: Cash (Asset) - Rs 40,000

10. Goods sold to Charu:

Debit: Accounts Receivable - Charu (Asset) - Rs 1,20,000

Credit: Sales (Income) - Rs 1,20,000

Credit: Cost of Goods Sold (Expense) - Rs 80,000

Credit: Profit on Sales (Income) - Rs 40,000

11. Rent paid for the building:

Debit: Rent Expense (Expense) - Rs 60,000

Credit: Bank (Liability) - Rs 60,000

12. Goods sold to Sunil:

Debit: Accounts Receivable - Sunil (Asset) - Rs 24,000

Credit: Sales (Income) - Rs 24,000

Credit: Cost of Goods Sold (Expense) - Rs 20,000

Credit: Profit on Sales (Income) - Rs 4,000

13. Goods purchased from Nanda and supplied to Helen:

Debit: Purchases (Expense) - Rs 1,000

Debit: Accounts Payable - Nanda (Liability) - Rs 1,000

Credit: Accounts Receivable - Helen (Asset) - Rs 1,300

Credit: Sales (Income) - Rs 1,300

14. Goods received from Rohit and Sons and supplied to Madan:

Debit: Purchases (Expense) - Rs 2,700 (after 10% trade discount)

Credit: Accounts Payable - Rohit and Sons (Liability) - Rs 2,700

Debit: Accounts Receivable - Madan (Asset) - Rs 3,000

Credit: Sales (Income) - Rs 3,000

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The length of wall in real life is 4 feet.

Given that on an architect's blueprint, 1 inch corresponds to 4 feet.

In this architect's plan, a wall represented by a line 1 inches long on the blue print.

So, the length of wall in real life is = 4 feet.

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Answer:

4ft

Step-by-step explanation:

If 1 inch corresponds to 4 ft on the blueprint. The ratio is 1 inch on the blue print for every 4 ft in real life.

Answer:

Step-by-step explanation:

(45 tulips)/(30 minutes) = (45/30 tulips)/minute = 1.5 tulips/minute

Answer:

>

Step-by-step explanation:

The result of 1/11 in the base 10 system is given as is 0.09.

To visualize the long division of 1 ÷ 11 in a base ten system:

Set up the long division: 11 outside and 1.00 inside the division bracket.

Place the decimal point in the quotient above the dividend's decimal point.

Bring down the first digit (0) and divide 11 into 10; it goes 0 times.

Subtract the product (0) from 10, leaving 10.

Bring down the next digit (0) and divide 11 into 100; it goes 9 times.

Subtract 99 from 100, leaving a remainder of 1.

The result is 0.09.

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