The solution to m + (-38) = 15 is 53. True False

Answer:

[tex]\text{Actual area of property}}=6400\text{ m}^2[/tex]

Step-by-step explanation:

Let x represent the actual side of square property.

We have been given that a map of square piece of property is drawn at a scale of 1 : 500.

We will use proportions to solve for our given problem.

[tex]\frac{\text{Actual length}}{\text{Map length}}=\frac{500}{1}[/tex]

[tex]\frac{x}{16\text{ cm}}=\frac{500}{1}[/tex]

[tex]\frac{x}{16\text{ cm}}*16\text{ cm}=500*16\text{ cm}}[/tex]

[tex]x=8000\text{ cm}[/tex]

[tex]x=80\text{ m}[/tex]

We know that area of a square is square of its side length.

[tex]\text{Actual area of property}}=(80\text{ m})^2[/tex]

[tex]\text{Actual area of property}}=6400\text{ m}^2[/tex]

Therefore, the actual area of property is 6400 square meters.

This is the answer that u seek

-5a^2b^5c^3

Final answer:

The value of x for the line with a slope of 3/5 that passes through (3,5) and (x,9) is 9 2/3 or 9.67 when expressed as a decimal.

Explanation:

To find the value of x for the point (x, 9) on a line with a slope of 3/5, we can use the slope formula which is (change in y) / (change in x) = slope. We know two points on this line, (3, 5) and (x, 9).

First, calculate the change in y which is 9 - 5 = 4. Then, use the slope of 3/5 and set it equal to the change in y (which is 4) over the change in x (which is x - 3):

3/5 = 4 / (x - 3)

To find x, we cross-multiply:

3 x (x - 3) = 5 x 4

3x - 9 = 20

Add 9 to both sides: 3x = 29

Divide both sides by 3: x = 29/3

Therefore, x = 9 2/3 or 9.67 when expressed as a decimal.

This calculation shows the x-coordinate for the point where the line passes through y-coordinate 9.

area = base x height

 base = 5

 and height = 7

 so area = 7*5 =35 square yards

The exact number of people who visited the art museum is; C) 352,483

We are told that about 400,000 people visited an art museum in December.

Now, the given statement makes it clear that the 400000 people is an approximate value. This means that it has been rounded up to the nearest 100000.

Now among the options, the only that when rounded up to the nearest hundred thousand will give 400000 will be Option C which is 353,483.

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Complete question is;

About 400,000 people visited an art museum in December. What could be the exact number of people who visited the art museum?

A) 478,051

B) 452,223

C) 352,483

D) 348,998

1.  The cost of the bicycle is $14.

2. Mike's weight is 109 pounds.

3. Mr. Washington has saved $6,000.

Let's solve each problem step by step:

1. The $70 selling price of a bicycle is the cost increased by 4 times the cost.

Let's assume the cost of the bicycle is C dollars.

According to the given information, the selling price of the bicycle is $70, and it is equal to the cost increased by 4 times the cost:

70 = C + 4C

Combine like terms:

70 = 5C

Now, divide both sides by 5 to solve for C:

C = 70 / 5

C = 14

So, the cost of the bicycle is $14.

2. Three pounds less than twice Mike's weight is 215 pounds.

Let's assume Mike's weight is W pounds.

According to the given information, three pounds less than twice Mike's weight is 215 pounds:

2W - 3 = 215

Add 3 to both sides to isolate 2W:

2W = 215 + 3

2W = 218

Now, divide both sides by 2 to solve for W:

W = 218 / 2

W = 109

So, Mike's weight is 109 pounds.

3. If Mr. Washington's savings were increased by 5 times his savings, he would then save $36,000.

Let's assume Mr. Washington's initial savings is S dollars.

According to the given information, if his savings were increased by 5 times, he would save $36,000:

S + 5S = 36000

Combine like terms:

6S = 36000

Now, divide both sides by 6 to solve for S:

S = 36000 / 6

S = 6000

So, Mr. Washington has saved $6,000.

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Answer:

B. 7.94 * 10^8 miles

Step-by-step explanation:

is the correct answer

31.7 -2.9 = 28.8

31.7 + 2.9 = 34.6

 between 28.8 & 34.6 hours

D is the correct answer

Answer:

13.5 in^2 is the answer

Step-by-step explanation:

The two possible original temperatures are 9 and - 9 degrees Celcius and the difference between the temperature is 18 degrees Celcius.

Any equation of the form [tex]\rm ax^2+bx+c=0[/tex]  where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

As we know, the formula for the roots of the quadratic equation is given by:

[tex]\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}[/tex]

It is given that:

A scientist finds the temperature of a sample at the beginning of an experiment is t degrees Celcius.

After 1 hour, the temperature is t² degrees Celcius. If the temperature after 1 hour is 81 degrees Celcius.

From the question:

t² = 81

Taking square root on both sides:

t = ±√81

t = ±9

t = 9 degrees Celcius

Or

t = -9 degrees Celcius

The difference between the temperature = (9) - (-9) = 18 degrees Celcius

Thus, the two possible original temperatures are 9 and - 9 degrees Celcius and the difference between the temperature is 18 degrees Celcius.

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a . 1 + 3 + 5 + 7 + 9 + ... + 99 = 2500

b. 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + ... - 100 = -50

c. -100 - 99 - 98 - .... -2 - 1 - 0 + 1 + 2 + ... + 48 + 49 + 50 = -3775

Let us learn about Arithmetic Progression.

Arithmetic Progression is a sequence of numbers in which each of adjacent numbers have a constant difference.

[tex]\large {\boxed {T_n = a + (n-1)d } }[/tex]

[tex]\large {\boxed {S_n = \frac{1}{2}n ( 2a + (n-1)d ) } }[/tex]

Tn = n-th term of the sequence

Sn = sum of the first n numbers of the sequence

a = the initial term of the sequence

d = common difference between adjacent numbers

Let us now tackle the problem!

1 + 3 + 5 + 7 + 9 + ... + 99

initial term = a = 1

common difference = d = ( 3 - 1 ) = 2

Firstly , we will find how many numbers ( n ) in this series.

[tex]T_n = a + (n-1)d[/tex]

[tex]99 = 1 + (n-1)2[/tex]

[tex]99-1 = (n-1)2[/tex]

[tex]98 = (n-1)2[/tex]

[tex]\frac{98}{2} = (n-1)[/tex]

[tex]49 = (n-1)[/tex]

[tex]n = 50[/tex]

At last , we could find the sum of the numbers in the series using the above formula.

[tex]S_n = \frac{1}{2}n ( 2a + (n-1)d )[/tex]

[tex]S_{50} = \frac{1}{2}(50) ( 2 \times 1 + (50-1) \times 2 )[/tex]

[tex]S_{50} = 25 ( 2 + 49 \times 2 )[/tex]

[tex]S_{50} = 25 ( 2 + 98 )[/tex]

[tex]S_{50} = 25 ( 100 )[/tex]

[tex]\large { \boxed { S_{50} = 2500 } }[/tex]

In this question let us find the series of even numbers first ,  such as :

2 + 4 + 6 + 8 + ... + 100

initial term = a = 2

common difference = d = ( 4 - 2 ) = 2

Firstly , we will find how many numbers ( n ) in this series.

[tex]T_n = a + (n-1)d[/tex]

[tex]100 = 2 + (n-1)2[/tex]

[tex]100-2 = (n-1)2[/tex]

[tex]98 = (n-1)2[/tex]

[tex]\frac{98}{2} = (n-1)[/tex]

[tex]49 = (n-1)[/tex]

[tex]n = 50[/tex]

We could find the sum of the numbers in the series using the above formula.

[tex]S_n = \frac{1}{2}n ( 2a + (n-1)d )[/tex]

[tex]S_{50} = \frac{1}{2}(50) ( 2 \times 2 + (50-1) \times 2 )[/tex]

[tex]S_{50} = 25 ( 4 + 49 \times 2 )[/tex]

[tex]S_{50} = 25 ( 4 + 98 )[/tex]

[tex]S_{50} = 25 ( 102 )[/tex]

[tex]\large { \boxed { S_{50} = 2550 } }[/tex]

At last , we could find the result of the series.

1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + ... - 100

= ( 1 + 3 + 5 + 7 + ... + 99 ) - ( 2 + 4 + 6 + 8 + ... + 100 )

= 2500 - 2550

= -50

1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + ... - 100 = -50

-100 - 99 - 98 - .... -2 - 1 - 0 + 1 + 2 + ... + 48 + 49 + 50

initial term = a = -100

common difference = d = ( -99 - (-100) ) = 1

Firstly , we will find how many numbers ( n ) in this series.

[tex]T_n = a + (n-1)d[/tex]

[tex]50 = -100 + (n-1)1[/tex]

[tex]50+100 = (n-1)[/tex]

[tex]150 = (n-1)[/tex]

[tex]n = 151[/tex]

We could find the sum of the numbers in the series using the above formula.

[tex]S_n = \frac{1}{2}n ( 2a + (n-1)d )[/tex]

[tex]S_{151} = \frac{1}{2}(151) ( 2 \times (-100) + (151-1) \times 1 )[/tex]

[tex]S_{151} = 75.5 ( -200 + 150 )[/tex]

[tex]S_{151} = 75.5 ( -50 )[/tex]

[tex]\large { \boxed { S_{151} = -3775 } }[/tex]

Grade: Middle School

Subject: Mathematics

Chapter: Arithmetic and Geometric Series

Keywords: Arithmetic , Geometric , Series , Sequence , Difference , Term

The sum of the series are:

Part(a): [tex]\fbox{\begin\\\ \math S=2500\\\end{minispace}}[/tex]

Part(b): [tex]\fbox{\begin\\\ \math S=-50\\\end{minispace}}[/tex]

Part(c): [tex]\fbox{\begin\\\ \math S=-3775\\\end{minispace}}[/tex]

Further explanation:

A series is defined as a sum of different numbers in which each term is obtained from a specific rule or pattern.

In this question we need to determine the sum of the series given in the part (a), part (b) and part (c).

Part(a):

The series given in part (a) is as follows:

[tex]1+3+5+7+9+...+99[/tex]

All the terms in the given series are odd numbers.

From the given series in part(a) it is observed that the series is an arithmetic series with the common difference of [tex]2[/tex].

An arithmetic series is a series in which each successive member of the series differs from its previous term by a constant quantity.

From the above series it is observed that the first term is [tex]1[/tex], second term is [tex]3[/tex], third term is [tex]5[/tex], fourth term is [tex]7[/tex], fifth term is [tex]9[/tex] and the last term is [tex]99[/tex].

The nth term in a arithmetic series is given as follows:

[tex]a_{n}=a+(n-1)d[/tex]           (1)

In the above equation a represents the first term, [tex]n[/tex] represents the total terms and [tex]d[/tex] represents the common difference.

Substitute [tex]99[/tex] for [tex]a_{n}[/tex], [tex]1[/tex] for [tex]a[/tex] and [tex]2[/tex] for [tex]d[/tex] in equation (1).

[tex]\begin{aligned}99&=1+2(n-1)\\2(n-1)&=98\\n-1&=49\\n&=50\end{aligned}[/tex]

Therefore, total number of terms in the series is [tex]50[/tex]. This implies that [tex]a_{50}=99[/tex].

The sum of an arithmetic series is calculated as follows:

[tex]S_{n}=\dfrac{n}{2}(a+a_{n})[/tex]          (2)

Substitute [tex]50[/tex] for [tex]n[/tex] in equation (2).

[tex]\begin{aligned}S_{50}&=\dfrac{50}{2}(a+a_{50})\\&=25\times (1+99)\\&=25\times 100\\&=2500\end{aligned}[/tex]

Therefore, the sum of the series for part(a) is [tex]\bf 2500[/tex].

Part(b):

The series given in part (b) is as follows:

[tex]1-2+3-4+5-6+7-8+….-100[/tex]

Express the given series as follows:

[tex]S=(1+3+5+7+...+99)-(2+4+6+8+...+100)\\S=S^{'}-S^{''}[/tex]

The series [tex]S^{'}[/tex] is as follows:

[tex]S^{'}=1+3+5+7+...+99[/tex]

It is observed that the above series [tex]S^{'}[/tex] is exactly same as the series given in the part(a) and the sum of the series of part(a) as calculated above is [tex]2500[/tex].

Therefore, sum of the series [tex]S^{'}[/tex] is [tex]2500[/tex] i.e., [tex]S^{'}=2500[/tex].

The series [tex]S^{"}[/tex] is as follows:

[tex]S^{"}=2+4+6+8+...+100[/tex]

From the above series it is observed that the series [tex]S^{"}[/tex] is an arithmetic series as the difference between each consecutive member is [tex]2[/tex] and the last term is [tex]100[/tex].

Substitute [tex]2[/tex] for [tex]a[/tex], [tex]2[/tex] for [tex]d[/tex] and [tex]100[/tex] for [tex]a_{n}[/tex] in equation (1).

[tex]\begin{aligned}100&=2+(n-1)2\\(n-1)2&=98\\n-1&=49\\n&=50\end{aligned}[/tex]

This implies that [tex]a_{50}=100[/tex].

To calculate the sum of  substitute [tex]50[/tex] for [tex]n[/tex] in equation (2).

[tex]\begin{aligned}S_{50}&=\dfrac{50}{2}(a+a_{50})\\&=(25)(2+102})\\ &=25\times 102\\&=2550\end{aligned}[/tex]

Therefore, sum of the series [tex]S^{"}[/tex] is [tex]2550[/tex].

Substitute [tex]2550[/tex] for [tex]S^{"}[/tex] and [tex]2500[/tex] for  in equation (3).

[tex]\begin{aligned}S&=S^{'}+S^{"}\\&=2500-2550\\&=-50\end{aligned}[/tex]

Therefore, the sum of the series for part(b) is [tex]\bf -50[/tex].

Part(c):

The series given in part(c) is as follows:

[tex]-100-99-9-...-2-1-0+1+2+...+48+49+50[/tex]

From the above series it is observed that it is an arithmetic series with common difference as [tex]1[/tex], first term as [tex]-100[/tex] and the last term as [tex]50[/tex].

Substitute [tex]-100[/tex] for [tex]a[/tex], [tex]1[/tex] for [tex]d[/tex] and [tex]50[/tex] for [tex]a_{n}[/tex] in equation (1).

[tex]\begin{aligned}50&=-100+(n-1)1\\n-1&=150\\n&=151\end{aligned}[/tex]

Substitute [tex]151[/tex] for [tex]n[/tex] in equation (2).

[tex]\begin{aligned}S_{151}&=\dfrac{151}{2}(a+a_{151})\\&=\dfrac{151}{2}(-100+50)\\&=-25\times 151\\&=-3775\end{aligned}[/tex]

Therefore, the sum of the series for part(c) is [tex]\bf -3775[/tex].

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Answer details:

Grade: High school

Subject: Mathematics

Chapter: Series

Keywords: Series, sequence, arithmetic sequence, arithmetic series, 1+3+5+7+9+….+99, 1-2+3-4+5-6+7-8+….-100, -100-99-98-….-2-1-0+1+2+…..+48+49+50, sum of series, first term, common difference.