Describe the transformations to the basic function. Then, write the equation of the function.

Answer:

y = (1/2)x + (3/2)

Step-by-step explanation:

(-5, -1), (-1, 1)

(x₁, y₁)  (x₂, y₂)

        y₂ - y₁         1 - (-1)          1 + 1            2            1

m = ------------- = ----------- = ------------ = --------- = --------

        x₂ - x₁        -1 - (-5)        -1 + 5           4            2

y - y₁ = m(x - x₁)

y - (-1) = (1/2)(x - (-5))

y + 1 = (1/2)(x + 5)

y + 1 = (1/2)x + (5/2)

    -1                 -1

---------------------------------

y = (1/2)x + (3/2)

I hope this helps!

The number of months where they will both have same amount in their accounts is 40 months.

From the information given, the function f(t) = 10t + 500 models the amount of money in Abby’s account over time and the function g(t) = 11.25t + 450 models the amount of money in Lexi’s account over time.

It should be noted that the number of years when there value will be the same is:

10t + 500 = 11.25t + 450

Collect like terms

11.25t - 10t = 500 - 450

1.25t = 50

Divide

t = 50/1.25

t = 40

In 40 months, Abby and Lexi will have same amount.

Note that an overview was given as the information is incomplete.

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The percentage of students in Brookland Middle School who are attending seventh grade is 54.54%

What is percentage?

In mathematics, a percentage is a number or ratio that is expressed as a fraction of 100. Although "pct," "pct," and occasionally "pc" are also used as abbreviations, the percent symbol "%" is most usually used to denote it. A% is a number that has no dimensions and is dimensionless (pure).

We know that, to get the value of percentage we have to divide the value for which percentage needs to be calculated by total value and then multiply the resultant by 100.

So, we will divide the number of student in seventh class by the total number of student and then multiply it with 100.

(750/1375)100 = 54.54 %

Therefore, percentage of students attending Brookland Middle School are in seventh grade is 54.54%.

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The equivalent expression using a positive exponent is (5/3)⁴ .

If an expression was created utilizing integer variables, constants, and algebraic operations, it is said to be algebraic (addition, subtraction, multiplication, division and exponentiation by an exponent that is a rational number)

A positive expression is given as (3/5)⁻⁸ ÷ (3/5)⁻⁴ .

Using the properties of exponents we can calculate the expression as:

(3/5)⁻⁸⁺⁴ = (3/5)⁻⁴ = (5/3)⁴

Hence the required expression is (5/3)⁴

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The given the initial equation

we can determine the part of the graph to be shaded if the equation changes to the inequality:

If we make y the subject of the formula, such that

Then we can put in the values of x and y at (0,0)

so that

we can see that the expression is false because 0 is not greater than 1

Hence, we will shade away from the origin this means that we will shade above the line

The graph is shown below

So for question B

We will shade above the line

Question C

The points on the line are not included because the inequality does not include an equal sign

Answer:

−9 < x < −1

Step-by-step explanation:

a) Let's find the critical points of the inequality.

x^2 + 10x + 9 = 0

(x + 1)(x + 9) = 0

x + 1 = 0 or x + 9 = 0

x = −1 or x = −9

x < −9 (Wrong)

−9 < x < −1 (Correct)

x > −1 (Wrong)

Your answer is −9 < x < −1

Answer:

8.13=3b

Step-by-step explanation:

8 13 is the total cost and the bags of oranges are three.

The length of the pole is 19.15 feet.

The angle of elevation is created by the horizontal line and the line of sight. If the line of sight is upward from the horizontal line, the angle formed is an angle of elevation.

Given:

Angle made by straight road = 15°

Angle of elevation of the sun = 57°

Length of the shadow = 75 feet

∠ACB = 90° - 57° = 33°

∠CAB = 57° - 15° = 42°

Using Sine Law to find BC i.e.length of the pole.

[tex]\frac{BC}{sin A} = \frac{AB}{sin C}[/tex]

[tex]\frac{BC}{sin 42} = \frac{75}{sin 33}[/tex]

[tex]BC= \frac{75sin 42}{sin 33}[/tex]

BC ≅ 19.15 feet (to the nearest hundredth.)

Therefore, the approximate length of the pole is 19.15 feet.

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The scale factor for the graph will be 3 and there will be enlargement.

The scale factor may be defined as the ratio of two dimensions such that one figure is large, and another is small. The scale factor is done due to the unpractical measurement of any figure. For example, if a figure is a scale copy of the scale factor of 6/2 it means 2 is converted into 6. We can use the following points to know about enlargement.

A(1, 1) ⇒ A'(3, 3)

B(2, 1) ⇒ B'(6, 3)

C(1, 2) ⇒ C'(3, 6)

D(2, 2) ⇒ D'(6, 6)

Now, to find the scale factor, simply divide the Point' with the original Point. Use any number.

A'(3, 3)/(A(1, 1)) = 3

B'(6, 3)/(B(2, 1)) = 3

C'(3, 6)/(C(1, 2)) = 3

D'(6, 6)/(D(2, 2)) = 3

which shows that the scale factor is 3.

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Answer:

your answer is 3.3

Step-by-step explanation:

Answer:

The answer is to this problem is five over three

Hey there!

[tex]x+ay=b\\[/tex] for y

Add -x to both sides:

[tex]ay+x+-x=b+-x\\ay=b-x[/tex]

Now, let's divide both sides by a:

[tex]\frac{ay}{a} =\frac{b-x}{a}[/tex]

[tex]y=\frac{b-x}{a}[/tex]

Hope this helps!

Ok, so

The shape of a distribution is described by its number of peaks and by its possession of symmetry, its tendency to skew, or its uniformity.

The problem is in the parenthesis

Explaination: plus and minus is minus. That means -0.46-0.54=-1.

Answer:

£20 note, £5 note, 50p,50p,50p, 5p

Step-by-step explanation:

Liam buys some fruit for £26.55 by paying 2 notes of £20 and £5 and 3 coins of 50 cent and one coin of 5 cent.

Combining objects and counting them as one big group is done through addition. In arithmetic, addition is the process of adding two or more integers together. Addends are the numbers that are added, and the sum refers to the outcome of the operation.

Given:

Liam buys some fruit for £26.55.

He pays the exact amount using two notes and four coins.

So, possible notes are £1, £5, £10 and £20.

And the possible coins are 1¢, 5¢, 10¢, 25¢, 50¢, and £1.

So, through trial and error,

we get notes of £20 and £5.

And the coins are three coins of 50 cent and 1 coin of 5 cent.

The total amount = adding the amount of notes and coins.

£26.55 =  £20 + £5 + 3(50¢) + 5¢

Therefore, notes of £20 and £5 and the coins are three coins of 50 cent and 1 coin of 5 cent, Liam used.

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The least volume to greatest volume will be sphere B, sphere C, sphere A and sphere D.

The volume of the sphere is calculated as:

= 4/3 πr³

where r = radius

The volume of sphere A will be:

= 4/3 πr³

= 4/3 × π × 4³

= 85.3π

The volume for sphere B will be:

= 4/3 πr³

= 4/3 × π × (6/2)³

= 4/3 × π × 3³

= 36π

The volume for C is given as 64π.

The volume for D has a radius double that of B. This will be Radius = 3 × 2 = 6

= 4/3 πr³

= 4/3 × π × 6³

= 288π

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Answer:

9

Step-by-step explanation:

15-2*3

15-6

9

Jesus loves u hope it helps

Step-by-step explanation:

that means "calculate the result with these values".

the expression is

15 - ab

or

15 - a×b

for a = 2, b = 3 this gives us

15 - 2×3 = 15 - 6 = 9

Answer:

X is 9

Step-by-step explanation:

Answer:

On the smaller figure the measures of it are 6 and 3. On the large figure one of the measures is 18. The number 6 and 18 are congruent, and 6 has be tripled in size, so in order to finding the missing measurement, you will simply triple the 3:

3 times 3 = 9

The missing measure is 9.

Step-by-step explanation:

Hope it helps! =D

Answer:

3

Step-by-step explanation:

1. Subtract 2 y-coordinates.

11 - 8 = 3

2. Subtract the corresponding x-coordinates in the same order.

-1 - (-2) = -1 + 2 = 1

3. slope = (difference in y)/(difference in x) = 3/1 = 3

[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \:x = 0\:\; or \:\: x = -7 [/tex]

____________________________________

[tex] \large \tt Solution \: : [/tex]

[tex]\qquad\displaystyle \tt \rightarrow \: {x}^{2} + 7x = 0 [/tex]

[tex]\qquad\displaystyle \tt \rightarrow \: {x}(x + 7) = 0 [/tex]

There are two cases now,

Case 1 :

[tex]\qquad\displaystyle \tt \rightarrow \: x = 0[/tex]

Case 2 :

[tex]\qquad\displaystyle \tt \rightarrow \: x + 7 = 0[/tex]

[tex]\qquad\displaystyle \tt \rightarrow \: x = - 7[/tex]

So, the possible values of x are :

[tex]\qquad\displaystyle \tt \rightarrow \: 0 \: \: and \: \: - 7[/tex]

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

Answer:

[tex]x=0, \quad x=-7[/tex]

Step-by-step explanation:

Given equation:

[tex]x^2+7x=0[/tex]

Factor x from the left side of the equation:

[tex]\implies x(x+7)=0[/tex]

Using the Zero Product Property, set each factor equal to zero and solve for x:

[tex]\implies x=0[/tex]

[tex]\implies x+7=0 \implies x=-7[/tex]

Therefore, the solutions of the equation are:

[tex]x=0, \quad x=-7[/tex]

Zero Product Property: If a⋅b = 0 then either a = 0 or b = 0 (or both).

Using point-slope form, the equation is [tex]y+7=-\frac{4}{3}(x-3)[/tex].

Answer:

y = - [tex]\frac{4}{3}[/tex] x - 3

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

here m = - [tex]\frac{4}{3}[/tex] , then

y = - [tex]\frac{4}{3}[/tex] x + c ← is the partial equation

to find c substitute (3, - 7 ) into the partial equation

- 7 = - 4 + c ⇒ c = - 7 + 4 = - 3

y = - [tex]\frac{4}{3}[/tex] x - 3 ← equation of line

The data in the question will be represented by a geometric sequence given by 3 (2) ⁿ⁻¹.

A geometric sequence is defined as the sequence in which the ratio between the consecutive terms of the sequence remains constant. The ratio is known as common ratio. The nth term of geometric sequence is written as a (r)ⁿ⁻¹ where a is the first term, r is the common ratio and n is the number of terms. We know from the question that the time in minutes Aurora takes to complete the task forms a sequence as 3, 6, 12, 24.....

The common ratio of this sequence is given by 6/3 = 2. The first term being equal to 3. Now, for the nth term the sequence will be represented as

a (r)ⁿ⁻¹

=> 3 (2)ⁿ⁻¹

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Answer:

Isosceles

Isosceles triangles have 2 side the same length and one side different

multiplication tables from 2 to 20

table of 2

2 x 1 = 2      

2 x 2 = 4

2 x 3 = 6

2 x 4 = 8

2 x 5 = 10

2 x 6 = 12

2 x 7 = 14

2 x 8 = 16

2 x 9 = 18

2 x 10 = 20

table of 3

3 x 1 = 3

3 x 2 = 6

3 x 3 = 9

3 x 4 = 12

3 x 5 = 15

3 x 6 = 18

3 x 7 = 21

3 x 8 = 24

3 x 9 = 27

3 x 10 = 30

table of 4

4 X 1 = 4

4 X 2 = 8

4 X 3 = 12

4 X 4 = 16

4 X 5 = 20

4 X 6 = 24

4 X 7 = 28

4 X 8 = 32

4 X 9 = 36

4 X 10 = 40

table of 5

5 x 1 = 5

5 x 2 = 10

5 x 3 = 15

5 x 4 = 20

5 x 5 = 25

5 x 6 = 30

5 x 7 = 35

5 x 8 = 40

5 x 9 = 45

5 x 10 = 50

table of 6

6 x 1 = 6

6 x 2 = 12

6 x 3 = 18

6 x 4 = 24

6 x 5 = 30

6 x 6 = 36

6 x 7 = 42

6 x 8 = 48

6 x 9 = 54

6 x 10 = 60

table of 7

7 x 1 = 7

7 x 2 = 14

7 x 3 = 21

7 x 4 = 28

7 x 5 = 35

7 x 6 = 42

7 x 7 = 49

7 x 8 = 56

7 x 9 = 63

7 x 10 = 70

table of 8

8 x 1 = 8

8 x 2 = 16

8 x 3 = 24

8 x 4 = 32

8 x 5 = 40

8 x 6 = 48

8 x 7 = 56

8 x 8 = 64

8 x 9 = 72

8 x 10 = 80

table of 9

9 x 1 = 9

9 x 2 = 18

9 x 3 = 27

9 x 4 = 36

9 x 5 = 45

9 x 6 = 54

9 x 7 = 63

9 x 8 = 72

9 x 9 = 81

9 x 10 = 90

table of 10

10 x 1 = 10

10 x 2 = 20

10 x 3 = 30

10 x 4 = 40

10 x 5 = 50

10 x 6 = 60

10 x 7 = 70

10 x 8 = 80

10 x 9 = 90

10 x 10 = 100

table of 11

11 x 1 = 11

11 x 2 = 22

11 x 3 = 33

11 x 4 = 44

11 x 5 = 55

11 x 6 = 66

11 x 7 = 77

11 x 8 = 88

11 x 9 = 99

11 x 10 = 110

table of 12

12 x 1 = 12

12 x 2 = 24

12 x 3 = 36

12 x 4 = 48

12 x 5 = 60

12 x 6 = 72

12 x 7 = 84

12 x 8 = 96

12 x 9 = 108

12 x 10 = 120

table of 13

13 x 1 = 13

13 x 2 = 26

13 x 3 = 39

13 x 4 = 52

13 x 5 = 65

13 x 6 = 78

13 x 7 = 91

13 x 8 = 104

13 x 9 = 117

13 x 10 = 130

table of 14

14 x 1 = 14

14 x 2 = 28

14 x 3 = 42

14 x 4 = 56

14 x 5 = 70

14 x 6 = 84

14 x 7 = 98

14 x 8 = 112

14 x 9 = 126

14 x 10 = 140

table of 15

15 x 1 = 15

15 x 2 = 30

15 x 3 = 45

15 x 4 = 60

15 x 5 = 75

15 x 6 = 90

15 x 7 = 105

15 x 8 = 120

15 x 9 = 135

15 x 10 = 150

table of 16

16    X 1 = 16

16 X 2 = 32

16 X 3 = 48

16 X 4 = 64

16 X 5 = 80

16 X 6 = 96

16 X 7 = 112

16 X 8 = 128

16 X 9 = 144

16 X 10 = 160

table of 17

17 x 1 = 17

17 x 2 = 34

17 x 3 = 51

17 x 4 = 68

17 x 5 = 85

17 x 6 = 102

17 x 7 = 119

17 x 8 = 136

17 x 9 = 153

17 x 10 = 170

table of 18

18 x 1 = 18

18 x 2 = 36

18 x 3 = 54

18 x 4 = 72

18 x 5 = 90

18 x 6 = 108

18 x 7 = 126

18 x 8 = 144

18 x 9 = 162

18 x 10 = 180

table of 19

19 x 1 = 19

19 x 2 = 38

19 x 3 = 57

19 x 4 = 76

19 x 5 = 95

19 x 6 = 114

19 x 7 = 133

19 x 8 = 152

19 x 9 = 171

19 x 10 = 190

table of 20

20 * 1=20

20*2=40

20*3=60

20*4=80

20*5=100

20*6=120

20*7=140

20*8=160

20*9=180

20*10=200

what is the use of multiplication chart?

A grid including all the times tables is plainly displayed on a multiplication chart. It's a useful tool for remembering multiplication facts. Make a chart of the math multiplication tables and use it for calculations or daily reading. This will aid in your memorization.

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Answer:

The answer is C.