in exponential growth functions. the base of the exponent must be greater than 1? how would the function change if the base of the exponent were between 0 and 1?

The range of the given function will lie from ( -0.333,0) to ( 0,1).

A function's range is the collection of values it can take as input. After we enter an x value, the function outputs this sequence of values.

Given that:-

When we plot the graph of the function we will see that the straight line will cut the x-axis at point (-0.333 ) and the y-axis cut at ( 0,1 ) the points between the two points will be the solutions of the equation.

Therefore range of the given function will lie from ( -0.333,0) to ( 0,1).

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Answer:i think the answer is B

Step-by-step explanation:

Answer:

3:4

Step-by-step explanation:

We know this because the GCD of 21 and 28 is 7

so we divide both numbers by 7 and get 3 and 4

so the answer is 3:4

Answer:

(-1,3)

y=5x+8 , y=-4x-1y=5x+8,y=−4x−1

Both sides have y on the left hand side so make the right hand side equal and solve for x

5x+8= -4x-15x+8=−4x−1

To solve for x , add 4x on both sides

9x+8=-19x+8=−1

Subtract 8 on both sides

9x=-99x=−9

Divide by 9 on both sides

x=-1x=−1

Now plug in -1 for x in the first equation and find out y

y=5x+8y=5x+8

y=5(-1)+8y=5(−1)+8

y=-5+8y=−5+8

y=3y=3

Ordered pair is (x,y) that is (-1,3)

Step-by-step explanation:

Mark me as brainliest plsss

All the transformations for which A and B would be congruent are; Translation, Reflection, and Rotation.

A congruence transformation is defined as a movement of shape such that it can produce a shape that is congruent to the original.

There are three types of congruence transformation;

1. Translation

2. Reflection

3. Rotation

Given;

Shape B is made by applying a single transformation to shape A.

Hence, All the transformations for which A and B would be congruent are; Translation, Reflection, and Rotation.

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The equation given is matched with the respective vertex forms.

The standard equation for a circle is given by

(x-h)² + (y-k)² = r²

The equation given in the question are

x² + y²  − 4x + 12y − 20 = 0

(x − 6)²  + (y − 4)²  = 56

x²  + y²  + 6x − 8y − 10 = 0

(x − 2)²  + (y + 6)²  = 60

3x²  + 3y²  + 12x + 18y − 15 = 0

(x + 2)²  + (y + 3)²  = 18

5x²  + 5y²  − 10x + 20y − 30 = 0

(x+3)²  + (y-4)²  = 35

2x²  + 2y²  − 24x − 16y − 8 = 0

(x-1)²  + (y+2)²   = 11

The first equation is

x² + y²  − 4x + 12y − 20 = 0

It can be written ad

x² -4x + 4 - 4 +y²+12y +36-36 -20 = 0

(x-2)² +(y+6)²= 60

The second equation is

x²  + y²  + 6x − 8y − 10 = 0

x² +6x +9-9 +y²  -8y +16-16 -10 = 0

(x+3)²  + (y-4)²  = 35

The third equation is

3x²  + 3y²  + 12x + 18y − 15 = 0

(x + 2)²  + (y + 3)²  = 18

The fourth equation is

5x²  + 5y²  − 10x + 20y − 30 = 0

(x-1)²  + (y+2)²   = 11

The fifth equation is

2x²  + 2y²  − 24x − 16y − 8 = 0

(x − 6)²  + (y − 4)²  = 56

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Answer:

1073

Step-by-step explanation:

it's simple addition

Answer: A

Step-by-step explanation:

[tex]\frac{dy}{dx}=\frac{1+x}{xy} \\ \\ y \text{ } dy=\frac{1+x}{x} \text{ } dx \\ \\ y \text{ } dy= \left(\frac{1}{x}+1 \right) \text{ } dx \\ \\ \int y \text{ } dy=\int \left(\frac{1}{x}+1 \right) \text{ } dx \\ \\ \frac{y^{2}}{2}=\ln \left| x \right|+x+C[/tex]

Substituting in the initial condition,

[tex]\frac{(-2)^{2}}{2}=\ln \left|1 \right|+1+C\\\\2=1+C\\\\C=1[/tex]

So,

[tex]\boxed{\frac{1}{2}y^{2}=\ln \left|x \right|+x+1}[/tex]

Answer:

[tex]\textsf{A.} \quad \dfrac{1}{2}y^2=\ln|x|+x+1[/tex]

Step-by-step explanation:

To solve the differential equation:

Rearrange the given equation to get all the terms containing y on the left side, and all the terms containing x on the right side:

         [tex]\dfrac{dy}{dx}=\dfrac{1+x}{xy}[/tex]

[tex]\implies y\dfrac{dy}{dx}=\dfrac{1+x}{x}[/tex]

[tex]\implies y\:dy=\dfrac{1+x}{x}\:dx[/tex]

[tex]\implies y\:dy=\dfrac{1}{x}+\dfrac{x}{x}\:dx[/tex]

[tex]\implies y\:dy=\left(\dfrac{1}{x}+1\right)\:dx[/tex]

Integrate both sides:

[tex]\displaystyle \implies \int y\:dy=\int \left(\dfrac{1}{x}+1\right)\:dx[/tex]

[tex]\implies \dfrac{1}{2}y^2=\ln|x|+x+C[/tex]

Find the value of C using the given values of x and y:

when x = 1, y = -2

[tex]\implies \dfrac{1}{2}(-2)^2=\ln|1|+1+C[/tex]

[tex]\implies 2=0+1+C[/tex]

[tex]\implies C=1[/tex]

Substitute the found value of C:

[tex]\implies \dfrac{1}{2}y^2=\ln|x|+x+1[/tex]

[tex]2^x-3^x=\sqrt{6^x-9^x}\\\\D:6^x-9^x\geq0 \wedge 2^x-3^x\geq0\\D: 6^x\geq 9^x \wedge 2^x\geq3^x\\D:x\leq0 \wedge x\leq 0\qquad(\text{if }a < b \text{ then } a^x < b^x\text{ for } x > 0 \text{ and } a^x > b^x \text{ for } x < 0 )\\D:x\leq 0\\\\(2^x-3^x)^2=6^x-9^x\\(2^x-3^x)^2=3^x(2^x-3^x)\\\\\text{We divide both sides by } 2^x-3^x \text{ assuming }2^x-3^x\not=0\Leftrightarrow x\not=0.\\\\2^x-3^x=3^x\\2^x=2\cdot3^x\\\ln 2^x=\ln (2\cdot3^x )\\x\ln 2=\ln 2+\ln 3^x\\x\ln 2=\ln 2+x\ln 3[/tex]

[tex]x\ln 2-x\ln 3=\ln 2\\x(\ln 2-\ln 3)=\ln 2\\x=\dfrac{\ln 2}{\ln 2-\ln 3}=-\dfrac{\ln 2}{\ln 3-\ln 2}=-\dfrac{\ln 2}{\ln \frac{3}{2}}\approx-1.7[/tex]

For [tex]x=0[/tex], our initial equation is obviously true.

Therefore [tex]x\in\left\{-\dfrac{\ln 2}{\ln \frac{3}{2}},0\right\}[/tex].

Answer to the miscellaneous equation, is x=0 and x=-1.71

Miscellaneous equation are the equations which are not polynomial

The question can be solved by:

Factorising Splitting the terms to find the required solution

Completing the squares, etc

2ˣ - 3ˣ = √(6ˣ - 9ˣ)

Squaring both sides,

2²ˣ+3²ˣ -2.2ˣ.3ˣ=2ˣ.3ˣ - 3²ˣ

2²ˣ+2.3²ˣ -3.2ˣ.3ˣ=0

Factorising the terms,

2ˣ(2ˣ-3ˣ) -2.3ˣ(2ˣ-3ˣ)=0

(2ˣ-3ˣ)(2ˣ-2.3ˣ)=0

Equating the braces to zero,

Either,

2ˣ=3ˣ

x=0

or

2ˣ=2.3ˣ

Taking log on both sides

xln2=ln2+xln3

x=ln2/(ln2-ln3)

  =-1.71

Therefore, x=0 and x=-1.71 is the only solution

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See the graph in the image.

Answer:

17.50 × 0.35 = 6.13

Step-by-step explanation:

Answer:

That would be $11.37

Step-by-step explanation:

First Multiply the discount and the original price

17.50(0.35)=  6.125 = 6.13

Use that answer and subtract it from the original price

17.50 - 6.13 = 11.37

Using the unit rate, the missing values that completes the table are:

A = 5; B = 15; C = 40

Unit rate (m) = change in y/change in x.

5 minutes for 8 Blocks (5, 8) and 10 minutes for 16 blocks (10, 16)are given.

Unit rate (m) = (16 - 8)/(10 - 5) = 8/5

An equation that will define the function is, y = 8/5x. Use it to complete the table.

Find A (y) when x is 5:

y = 8/5(5) = 8

The value of A is: 5

Find B (x) when y is 24:

24 = 8/5(x)

5(24) = 8x

120 = 8x

120/8 = x

15 = x

The value of B is: 15

Find A (y) when x is 25:

y = 8/5(25) = 40

The value of C is: 40

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Assessing the given Venn diagram and the data related to it, we can answer the required question:

Q1. How many people were surveyed = 86.

Q2. What is the probability that a customer chosen at random:

a) takes salt P(S) = 46/85.

b) takes both salt and vinegar P(S ∩ V) = 13/85.

c) takes salt or vinegar or both P(S ∪ V) = 66/85.

The number of people who take salt chips n(S) = 46.

The number of people who take vinegar chips n(V) = 33.

The number of people who takes both n(S ∩ V) = 13.

The number of people who take neither n((S ∪ V)') = 19.

Therefore, the total number of people surveyed n(U) = n(S) + n(V) - n(S ∩ V) + n((S ∪ V)') = 46 + 33 - 13 + 19 = 85.

The number of people who takes salt or vinegar or both n(S ∪ V) = n(U) - n((S ∪ V)') = 85 - 19 = 66.

The probability that a customer chosen at random takes salt P(S) = n(S)/n(U) = 46/85.

The probability that a customer chosen at random takes both salt and vinegar P(S ∩ V) = n(S ∩ V)/n(U) = 13/85.

The probability that a customer chosen at random takes salt or vinegar or both P(S ∪ V) = n(S ∪ V)/n(U) = 66/85.

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The correct answer is option B which is ( 5 / 3)

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication, and division

The value will be calculated as:-

[tex]5\times \dfrac{1}{3}[/tex]  =  [tex]\dfrac{5}{3}[/tex]

1.67  = 1.67

Therefore the correct answer is option B which is ( 5 / 3)

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Inequalities help us to compare two unequal expressions. The correct option is B.

Inequalities help us to compare two unequal expressions. Also, it helps us to compare the non-equal expressions so that an equation can be formed.

It is mostly denoted by the symbol <, >, ≤, and ≥.

The solution to two systems of inequalities in the region where the area of the two inequalities overlaps. As it can be observed in this case that the overlap area is B.

Hence, the correct option is B.

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The equation is y = 250(0.89)^x and the decay rate is 11%.

The initial value will be:

= 250(0.89)^x

= 250(0.89^0)

= 250

The decay rate will be:

= 1 - 0.89

= 0.11

= 11%

In conclusion, the equation is y = 250(0.89)^x and the decay rate is 11%.

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Answer:

[tex]\fbox {See below} \downarrow[/tex]

Step-by-step explanation:

Question 1 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(4x^{3})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 4(3)x^{3-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 12x^{2}}[/tex]

Question 2 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(\frac{2}{x^{4}})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(2x^{-4})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 2(-4)x^{-4-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = -8x^{-5}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = -\frac{8}{x^{5}}}[/tex]

Question 3 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(3\sqrt{x})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(3(x)^{\frac{1}{2}})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 3(\frac{1}{2})x^{\frac{1}{2}-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{3}{2}x^{-\frac{1}{2}}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{3}{2\sqrt{x}}}[/tex]

Question 4 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(\frac{3}{\sqrt{x}})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(3x^{-\frac{1}{2}})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 3(-\frac{1}{2})x^{-\frac{1}{2}-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = -\frac{3}{2}x^{-\frac{3}{2}}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = -\frac{3}{2x\sqrt{x}}}[/tex]

Question 5 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(\frac{x^{3}}{4})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{1}{4}(3)x^{3-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{3}{4}x^{2}}[/tex]

Question 6 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(3x^{5})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 3(5)x^{5-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 15x^{4}}[/tex]

Answer:

8.2 minutes

Step-by-step explanation:

[tex] \ \frac{492}{60} = 8.2[/tex]

492 seconds = 8.2 minutes

Step-by-step explanation:

0.000672, 6.72×10⁵, 67.2×10‐⁴, 672×10⁴

Scientific notation is a way of writing numbers that looks like the following:

[tex]a\times10^n[/tex]

For instance, we can write 2930 in scientific notation.

First, write all the given values in scientific notation:

Now, let's compare the values of n, and organize the numbers from least to greatest:

[tex]6.72\times10^{-4}[/tex]

[tex]6.72\times 10^{-3}[/tex]

[tex]6.72\times10^5[/tex]

[tex]6.72\times 10^6[/tex]

Finally, rewrite all the numbers as their given format:

[tex]0.000672[/tex]

[tex]67.2\times 10^{-4}[/tex]

[tex]6.72\times10^5[/tex]

[tex]672\times 10^4[/tex]

[tex]0.000672[/tex]

[tex]67.2\times 10^{-4}[/tex]

[tex]6.72\times10^5[/tex]

[tex]672\times 10^4[/tex]

Answer:

m∠M=98º

Step-by-step explanation

Sum of interior angles;

  3x-16+x+6+x = 180

                   5x = 180-6+16

                   5x = 190

                      x = 38º

Angle M;

      m<M= 3x - 16 = 3(38)-16= 98º

      m∠M=98º

Answer:

<M = 98

Step-by-step explanation:

The sum of the angles of a triangle is 180 degrees

< M + < N + < O = 180

3x-16 + x + x+6 = 180

Combine like terms

5x - 10 = 180

Add 10 to each side

5x-10 + 10 = 180+10

5x = 190

Divide by 5

5x/5 = 190/5

x=38

We want to find <M

<M = 3x-16  = 3*38 -16 = 114-16 =98

Answer:

0.4 bottles can be filled

Step-by-step explanation:

divide the amount that is present by the max amount

the bottle can only hold 5.55 millimeters; therefore the max amount would be 5.55 millimeters.

the amount that is present is 2.22 millimeters.

[tex]\frac{222 millimeters}{555 millimeters} = 0.4[/tex]

0.4 bottles

if the given values are 555 and 222 then the answer is still .4

Answer:

0.4 bottles can be filled

Gracie’s overtime rate of pay would be $635.06.

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

Gracie earns 17.25/h for a 37.5h week, plus extra for overtime.

For one hour she earns $17.25

For 37.5h she earns = 37.5h x 17.25

                                   = $646.87

The total money she earns = $646.87 + overtime pay

During store inventory, she worked 48.5h. Her gross earnings were $1,471.88.

= 48.5h x 17.25

= $836.625

So, $836.625 + overtime pay = $1,471.88

overtime pay = $1,471.88 - $836.625 = $635.06

Hence, Gracie’s overtime rate of pay would be $635.06.

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Answer:

Step-by-step explanation:

Negative numbers don't have real square roots since a square is either positive or 0. so it doesn't exist

Answer:

i

Step-by-step explanation:

the square root of  negative one is not real.

It is classed as an imaginary number denoted by  i , that is

[tex]\sqrt{-1}[/tex] = i

Answer:

14 hours

Step-by-step explanation:

7h - 5(3h-8) = -72

7h - 15h + 40 = -72

-8h +40 = -72

-8h = -72 -40

-8h = -112

h = -112/ -8

h = 14

Answer: 23.476 = 20+3+0.4+0.07+0.006

Step-by-step explanation:

I think you were supposed to write down the choices but I will guess the answer is this one:

23.476 = 20+3+0.4+0.07+0.006

Hope I helped!

Answer: In the first week, Micah bought 11 gallons of gas,

and in the second week, he bought 8 gallons of gas

Step-by-step explanation:

In the first week, Micah bought x gallons of fuel at $2.39/gallon

In the second week, he bought (x - 3) gallons at $2.49/gallon

Formulating an equation and solving for x gallons of fuel:

x × 2.39 + (x - 3) × 2.49 = 46.21

2.39x + x × 2.49 − 3 × 2.49 = 46.21

2.39x + 2.49x − 7.47 = 46.21

4.88x − 7.47 = 46.21

4.88x = 53.68

[tex]\frac{4.88x}{4.88} = \frac{53.68}{4.88}[/tex]

x = 11

Therefore,

In the first week, Micah bought 11 gallons of gas

In the second week, he bought (11 - 3) = 8 gallons of gas

Option D is correct the face value is the price that must be paid to the issuer to buy it at the time it is issued.

If the bond issuer doesn't default, face value, also known as par value, is the sum that is paid to a bondholder at the bond's maturity date. Bonds offered on the secondary market do, however, change in response to interest rates.

Therefore Option D is correct the face value is the price that must be paid to the issuer to buy it at the time it is issued.

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[tex]15 \leq X \leq 25[/tex] is the equation represents the drivers allotted time (x) to receive a bonus.

An absolute value inequality is an expression with absolute functions as well as inequality signs.

According to the question:

Mathematically absolute value inequality is represented as :

|X -20|≤ 5

Solving the equation using

|z| ≤ a→ -a ≤ z  ≤ a :

here, z = X- 20 and a = 5:

[tex]| X-20| \leq 5\\\\ - 5\leq X-20 \leq 5[/tex]

By Solving X adding 20 on the three sides of the inequality

[tex]-5 + 20 \leq X -20 +20 \leq 5 + 20[/tex]

By simplifying it:

[tex]15 \leq X \leq 25[/tex]

Therefore the equation or inequality which represents the drivers allotted time (x) to receive a bonus [tex]15 \leq X \leq 25[/tex]

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