What is the distance between the y-intercept of the function f(x)=2(3)^x and the y-intercept of the function g(x) represented by the table below?

Answer:

x = 80

Step-by-step explanation:

The diagram shows that arc AB is 80 degrees, and the measure of an arc defined as being equal to the central angle(the minor angle of the arc). Because the arc AB is 80, the central angle x is 80 degrees as well. Although if point O wasn't at the center of the circle, x might not be 80.

Answer:

Tank Volume/1 cubic centimeter.

Step-by-step explanation:

We dont know how big the tank is.

Answer:

Step-by-step explanation:

[tex]3^2=3\cdot3=9\\\\2^3=2\cdot2\cdot2=8\\\\3^2+2^3=9+8=17[/tex]

Answer:

17

Step-by-step explanation:

Solving

Answer:

$5

Step-by-step explanation:

Conditions

In the currency bills are printed as : $1, $5, $10, $20, $50, $100

Hence, the largest amount he could give the woman is $5

Answer:

37

Step-by-step explanation:

90 - 53 = 37

Answer:

x = 37°

Step-by-step explanation:

Given angle AOB = 53°, angle AOC = 90°

To find x°

From above figure, angle AOC = angle AOB + angle BOC

Now angle AOB = 53°, angle AOC = 90°

90° = 53° + angle BOC

=> angle BOC = x° = 90 - 53

=> x° = 37°

HOPE THIS HELPS AND HAVE A NICE DAY <3

The volume of the actual ramp including the underside is 16 cubic ft.

The height of the actual ramp = 15 feet  

The height of the model = 6 cm

A right triangular prism is a 3-dimensional triangle with two congruent triangular faces and 3 rectangular faces that connect the triangular.

The volume of a right triangular prism = ( height× Base Area)

Scale factor = 1.2 feet / 6 cm = 0.2 ft/cm

Actual base = 16 cm × 0.2 ft/cm = 3.2 ft

The volume of a right triangular prism = ( height× Base Area)

= [tex]5 \times 3.2\\ 16.0[/tex]

Hence, the volume of the actual ramp including the underside is 16 cubic ft.

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Answer:

B. -6 and D. 5

Step-by-step explanation:

x^2 - 16 > 0

y = (x - 4)(x + 4)  > 0

The critical point are x = -4 and x = 4

When x  < -4,   y > 0

When x > 4, y > 0

When -4 < x < 4, y < 0

When x = -4 and x = 4 , y = 0.

So the values that apply are

B. -6 and D. 5

Answer:

5 pi / 9   &    23 pi / 9

Step-by-step explanation:

adding or subtracting multiples of   2pi are coterminal

   2pi =   18pi/9

5 pi / 9   +  18pi/9 = 23 pi/9

If the radius of the circle is 6 feet. Then the area of the circle will be 113.097 square feet.

It is the centre of an equidistant point drawn from the centre. The radius of a circle is the distance between the centre and the circumference.

The radius of the circle is 6 feet.

Then the area of the circle will be

Area = πr²

where r is the radius of the circle. Then we have

Area = π × 6²

Area = 36π

Area = 113.097 square feet

More about the circle link is given below.

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Answer:

what r u supposed to do with that perimeter or area pls state question well

Answer:

The formula for exponential growth is: Total = Starting value(1 + rate)^time

2018 to 2025 is 7 years:

Step-by-step explanation:

Total = 50,000(1+0.037)^7

Total = 50,000(1.037)^7

Total = 64,479 -(Answer)

Answer:

None of the answer choices.

Step-by-step explanation:

The pattern to this problem is multiply by 4, and subtract 3. However when doing 17 * 4, you get 68. Look how both of the answer chouces start with 68, and we haven't even subtracted 3 yet. This shows that the answer choices are wrong.

Correct Answer: 65, 257

The answer: ———————> 68, 257

Answer:

B: π/6

Step-by-step explanation:

A: 0

B: π / 6

C: π / 4

D: π / 3

E: π / 2

F: π

G: 3π / 2

Here is a complete unit circle...

Answer: 8

Step-by-step explanation: divion

Answer:

72 degrees

Step-by-step explanation:

180-108=72

Hope this helps!

Answer:

Step-by-step explanation:

m<2 + 108 = 180

m<2 = 72

Answer:

C. Segment AB is congruent to segment A'B'.

Step-by-step explanation:

Since ABC is being dilated by 1.5, it means that A'B'C will be bigger than ABC.

So, since A'B'C will be bigger than ABC, it sides  will also be bigger than the sides in ABC.

And since the sides/segments are bigger than the original, they are not congruent (congruent means equal).

Answer:

so on the bottom you will see an x and on the left side you will see an y, for 1,4 the 1 will be on the x and the 4, just go up to the four for 2,8 on the x you will go on two on the bottom, for y you will go up to the 8, for 4,16 the x is four and you will go up to 16, for 5,20 the x will be 5 and then go up to 20. I really hope that this helped

Step-by-step explanation:

The second box is less expensive to make since it has a smaller surface area and hence requires less wood to construct the sides.

It is defined as the six-faced shape, a type of hexahedron in geometry.

It is a three-dimensional shape.

Here some data is not given, so we are assuming the volume of the box is 24 cubic feet.

If the volume is 24 cubic feet, the box's three dimensions must multiply to equal 24.

We can make two designs:

2 × 2 × 6 and 2 × 3 × 4.

Now, finding the surface area for each design.

First box

SA = 2wl + 2wh + 2hl

SA = 2×2×2 + 2×2×6 + 2×2×6

SA = 56 square feet

Second box

SA = 2×2×3 + 2×2×4 + 2×3×4

SA = 52 square feet

At $4 per square foot

For 56 ft², the price is $224

For 52 ft², the price is $208

Thus, the second box is less expensive to make since it has a smaller surface area and hence requires less wood to construct the sides.

Learn more about the cuboid here:

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Problem 9

The instructions aren't stated anywhere, but I'm assuming your teacher wants you to find the equation of each parabola.

The vertex here is (h,k) = (-6,-4) which you have correctly determined.

This means

[tex]y = a(x-h)^2 + k\\\\y = a(x-(-6))^2 +(-4)\\\\y = a(x+6)^2 - 4\\\\[/tex]

Next we plug in one of the other points on the parabola. We cannot pick the vertex again. Let's pick the point (-4,0) which is one of the x intercepts. We'll do this to solve for 'a'

[tex]y = a(x+6)^2 - 4\\\\0 = a(-4+6)^2 - 4\\\\0 = a(2)^2 - 4\\\\0 = a(4) - 4\\\\0 = 4a-4\\\\4a-4 = 0\\\\4a = 4\\\\a = 4/4\\\\a = 1\\\\[/tex]

This means

[tex]y = a(x+6)^2 - 4\\\\y = 1(x+6)^2 - 4\\\\y = (x+6)^2 - 4\\\\[/tex]

represents the equation of the parabola in vertex form.

========================================================

Problem 10

The vertex is (h,k) = (-2,-6)

So,

[tex]y = a(x-h)^2 + k\\\\y = a(x-(-2))^2 + (-6)\\\\y = a(x+2)^2 - 6\\\\[/tex]

Now plug in another point on the parabola like (-1,-8) and solve for 'a'

[tex]y = a(x+2)^2 - 6\\\\-8 = a(-1+2)^2 - 6\\\\-8 = a(1)^2 - 6\\\\-8 = a(1) - 6\\\\a-6 = -8\\\\a = -8+6\\\\a = -2\\\\[/tex]

For each equation, you could optionally expand things out to get it into y = ax^2+bx+c form, but I think it's fine to leave it as vertex form.

Answer:

[tex]\textsf{9)} \quad y=(x+6)^2-4[/tex]

[tex]\textsf{10)} \quad y=-2(x+2)^2-6[/tex]

Step-by-step explanation:

Vertex form of a parabola:  

[tex]y=a(x-h)^2+k[/tex]  where (h, k) is the vertex

From inspection of the graph, the vertex is (-6, -4)

[tex]\implies y=a(x+6)^2-4[/tex]

To find [tex]a[/tex], substitute the coordinates of a point on the curve into the equation.

Using point (-4, 0):

[tex]\implies a(-4+6)^2-4=0[/tex]

[tex]\implies a(2)^2-4=0[/tex]

[tex]\implies 4a=4[/tex]

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

Therefore, the equation of the parabola in vertex form is:

[tex]y=(x+6)^2-4[/tex]

From inspection of the graph, the vertex is (-2, -6)

[tex]\implies y=a(x+2)^2-6[/tex]

To find [tex]a[/tex], substitute the coordinates of a point on the curve into the equation.

Using point (-1, -8):

[tex]\implies a(-1+2)^2-6=-8[/tex]

[tex]\implies a(1)^2-6=-8[/tex]

[tex]\implies a-6=-8[/tex]

[tex]\implies a=-2[/tex]

Therefore, the equation of the parabola in vertex form is:

[tex]\implies y=-2(x+2)^2-6[/tex]

Answer:

  (a)  3 ft, 4 ft, 5 ft

Step-by-step explanation:

The triangle inequality requires the sum of the two short sides exceed the long side.

__

a) 3 ft + 4 ft = 7 ft > 5 ft . . . . . triangle is possible

b) 4 yd + 1 2/3 yd = 5 2/3 yd < 10 yd . . . . not a possible triangle

c) 3 ft + 3 ft = 6 ft < 7 ft . . . . not a possible triangle

d) 4 in + 4 in = 8 in . . . . not a possible triangle (not greater than 8 in)

__

The first set of side lengths can form a triangle: 3 ft, 4 ft, 5 ft.

Answer:40

Step-by-step explanation: i may be wrong but i subtracted 136 from 95 and i got 40 im sorry if it is wrong

Answer:

11 + 16n

Step-by-step explanation:

Collect the like terms

2 + 9 = 11
10n + 6n = 16n

11 + 16n is the final answer, and cannot be simplified further

Answer:

16n + 11

Step-by-step explanation:

Simpifying

Combine like terms (n and number terms)

Answer:

Ted is wrong

The triangle is a right triangle because of the right 90° angle formed at the bottom.

Answer:

0.7453

Step-by-step explanation:

[tex]arcsin( \frac{4}{6} ) = 0.7297 \\ cos(0.7297) = 0.7453[/tex]

Answer:

5

Step-by-step explanation:

a = 3

ab = 6

so b must be 2

3 squared is 9 and 2 squared is 4

so 9 - 4 is equal to 5

Answer:

Step-by-step explanation:

If a =3 and ab =6 then a²-b² =

a = 3

ab = 6

find b

b = 6:3 = 2

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

3² - 2² =

9 - 4 =

5

[tex]25 {x}^{2} - 49 {y}^{2} [/tex]

[tex] {a}^{2} - {b}^{2} = (a + b)(a - b)[/tex]

[tex]a = 5x[/tex]

[tex]b = 7y[/tex]

[tex]25 {x}^{2} - 49 {y}^{2} [/tex]

[tex] = (5x - 7y) {}^{2} [/tex]

[tex] = (5x - 7y)(5x + 7y)[/tex]

hence, the correct answer is option B

Answer:

105cm

Step-by-step explanation:

please click crown below my answer.

since we have the area of the front side, to get its volume we can simple get the product of the area and the length, let's firstly change the mixed fractions to improper fractions.

[tex]\stackrel{mixed}{23\frac{2}{3}}\implies \cfrac{23\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{71}{3}} ~\hfill \stackrel{mixed}{4\frac{7}{8}}\implies \cfrac{4\cdot 8+7}{8}\implies \stackrel{improper}{\cfrac{39}{8}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{71}{3}\cdot \cfrac{39}{8}\implies \cfrac{71}{8}\cdot \cfrac{39}{3}\implies \cfrac{71}{8}\cdot 13\implies \cfrac{923}{8}\implies 115\frac{3}{8}~in^3[/tex]

Answer:

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

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{6 cm}\underline{Product\:Rule}\\\\$\dfrac{d}{dx}[f(x)g(x)]=f(x)g'(x)+f'(x)g(x)$\\\\\\\underline{Chain\:Rule}\\\\$\dfrac{dy}{dx}=\dfrac{dy}{du} \times \dfrac{du}{dx}$\\\end{minipage}}[/tex]

[tex]\textsf{Given}:y=\log (x \sin x)[/tex]

[tex]\textsf{Let }u=x \sin x \implies y=\log(u)[/tex]

Using the product rule, differentiate y with respect to [tex]u[/tex]:

[tex]\textsf{If }y=\log(u) \implies \dfrac{dy}{du}=\dfrac{1}{u}\implies \dfrac{dy}{du}=\dfrac{1}{x \sin x}[/tex]

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

Differentiate [tex]u=x \sin x[/tex] with respect to x using the product rule:

[tex]\implies \textsf{Let }f(x)=x \implies f'(x)=1[/tex]

[tex]\implies \textsf{Let }g(x)=\sin x \implies g'(x)=\cos x[/tex]

[tex]\implies \dfrac{du}{dx}=x \cos x + \sin x[/tex]

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

Use the chain rule to differentiate y with respect to x:

[tex]\implies \dfrac{dy}{dx}=\dfrac{dy}{du} \times \dfrac{du}{dx}[/tex]

[tex]\implies \dfrac{dy}{dx}=\dfrac{1}{x \sin x} \times (x \cos x + \sin x)[/tex]

[tex]\implies \dfrac{dy}{dx}=\dfrac{x \cos x + \sin x}{x \sin x}[/tex]

[tex]\implies \dfrac{dy}{dx}=\dfrac{x \cos x}{x \sin x}+\dfrac{\sin x}{x \sin x}[/tex]

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

Answer:

[tex]\displaystyle \large{\dfrac{d}{dx}[\log(x\sin x)] = \dfrac{1}{x\ln 10} + \dfrac{\cot x}{\ln 10}}[/tex]

Step-by-step explanation:

Given:

[tex]\displaystyle \large{y=\log (x\sin x)}[/tex]

Recall the (common) logarithmic differentiation & chain rules:

[tex]\displaystyle \large{\dfrac{d}{dx}[\log (u)] = \dfrac{u'}{u\ln 10}}[/tex]

Let [tex]\displaystyle \large{u=x\sin x}[/tex] then:

[tex]\displaystyle \large{\dfrac{d}{dx}[\log(x\sin x)] = \dfrac{(x\sin x)'}{x\sin x \ln 10}}[/tex]

Recall product rules:

[tex]\displaystyle \large{[f(x)g(x)]' = f'(x)g(x)+f(x)g'(x)}[/tex]

Hence:

[tex]\displaystyle \large{\dfrac{d}{dx}[\log(x\sin x)] = \dfrac{(x)'\sin x + x(\sin x)'}{x\sin x \ln 10}}\\\\\displaystyle \large{\dfrac{d}{dx}[\log(x\sin x)] = \dfrac{\sin x + x\cos x}{x\sin x \ln 10}}\\\\\displaystyle \large{\dfrac{d}{dx}[\log(x\sin x)] = \dfrac{\sin x}{x \sin x \ln 10} + \dfrac{x \cos x}{x\sin x \ln 10}}\\\\\displaystyle \large{\dfrac{d}{dx}[\log(x\sin x)] = \dfrac{1}{x\ln 10} + \dfrac{\cot x}{\ln 10}}[/tex]

From above, recall that:

[tex]\displaystyle \large{\dfrac{d}{dx}(\sin x) = \cos x}\\\\\displaystyle \large{\dfrac{A\pm B}{C} = \dfrac{A}{C} \pm \dfrac{B}{C}}\\\\\displaystyle \large{\dfrac{\cos x}{\sin x} =\cot x}[/tex]

Hence, the answer is:

[tex]\displaystyle \large{\dfrac{d}{dx}[\log(x\sin x)] = \dfrac{1}{x\ln 10} + \dfrac{\cot x}{\ln 10}}[/tex]