please help!!!!!

5) A box of chocolates contains four milk
chocolates and seven dark chocolates.
You randomly pick a chocolate and eat it.
Then you randomly pick another piece.
The first piece is milk chocolate and the
second piece is dark chocolate.
24
25
A)
≈ 0.264
B)
≈ 0.309
91
81
1-14-4
D) None of these
4
15
C)
E)
= 0.25
≈ 0.267
E

Answer:

80

Step-by-step explanation:

10x7=70

2x5=10

70+10=80

Answer:

7/8 (seven eighths)

Step-by-step explanation:

add the numerators (top numbers) together, to get 7. Leave the denominator (bottom number) alone, since they are the same. Think about it like you're adding two different parts of a whole.

you should get 7/8

hope this helps

Answer:

the independent variable is s

the dependent one is m

hope this helps <3

Answer:

42 ft ^ 2

Step-by-step explanation:

All we have to do is multiply 14 and 3. That gives us 42.

Answer:

3/4 sq ft

Step-by-step explanation:

1/4*3=3/4

im pretty sure u meant 1/4 not 14 bc it says "mixed number"

The true correct statements about the box-and-whiskers plot are:

In a box-and-whiskers plot, the interquartile range (IQR) is a measure of the spread of the data which shows the variability of the data.

Median of a data is the value at the point where the vertical line divides the rectangular box.

Median for Group A is between: 45 and 50

Median time for Group B is: 40.

Therefore, the three statements are:

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The different angle measures are

1. [tex]m\angle EBD = 34^\circ[/tex]

2. [tex]m\angle ACE = 52^{\circ}[/tex]

3. [tex]m \overset{\LARGE\frown}{GB} = 76^\circ[/tex]

4. [tex]m \overset{\LARGE\frown}{GBD} = 256^\circ[/tex]

5. [tex]m \angle DBA= 90^\circ[/tex]

6. [tex]m \overset{\LARGE\frown}{GD} = 104^\circ[/tex]

7. [tex]m\angle DFC= 56^\circ[/tex]

From the question, we are to determine the measures of the different angles

1.

From the given information,

[tex]m \overset{\LARGE\frown}{HD} = 68^\circ[/tex]

This implies that the measure of the central angle is 68°

Then,

[tex]m\angle EBD = \frac{1}{2} \times 68^\circ[/tex] (Angle at the center is twice the angle at the circumference)

∴ [tex]m\angle EBD = 34^\circ[/tex]

2.

From the diagram

[tex]m\angle ACE + m\angle BED + 90^\circ = 180^\circ[/tex]

From the given information

[tex]m\angle BED = 38^\circ[/tex]

Then,

[tex]m\angle ACE + 38^\circ + 90^\circ = 180^\circ[/tex]

[tex]m\angle ACE + 128^{\circ} = 180^\circ[/tex]

[tex]m\angle ACE = 180^\circ-128^{\circ}[/tex]

[tex]m\angle ACE = 52^{\circ}[/tex]

3.

First, we will determine [tex]m \overset{\LARGE\frown}{GD}[/tex]

[tex]m \overset{\LARGE\frown}{GD} = 2 \times m\angle ACE[/tex] (Angle at the center is twice the angle at the circumference)

[tex]m \overset{\LARGE\frown}{GD} = 2 \times 52^\circ[/tex]

[tex]m \overset{\LARGE\frown}{GD} = 104^\circ[/tex]

But,

[tex]m \overset{\LARGE\frown}{GD} + m \overset{\LARGE\frown}{GB} = 180^\circ[/tex]

[tex]m \overset{\LARGE\frown}{GB} = 180^\circ - m \overset{\LARGE\frown}{GD}[/tex]

[tex]m \overset{\LARGE\frown}{GB} = 180^\circ - 104^\circ[/tex]

[tex]m \overset{\LARGE\frown}{GB} = 76^\circ[/tex]

4.

[tex]m \overset{\LARGE\frown}{GBD} = m \overset{\LARGE\frown}{GB} + 180^\circ[/tex]

[tex]m \overset{\LARGE\frown}{GBD} = 76^\circ +180^\circ[/tex]

[tex]m \overset{\LARGE\frown}{GBD} = 256^\circ[/tex]

5.

[tex]m \angle DBA= 90^\circ[/tex] (Tangent and diameter/ radius theorem)

If a tangent and a diameter meet at the point of tangency, then they are perpendicular to one another

6.

[tex]m \overset{\LARGE\frown}{GD} = 104^\circ[/tex] (As determined above)

7.

First, we will determine [tex]m\angle GFB[/tex]

[tex]m\angle GFB + m\angle EBD + 90^\circ = 180^\circ[/tex]

[tex]m\angle GFB = 180^\circ -(m\angle EBD + 90^\circ)[/tex]

[tex]m\angle GFB = 180^\circ - (34^\circ+90^\circ)[/tex]

[tex]m\angle GFB = 180^\circ - 124^\circ[/tex]

[tex]m\angle GFB = 56^\circ[/tex]

[tex]m\angle GFB = m\angle DFC[/tex] (Vertically opposite angles)

[tex]m\angle DFC= 56^\circ[/tex]

Hence, the different angle measures are

1. [tex]m\angle EBD = 34^\circ[/tex]

2. [tex]m\angle ACE = 52^{\circ}[/tex]

3. [tex]m \overset{\LARGE\frown}{GB} = 76^\circ[/tex]

4. [tex]m \overset{\LARGE\frown}{GBD} = 256^\circ[/tex]

5. [tex]m \angle DBA= 90^\circ[/tex]

6. [tex]m \overset{\LARGE\frown}{GD} = 104^\circ[/tex]

7. [tex]m\angle DFC= 56^\circ[/tex]

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The adjusting entry for the Software of Northland Physical Therapy is as follows:

December 31, 2018:

Debit Amortization Expense $3

Credit Accumulated Amortization $3

An adjusting entry for Amortization records the amortization expense for the period, which is debited.  It also credits the Accumulated Amortization, increasing the amount of the amortization expense.

For instance, the Amortization Expense for Northland Physical Therapy's Software is recorded as above.

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Answer:

The answer should be $1127.74

Step-by-step explanation:

ask me if you don't understand

Answer:

B) - 6 + 3i

Step-by-step explanation:

   Complex number notation:

i² = -1

[tex]\sf \sqrt{-9}-6=\sqrt{3^2*i^2}-6[/tex]

               [tex]\sf = 3i - 6[/tex]

Answer:

35.66

Step-by-step explanation:

Don't work with the mean given at first

1. Multiply the mean of class A by the number of students in class A

35 times 15=525

2. Multiply the mean of class B by the number of pupils in class B

36 times 29=1044

3. Add the answers you got

1044+525

4.Add the pupils of both classes

29+15=44

5.Then divide the total grades and pupils

1569/44 =35.659

6 Then round off

=35.66

Answer:

A

Step-by-step explanation:

There is an even number of values.
Hope it helps

Sorry if I’m wrong

Answer:

True, (-∞, ∞)           0 > -12

Step-by-step explanation:

8 > −4

8 - 8 > −4 - 8

0 > -12

Hope it helps :)

If you have any questions or anymore help lmk, happy to help with any school work!!

Have a nice day, you’ll make it thru :)

(shhhh i added your deleted question to this one as well, Image below)

The value of x will be equal to 3 for the given expression.

The fraction is defined as the division of the whole part into an equal number of parts.

Given expression is solved as:-

[tex]\dfrac{\dfrac{3}{4} }{\dfrac{6}{x}} = \dfrac{3}{8}[/tex]

[tex]\dfrac{3}{4}\times \dfrac{x}{6}=\dfrac{3}{8}[/tex]

[tex]\dfrac{x}{8}=\dfrac{3}{8}[/tex]

x = 3

The value of x will be equal to 3 for the given expression.

Therefore the value of x will be equal to 3 for the given expression.

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The value of two arithmetic means which are inserted between 3 and 24 are 24/9 and 75/9.

Arithmetic mean is the mean or average which is equal to the ratio of sum of all the group numbers to the total numbers.  

The two arithmetic means between 3 and 24 are has to be inserted.

3, A₂, A₃, 24

All the four numbers are in arithmetic progression. The nth terms of AM can be found using the following formula:

t(n)=a(n-1)d

Here, d is the common difference a is the first terms and n is the total term.  The first term, a=3 and t₄=24. Thus, the common difference is;

[tex]t(n)=a(n-1)d\\t(4)=3(4-1)d\\24=3(3)d\\24=9d\\d=\dfrac{24}{9}[/tex]

The second and 3rd term are:

[tex]A₂=3+\dfrac{24}{9}=\dfrac{51}{9}\\ A₃=\dfrac{51}{9}+\dfrac{24}{9}=\dfrac{75}{9}[/tex]

Thus, the value of two arithmetic means which are inserted between 3 and 24 are 24/9 and 75/9.

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Answer:

[tex]\displaystyle \int {x^{-11}(1 + x^4)^\Big{- \frac{1}{2}}} \, dx = \boxed{ - \frac{\sqrt{x^4 + 1} (8x^8 - 4x^4 + 3)}{30x^{10}} + C }[/tex]

General Formulas and Concepts:

Calculus

Differentiation

Derivative Rule [Basic Power Rule]:

Integration

Integration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

Integration Methods: U-Substitution, U-Solve, Trigonometric Substitution

Step-by-step explanation:

*Note:

The problem is too big to fit all work. I will assume that you know how to do basic calculus.

Step 1: Define

Identify given.

[tex]\displaystyle \int {x^{-11}(1 + x^4)^\Big{- \frac{1}{2}}} \, dx[/tex]

Step 2: Integrate Pt. 1

Step 3: Integrate Pt. 2

Identify variables for u-substitution/u-solve.

Step 4: Integrate Pt. 3

Start solving the integral using u-solve:

[tex]\displaystyle \begin{aligned}\int {x^{-11}(1 + x^4)^\Big{- \frac{1}{2}}} \, dx & = \int {\frac{1}{x^{11}\sqrt{x^4 + 1}}} \, dx \\& = \int {\frac{1}{2u^6 \sqrt{u^2 + 1}}} \, du \\& = \frac{1}{2} \int {\frac{1}{u^6 \sqrt{u^2 + 1}}} \, du \\\end{aligned}[/tex]

Step 5: Integrate Pt. 4

Identify variables for trigonometric substitution.

Step 6: Integrate Pt. 5

Let's focus on just the integral itself. Apply the Trigonometric Substitution Integration Method and other basic integration techniques listed under "Calculus":

[tex]\displaystyle \begin{aligned}\int {\frac{1}{u^6 \sqrt{u^2 + 1} }} \, du & = \int {\frac{\sec^2 v}{\tan^6 v \sqrt{\tan^2 v + 1} }} \, dv \\& = \int {\frac{\sec v}{\tan^6 v}} \, dv \\& = \int {\cot v \csc v (\csc^2 v - 1)^2} \, dv \\\end{aligned}[/tex]

Step 7: Integrate Pt. 6

Identify variables for u-substitution/u-solve again.

Use another variable besides u to avoid confusion with earlier substitutions:

Step 8: Integrate Pt. 7

Reduce the integral using the U-Solve Integration Method and other basic integration techniques listed under "Calculus":

[tex]\displaystyle \begin{aligned}\int {\frac{1}{u^6 \sqrt{u^2 + 1} }} \, du & = \int {\frac{\sec^2 v}{\tan^6 v \sqrt{\tan^2 v + 1} }} \, dv \\& = \int {\frac{\sec v}{\tan^6 v}} \, dv \\& = \int {\cot v \csc v (\csc^2 v - 1)^2} \, dv \\& = \int {-(w^2 - 1)^2} \, dw \\& = - \int {(w^2 - 1)^2} \, dw \\& = - \int {\bigg( w^4 - 2w^2 + 1 \bigg)} \, dw \\\end{aligned}[/tex]

Step 9: Integrate Pt. 8

Solve the integral using basic integration techniques listed under "Calculus":

[tex]\displaystyle\begin{aligned}- \int {\bigg( w^4 - 2w^2 + 1 \bigg)} \, dw & = - \Bigg[ \int {w^4} \, dx - \int {2w^2} \, dx + \int {} \, dx \Bigg] \\& = - \Bigg[ \int {w^4} \, dx - 2 \int {w^2} \, dx + \int {} \, dx \Bigg] \\& = - \Bigg[ \frac{w^5}{5} - \frac{2w^3}{3} + w + C \Bigg] \\& = - \frac{w^5}{5} + \frac{2w^3}{3} - w + C \\& = - \frac{\csc^5 v}{5} + \frac{2\csc^3 v}{3} - \csc v + C \\\end{aligned}[/tex]

[tex]\displaystyle\begin{aligned}- \int {\bigg( w^4 - 2w^2 + 1 \bigg)} \, dw & = - \frac{(u^2 + 1)^\Big{\frac{5}{2}}}{5u^5} + \frac{2(u^2 + 1)^\Big{\frac{3}{2}}}{3u^3} - \frac{\sqrt{u^2 + 1}}{u} + C \\\end{aligned}[/tex]

Step 10: Integrate Pt. 9

Let's substitute our integral value into our integral from "Step 4":

[tex]\displaystyle\begin{aligned}\frac{1}{2} \int {\frac{1}{u^6 \sqrt{u^2 + 1}}} \, du & = - \frac{(u^2 + 1)^\Big{\frac{5}{2}}}{10u^5} + \frac{(u^2 + 1)^\Big{\frac{3}{2}}}{3u^3} - \frac{\sqrt{u^2 + 1}}{2u} + C \\& = - \frac{(x^4 + 1)^\Big{\frac{5}{2}}}{10x^{10}} + \frac{(x^4 + 1)^\Big{\frac{3}{2}}}{3x^6} - \frac{\sqrt{x^4 + 1}}{2x^2} + C \\& = \boxed{ - \frac{\sqrt{x^4 + 1}(8x^8 - 4x^4 + 3)}{30x^{10}} + C } \\\end{aligned}[/tex]

∴ we have found the indefinite integral.

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Topic: Calculus

Step-by-step explanation:

y=2x find the value of y and x

Step-by-step explanation:

oh, come on ! one is finding the inequality itself, the other is solving it. you don't know the difference of that ?

if so, please let me know, what you don't understand of the following explanations, so that I can add whatever is necessary.

you know, what an equality (ahem, equation) is ?

both, equations and inequalities are like a balancing scale with usual different things on the two pans left and right.

an equation is when both sides are in balance. their weight is equal.

an inequality is, when one side is heavier than the other.

and you know, what a variable is ? it is a kind of placeholder for actual values. it is like keeping some space open on the scale pan with a list of conditions for what kind of things you can put in that open space, so that the status of the scale stays unchanged.

A

twenty less than a number is at least 15

looks mathematically

x - 20 >= 15 (larger or equal = "at least")

B

x - 20 >= 15

x >= 35

so, this is true for

{x € R | x >= 35}

or

{x € R | x = 35 or x > 35}

so, this is true for all real values of x that are at least 35.

Answer:

f(x) = 8x^6 + 1

Step-by-step explanation:

We want to examine which polynomial function grows larger and larger as the negative variable rises. We'll see that option d is correct: f(x) = 8x6 + 1

There are two things to keep in mind here.

The polynomial grows quicker as the exponent increases.

When a negative integer is evaluated with an even exponent, the result is always positive.

Then all the polynomials with an even leading exponent become larger and larger as x grows in the negative direction, especially the one with the biggest leading exponent, which is:

The density of the material that this figurine was made from in pounds/inch³ is 0.183 lb/in³

To answer the question, we need to know what density is

Density is the mass per unit volume of a material.

It is given by ρ = m/v where

Now, given that the toy is dropped into a basin filled with water and found to displace 125.4 mL of water. It's volume is v = 125.4 mL

= 125.4 × 1 mL

= 125.4 × 0.0610237 in³

= 7.65237198 in³

Also, given that the figurine is then put on a scale and found to be 1.4 pounds in weight. It's mass is m = 1.4 lb

So, the density of the material ρ = m/v

ρ = 1.4 lb/7.65237198 in³

ρ =  0.183 lb/in³

So, the density of the material that this figurine was made from in pounds/inch³ is 0.183 lb/in³

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The codomain of the function [tex]g(x) = 2x^2 + 4x - 1[/tex] is [tex]g(x) = 2x^2 + 4x - 1[/tex]

The equation of the function is given as:

[tex]g(x) = 2x^2 + 4x - 1[/tex]

The above function is a quadratic function and it has the following features:

Domain = [tex](-\infty, \infty)[/tex]

Range: [tex][-3, \infty)[/tex]

When the range and the domain are combined, we have:

[tex][-3, \infty)\ and\ (-\infty, \infty)[/tex]

This gives

[tex][-3, \infty[[/tex]

Hence, the codomain of the function [tex]g(x) = 2x^2 + 4x - 1[/tex] is [tex]g(x) = 2x^2 + 4x - 1[/tex]

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Answer:

-2

Step-by-step explanation:

Slope of a line, given two points : formatslope=y2−y1x2−x1=4−2−2−(−1)=2−1=−2

In the given question, all the equations are fourth degree polynomial because they all have the highest power of 4

A polynomial's degree is the highest or the greatest power of a variable in a polynomial equation.

In the given question, all the equation are fourth degree polynomial

What closure property relates here is that when we add two different polynomial, the result will definitely be a polynomial.

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Step-by-step explanation:

6x-12=x+3

6x-x=3+12

5x=15

x=15/5

x=3

Answer:

Step-by-step explanation:

What is value of m in the equation

3 (2x - 4) = x +3​?

3 (2x - 4) = x + 3​

6x - 12 = x + 3

5x = 15

x = 15 : 5

x = 3

Answer:

A, 3

Step-by-step explanation:

Just look at x axis, find 6, and then start going up until you hit the line. Then record the point you hit.

Answer:

-2

Step-by-step explanation:

Find two points which lie on the line :

Formula for slope (m) :

Answer: [tex]-2[/tex]

Step-by-step explanation:

[tex]Slope:m=0[/tex]

[tex]y=mx+b[/tex]

[tex]-2*10^0[/tex]

[tex]-2~is~a~rational~number[/tex]

[tex]\frac{180}{pi}[/tex]°

Answer:

60/13

Step-by-step explanation: