Suppose we choose7 objects without replacement from 9 objects

Answer:

Graph 2

Step-by-step explanation:

We are given the inequality x≤2

First off, with the inequality sign being ≤ rather than <, we know that 2 is also included. This means that x can be equal to 2 or less than 2.

Since 2 is included, we know we are looking for a solid line.

That said, we can eliminate the 1st and 4th graphs because there is a dotted line at 2, meaning that 2 is not included.

Next, based on the inequality sign, we know that values less than 2 satisfy it. So judging by the shaded region on each graph, we see that Graph 2 is the one that satisfies it.

For clarity's sake, one trick you can use is thinking of the inequality sign as an arrow. If it points left, values toward the left should be shaded and vice versa.

Lastly, I'll give you the inequality for each graph.

Graph 1: x<2

Graph 2: x≤2

Graph 3: x≥2

Graph 4: x>2

Hope that helps, let me know if you have any questions!

Answer: [tex]4\sqrt{7}[/tex]

Step-by-step explanation:

If [tex]AC=16, DC=7[/tex], then this means [tex]AD=9[/tex].

So, by the geometric mean theorem, [tex]DB=\sqrt{(9)(7)}=\sqrt{63}[/tex].

This means that by the Pythagorean theorem, [tex]BC=\sqrt{\left(\sqrt{63} \right)^{2}+7^{2}}=\boxed{4\sqrt{7}}[/tex]

Answer:

7 7^10

1 5^9

yan po yung sagot

sana makatulong

Answer:

Slope = undefined.

Step-by-step explanation:

Given two points:

To Find:

Solution:

Using slope's formulae,

[m denotes slope]

[tex] \boxed{ \rm{m = \cfrac{y_2 -y_1 }{x_2 - x_1} }}[/tex]

According to the question,here:

Substitute them onto the formulae:

[tex] \rm \: m = \cfrac{9 - 3}{2 - 2} [/tex]

Simplify.

[tex] \rm \: m = \cfrac{6}{2 - 2} [/tex]

[tex]\rm \: m = \cfrac{6}{0} = \boxed{ \rm \: undefined}[/tex]

if an expression contains division by 0 , then It's undefined.

So, slope(m) = undefined.

Answer:

Step-by-step explanation:

Problem a

a is an isosceles triangle. It has 2 sides marked as equal. Therefore the two angles marked in red are also equal.  

The top angle = x

x + 70.5 + 70.5 = 180      All triangles have 180 degrees.

x + 141 = 180                    Subtract 141 from both sides

x + 141-141 =180 - 141       Combine

x = 39

c + 39 + 87.2 =180           Combine the left

c + 126.2 = 180                 Subtract 126.2 from both sides

c+126.2-126.2 = 180-126.2

c = 53.45

Answer: 53.45

Problem B

Comment

Problem b can only be found by taking the supplements of the two given angles. After that is done, the vertically opposite angle to d can be found because d is opposite the third angle.

Supplement of 115.6

Call the supplement = y

y + 115.6 = 180                         Supplementary angles = 180 degrees

y +115.6-115.6 = 180 - 115.6     Combine

y = 64.2

Now we need the supplement of 99.9

Call the supplement = z

z + 99.9 = 180                           Supplementary angles = 180

z+99.9-99.9 = 180-99.9

z = 80.1

Now subtract y + z from 180

180 - 64.2 - 80.1 = x

x = 35.7

x is opposite d

d = x

d = 35.7 degrees.

Answer

d = 35.7

[tex]\begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ \underset{\textit{we'll use this one}}{log_a a^x = x}\qquad \qquad a^{log_a x}=x \end{array} \\\\[-0.35em] ~\dotfill\\\\ 2.7e^x-5 ~~ = ~~33.6\implies 2.7e^x~~ = ~~38.6\implies e^x=\cfrac{38.6}{2.7} \\\\\\ \log_e(e^x)=\log_e\left( \cfrac{38.6}{2.7} \right)\implies x=\ln\left( \cfrac{38.6}{2.7} \right)\implies x\approx 2.66[/tex]

Step-by-step explanation:

the thing you need to do here is put the value of x and just simply the expression.....

= 3y + 78 + 6y + 13

= 3 × 3 + 78 + 6 × 3 + 13

= 9 + 78 + 18 + 13

= 118

[tex]....[/tex]

Answer:

3×3+78+6×3÷13

9+78+18÷13

88.38

Step-by-step explanation:

i m not sure if it's correct since i didn't follow tye bodmas rule properly and it was bit tricky and it comes in decimal so not sure if it's correct

Answer:

the independent variable is s

the dependent one is m

hope this helps <3

Answer:

thousandths

Step-by-step explanation:

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:

slope = 1

y intercept = (0,8)

Step-by-step explanation:

the equation y = x + 8 is put in y = mx + b form

where m = slope and b = y intercept

in the equation y = x + 8 , the value of "m" is 1 meaning the slope is 1 and the value of "b" is 8 so the y intercept is (0,8)

Answer:

m = 1 and b = 8

Step-by-step explanation:

y = x + 8

Here the slope is 1, since y = mx+ b. m is the slope so m = 1. b is the

y-intercept so b=8

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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Answer:

A polygon is a plane figure.

Step-by-step explanation:

This means it is a 2 dimensional shape that lies flat, it has length and width, it has area but no thickness.

There is no max number of sides, so the first answer is not right.

In math, saying it is solid means its 3-d, so that is not right (it's only 2-d)

Saying its open, means that somewhere on the shape, the sides don't meet up at a corner, like an animal corral with the gate left open. So that's not right. Part of the definition of a polygon is that it is closed.

It is a 2-d shape with straight sides and closed, it lies flat in a plane. The last answer is correct.

Answer:

A polygon is a plane figure.

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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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.

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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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:

the H stands for hours. the E stands for money. because it says 10h that means she gets $10 per hour. that answers the first question. I don't understand the second part though. hope you get it!

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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Answer:

80

Step-by-step explanation:

10x7=70

2x5=10

70+10=80

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

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:

-2

Step-by-step explanation:

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

Answer:

A

Step-by-step explanation:

There is an even number of values.
Hope it helps

Sorry if I’m wrong

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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Answer:

Step-by-step explanation:

Making the equation of the parabola :

⇒ y = a(x - h)² + k

⇒ y = (x + 1)² - 17 - 13

⇒ y = (x + 1)² - 30

x-intercepts have y = 0 :

Taking the square root on each side :

The x-intercepts are :

Answer:

1) Answer is D

2) Answer is C

3) Answer is D

Step-by-step explanation:

Problem 1

Since we are given 12 numbers to choose from, there's a 2/12 chance of selecting a 2. Additionally, there's a 3/11 chance of selecting a number divisible by 4 AFTER selecting a 2. Thus, since the two events are independent, their probabilities are multiplied, so the probability of selecting a 2 and then a number divisible by 4 is (2/12)(3/11) = 6/132 = 1/22, so option D is correct.

Problem 2

There's a 4/12 chance of selecting an odd number and a 6/11 chance of selecting an even number AFTER selecting an odd number (remember that 0 is neither even nor odd). Multiplying the probabilities, the probability of drawing an odd number and then drawing an even number is (4/12)(6/11) = 24/132 = 2/11, so option C is correct.

Problem 3

There's a 2/12 chance of selecting a zero and a 1/11 chance of selecting a second zero. Thus, multiplying their probabilities, the probability of drawing 2 zeroes is (2/12)(1/11) = 2/132 = 1/66, so option D is correct.

Step-by-step explanation:

y=2x find the value of y and x

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: