match the range of the function f(x)=x^2+2x-1 to its domain
2
-2
3
-3

Answer:

3x-13+2x+4=116( the sum of two interior angle is epual to the sum of exterior angle )

5x =116-9

5x=107

x=21.4 degree

then

angle a= 3*21.4-13

a=51.2degree

angle b=2*21.4+2

b =44.8degree

Step-by-step explanation:

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73.1 A(C2) is a quadrilateral with vertices (0, 0), (1, 0), (1, 2), and (0, 2).

73.2 The area of A(C2) is 2.

73.3 The determinant of A is 0

To draw C2 and A(C2),  to understand the transformation A and to the unit square.

Given the information:

A:

| 1 0 |

| 2 0 |

A maps the vector [1, 0] to [2, 0], and it maps the vector [0, 1] to [1, 1]. This provides us with the information to draw C2 and A(C2).

73.1 Draw C2 and A(C2):

C2 represents the unit square with vertices (0, 0), (1, 0), (1, 1), and (0, 1).

To draw A(C2), the transformation A to each vertex of C2 and connect the transformed vertices to form the image of C2 under A.

Applying A to the vertices of C2:

A(0, 0) = (1 × 0, 2 × 0) = (0, 0)

A(1, 0) = (1 × 1, 2 × 0) = (1, 0)

A(1, 1) = (1 × 1, 2 × 1) = (1, 2)

A(0, 1) = (1 × 0, 2 × 1) = (0, 2)

Connecting the transformed vertices, we get the image of C2 under A:

73.2 Compute the area of A(C2):

To compute the area of A(C2),  divide it into two triangles and calculate their areas separately.

Triangle 1: (0, 0), (1, 0), (1, 2)

Base of Triangle 1 = 1 - 0 = 1

Height of Triangle 1 = 2 - 0 = 2

Area of Triangle 1 = (1/2) × Base × Height = (1/2) × 1× 2 = 1

Triangle 2: (0, 0), (0, 2), (1, 2)

Base of Triangle 2 = 1 - 0 = 1

Height of Triangle 2 = 2 - 0 = 2

Area of Triangle 2 = (1/2) × Base × Height = (1/2) × 1 × 2 = 1

Total Area of A(C2) = Area of Triangle 1 + Area of Triangle 2 = 1 + 1 = 2

73.3 Compute det(A):

To compute the determinant of A (denoted as det(A)), use the formula for a 2x2 matrix.

A:

| 1 0 |

| 2 0 |

det(A) = (1 ×0) - (0 × 2) = 0 - 0 = 0

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Answer:

Two solutions

Step-by-step explanation:

Given the functions Y = -x + 1 y = -x² + 4x − 2

Since they are both y values, we will equate them to have:

-x+1 = -x² + 4x − 2

Equate to zero

x²-x-4x+1+2=0

x²-5x+3 = 0

Since the resulting equation is quadratic (highest degree being 2), hence there will be two solutions to the equation

Answer:

No such a triangle can not exist.  

:-):-):-):-):-)

Step-by-step explanation:

1) 40÷10

2) 40÷2

Answer:

7,-7 are the solutions

Step-by-step explanation:

Not a perfect square so its

+-7

I would go with -7 since its the lowest amount

Answer:

-7

Step-by-step explanation:

Answer:

$97,894.33.

Step-by-step explanation

Answer:

4

Step-by-step explanation:

To find median, arrange the data set in ascending order. Then the middle term is the median

2 , 3 , 4 , 5 , 6

Median = 3rd term  = 4

Answer:

4 is the median

Step-by-step explanation:

From least to greatest:

2,3,4,5,6

Since the total values is odd, the middle number is the median

4 is the median

Answer:

rap music

Step-by-step explanation:

^

Answer:

i’m not sure if these are right or not but

Step-by-step explanation:

the first one is no

i think the second one is no

i think the third one is yes

i think the fourth one is yes

i am not sure and i’m REALLY sorry if its wrong

Answer:

r = dx - de

Step-by-step explanation:

x = e + r/d

x - e = r/d

d (x - e) = r

dx- de = r

r = dx - de

a. The area of the shaded region = (3x)(5x + 2y).

b. The area of shaded region = 2x(2xπ - 3x).

Given that,

We have to write an expression in factored form to represent the area of each shaded region.

We know that,

a. In the picture we can see that the shaded region.

The area of the shaded region is area of rectangle with vertical + area of rectangle Horizontal

The area of rectangle is length × width.

The area of the shaded region = (3x)(3x + y) + (2x + y)(3x)

The area of the shaded region = (3x)(3x + y + 2x + y)

The area of the shaded region = (3x)(5x + 2y)

Therefore, The area of the shaded region = (3x)(5x + 2y)

b. In the picture we can see that circle with shaded region,

The area of shaded region is area of the circle - area of the rectangle

The area of the circle is πr² and area of rectangle is length × width.

The area of shaded region = πr² - (l×w)

The area of shaded region = π(2x)² - (2x × 3x)

The area of shaded region = 4x²π - 6x²

The area of shaded region = 2x(2xπ - 3x)

Therefore, The area of shaded region = 2x(2xπ - 3x).

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(a) The top of the ladder is sliding down the wall at a rate of 3/2 m/sec.

(b) The angle between the ladder and the ground is changing at a rate of 1/2 rad/sec.

To solve this problem, we can use related rates, considering the ladder as a right triangle formed by the ladder itself, the wall, and the ground.

Let's denote:

x: the distance from the base of the ladder to the house (in meters)

y: the height of the ladder on the wall (in meters)

θ: the angle between the ladder and the ground (in radians)

Given:

dx/dt = -2 m/sec (the rate at which the base of the ladder is moving away from the house)

Using the Pythagorean theorem, we have:

x^2 + y^2 = 5^2 (since the ladder has a length of 5 meters)

Taking the derivative of both sides with respect to time (t), we get:

2x(dx/dt) + 2y(dy/dt) = 0

(a) To find dy/dt:

We can solve the equation above for dy/dt:

2x(dx/dt) + 2y(dy/dt) = 0

2(3)(-2) + 2y(dy/dt) = 0 (substituting x = 3 and dx/dt = -2)

-12 + 2y(dy/dt) = 0

2y(dy/dt) = 12

dy/dt = 12/(2y)

dy/dt = 6/y

Now, we need to find y. Using the Pythagorean theorem again:

x^2 + y^2 = 5^2

3^2 + y^2 = 5^2

9 + y^2 = 25

y^2 = 25 - 9

y^2 = 16

y = 4 (taking the positive value as y represents a length)

Substituting y = 4 into dy/dt = 6/y:

dy/dt = 6/4

dy/dt = 3/2 m/sec

Therefore, the top of the ladder is sliding down the wall at a rate of 3/2 m/sec.

(b) To find dθ/dt:

We can use trigonometry to relate θ, x, and y:

tan(θ) = y/x

Differentiating both sides with respect to time (t), we get:

sec^2(θ)dθ/dt = (x(dy/dt) - y(dx/dt))/x^2

Substituting the given values:

sec^2(θ)dθ/dt = (3(3/2) - 4(-2))/3^2

sec^2(θ)dθ/dt = (9/2 + 8)/9

sec^2(θ)dθ/dt = (25/2)/9

sec^2(θ)dθ/dt = 25/18

Since sec^2(θ) is equal to 1 + tan^2(θ) and tan(θ) = y/x:

sec^2(θ) = 1 + (y/x)^2

sec^2(θ) = 1 + (4/3)^2

sec^2(θ) = 1 + 16/9

sec^2(θ) = (9 + 16)/9

sec^2(θ) = 25/9

Substituting sec^2(θ) = 25/9 into the equation:

(25/9)dθ/dt = 25/18

Simplifying and solving for dθ/dt:

dθ/dt = (25/18) * (9/25)

dθ/dt = 1/2 rad/sec

Therefore, (b) the angle between the ladder and the ground is changing at a rate of 1/2 rad/sec.

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Answer:

A. 20.2 ft

Step-by-step explanation:

use the pythagorean theorem to get your answer.

a^2+b^2=c^2

17^2+11^2=c^2

289+121=410

***IMPORTANT, remember to square root the answer. 410 isn't your final answer, it's still in c^2 form. 20.2 is the correct answer after square rooting it. don't forget that for future problems

Answer:

A. 20.2 ft

Step-by-step explanation:

Hope it helps..

The derivatives of the vector function are r'(t) = (18 · t, 1, 0), r''(t) = (18, 0, 0) and r'''(t) = (0, 0, 0).

In this question we find the definition of a vector function in terms of time, whose first, second and third derivatives must be found. This can be done by using derivative rules several times. First, define the vector function:

r(t) = (9 · t² + 6, t + 5, 6)

Second, find the first derivative:

r'(t) = (18 · t, 1, 0)

Third, find the second derivative:

r''(t) = (18, 0, 0)

Fourth, find the third derivative:

r'''(t) = (0, 0, 0)

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Answer:

Step-by-step explanation:

In Problem 6, 3/x = tan 30 degrees, or x/3 = 1 / (tan 30 degrees), or

x =   3 / (tan 30 degrees).  Using a calculator:  x = 3 / (0.449) = 6.68 units

In Problem 9, we begin by finding the length of the hypotenuse of the upper triangle:  sin 60 degrees = 9√3 / hypotenuse, or

hypotenuse = 9√3/ (√3 / 2) = 9/2.  Then the hypotenuse of the 45-45-90 triangle is 9/2.  Thus, cos 45 degrees = x / (9/2), or

2x      √3

---- = -------

 9        2

                                                                                    9√3

Cross-multiplying, we get 4x = 9√3, so that x = -------------

                                                                                        4

The probability of getting four consecutive heads in four 2-sided coin tosses is 1/16 or 0.0625 or 6.25%.

The Probability is defined as a measure of the likelihood or chance that an event will occur, expressed as a value between 0 and 1, where 0 represents impossibility and 1 represents certainty.

To calculate the probability of getting four consecutive heads in a row, we multiply the probabilities of each individual toss.

The Probability of heads (H) is = 1/2,

So, The Probability of four consecutive heads (HHHH) is :

= (1/2) × (1/2) × (1/2) × (1/2) = (1/2)⁴ = 1/16,

Therefore, the required probability is 1/16 or 6.25%.

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Answer:

answer is 65 in²

hope this helps

Answer:

Rate of change is 1

Step-by-step explanation:

Well rate of change in a function is basically slope

theres 1 rise and.1 run

1/1=1

So rate of change is 1

25

porque al multiplicar 25×25 nos da 625

Answer:

25

Step-by-step explanation:

25x25=625

Answer: the answer is 3/10

Step-by-step explanation: because its 3/10

Answer:

es

3/ 10

Step-by-st

Answer:

Answer:

uhuhuhuhuhuhuhuhuhuhuh oh yeah

Step-by-step explanation:

Answer:

7 people

Step-by-step explanation:

Chocolate: 350 / 100 = 3.5 x 2 = 7

Butter: 45 / 10 = 4.5 x 2 = 9

Eggs: 8 / 2 = 4 x 2 = 8

The amount of chocolate is the limiting factor as you have enough eggs for 8 people and enough butter for 9 people, but only enough chocolate for 7 people

Answer:

2x^2-x-3

(2x)(-3)

(x)

(1)

Step-by-step explanation:

Distribute

(2x^2) (-3x)

(2x) (-3)

The question is incomplete. The compete question is :

Saul decides to use the IQR to measure the spread of data. Saul calculates the IQR of the data set to be 27. Saul asks Jasmine to check his work. Saul copies the data in numerical order from least to greatest for Jasmine. 25, 30, 50, 50, 50, 50, 56, n, 250. What is the value of n that will make the IQR of the data set equal to 27?

Solution :

The inter quartile range is being measured as : 3rd quartile - 1st quartile

The 1st quartile = 25% mark

The 3rd quartile = 75% mark

The data set as given in the question is :

25, 30, 50, 50, 50, 50, 56, n, 250

Therefore, the total number of the sample = 9

And the median is = 50

The median separates the given data sets into two equal halves-- that is the upper half and the lower half.

Let the middle of a lower half is the 1st quartile be [tex]$Q_1$[/tex]

And the middle of a 2nd half is the 3rd quartile be [tex]$Q_3$[/tex]

∴ [tex]$Q_1=\frac{30+50}{2}$[/tex]

        [tex]$=40$[/tex]

  [tex]$Q_3=\frac{56+n}{2}$[/tex]

Now for the IQR of the data set to be equal to [tex]$27$[/tex],

[tex]$\frac{56+n}{2}-40=27$[/tex]

[tex]$\frac{56+n}{2}=27+40$[/tex]

[tex]$\frac{56+n}{2}=67$[/tex]

[tex]$56 +n=134$[/tex]

[tex]$n=134-56$[/tex]

[tex]$n=78$[/tex]