Triangle P Q R is shown. Angle P Q R is a right angle. The length of hypotenuse P R is q, the length of P Q is r, and the length of Q R is p.,
Which cosine ratios are correct for ΔPQR? Check all that apply.

cos (P) = StartFraction r Over q EndFraction
cos (P) = StartFraction p Over q EndFraction
cos (Q) = StartFraction r Over p EndFraction
cos (R) = StartFraction r Over p EndFraction
cos (R) = StartFraction p Over q EndFraction

The value of x is calculated with the help of a similar triangle is 24.

A triangle is a three-sided polygon with three angles. The angles of the triangle add up to 180 degrees.

In a triangle, if there is a parallel line with any side of the triangle, then the line divides the other sides in the same ratio.

Then we have

x / 12 = 32 / 16

     x = 24

Then the value of x is 24.

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Answer:

-12

Step-by-step explanation:

Order of operations:

PEMDAS

First, do parenthesis, 2 x 2.2 = 4.4 then do -7.6 - 4.4 = -12

Answer:

(6,4)

Step-by-step explanation:

^^^

The zeroes of the quadratic function f(x) = 6x² + 12x – 7 will be given below. Then the correct option is A.

The quadratic function is given as ax² + bx + c = f(x).

For finding the zeroes of the polynomial we have to find the roots of the polynomial , The value of f(x) = 0

The given function is f(x) = 6x² + 12x – 7

Then the zeroes of the quadratic function will be

[tex]roots = \dfrac {-b \pm \sqrt{b^2 -4ac}}{2a}\\\\[/tex]

Here a =6 , b = 12 ,c = -7

Therefore the zeroes are

[tex]\rm roots = \dfrac {-12 \pm \sqrt{12^2 -4 * 6 * (-7)}}{2 *6}\\\\[/tex]

[tex]\rm root_1 = -1 + \sqrt{\dfrac {13}{6}}\\\\\rm root_2 = -1 - \sqrt{\dfrac {13}{6}}[/tex]

Therefore the correct option is A.

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Answer:

(-2,7)

Step-by-step explanation:

x=7-3y

2(7-3y)+4y=8

14-6y +4y=8

14-2y=8

-2y = -6

y=3

x + 3(3) = 7

x + 9 = 7

x = -2

(-2,7)

Answer: -2,3

Step-by-step explanation:

Answer:  (5/1, -3)

Hope this helps :)

Answer:

Look at the attachment.

Step-by-step explanation:

Look at the attachment.

Answer:

6.34 min to empty

Step-by-step explanation:

80 cm x 40 cm x 30 cm = 96000 cm^3

  this volume represents  25.36 gallons of water (US)

25.36 gal / 4 gal/min = 6.34 min  to empty

The stock worth now, to the nearest cent, is $58.14.

The current worth of stock = initial price of stock+increase in the price of stocks

If you bought stock last year for a price of $66, and it has gone down 13.9% since then,

Let X be the worth of stock;

66 ×  100 = (13.5% + 100%) × X

6600 = 113.5X

X = 6600/113.5

X = 58.14458

X = $58.14

Hence, The stock worth now, to the nearest cent is $58.14.

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Answer: There is no mode.

Step-by-step explanation:

The mode is the number that appears the most in a given set of data. Each number in this data set is repeated twice. Therefore, there is no mode.

I hope this helped!! :)

Answer:

1, 2, 3, 4

or none

Step-by-step explanation:

Mode is the value or values in the data set that occurs most frequently.

example:

For the data set 1, 1, 2, 5, 6, 6, 9 the mode is 1 and also 6.

in 1, 1, 2, 2, 3, 3, 4, 4 all numbers occur the same amount of times, therefore they all (or none) are the mode

hope this helps:)

Use the definitions of expectation and variance.

[tex]E(X) = \displaystyle \int_{-\infty}^\infty x f_X(x) \, dx = \frac14 \int_0^\infty x e^{-x/4} \, dx[/tex]

Integrate by parts,

[tex]\displaystyle \int_a^b u \, dv = uv \bigg|_a^b - \int_a^b v \, du[/tex]

with

[tex]u = x \implies du = dx \\\\ dv = e^{-x/4} \, dx \implies v = -4 e^{-x/4}[/tex]

Then

[tex]E(X) = \displaystyle \frac14 \left(\left(-4x e^{-x/4}\right)\bigg|_0^\infty + 4 \int_0^\infty e^{-x/4} \, dx\right)[/tex]

[tex]E(X) = \displaystyle \int_0^\infty e^{-x/4} \, dx = \boxed{4}[/tex]

(since the integral of the PDF is 1, and this integral is 4 times that)

[tex]V(X) = E\bigg((X - E(X))^2\bigg) = E(X^2) - E(X)^2[/tex]

Compute the so-called second moment.

[tex]E(X^2) = \displaystyle \int_{-\infty}^\infty x^2 f_X(x)\, dx = \frac14 \int_0^\infty x^2 e^{-x/4} \, dx[/tex]

Integrate by parts, with

[tex]u = x^2 \implies du = 2x \, dx \\\\ dv = e^{-x/4} \, dx \implies v = -4 e^{-x/4}[/tex]

Then

[tex]E(X^2) = \displaystyle \frac14 \left(\left(-4x^2 e^{-x/4}\right)\bigg|_0^\infty + 8 \int_0^\infty x e^{-x/4} \, dx\right)[/tex]

[tex]E(X^2) = 8 E(X) = 32[/tex]

and the variance is

[tex]V(X) = 32 - 4^2 = \boxed{16}[/tex]

Answer:

15

Step-by-step explanation:

0 + 1 = 1

1 + 2 = 3

3 + 3 = 6

6 + 4 = 10

10 + 5 = 15

The polynomial function with the given properties is; y = 2(x - 3)³(x + 4)²

Polynomials are defined as algebraic expressions that consist of variables, coefficients, and constants. The standard form of polynomials has mathematical operations such as addition, subtraction, and multiplication.

We need to form a polynomial expression with the following properties;

fifth-degree, 3 is a zero of multiplicity 3, −4 is the only other zero and ,the leading coefficient is 3. Thus, the polynomial will be;

y = 2(x - 3)³(x + 4)²

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Answer:

−135z

2

−2z
The 2 is representing Square

Step-by-step explanation:

(3z)
2

−4z+11−(12z)
2

+2z−11

Expand (3z)

2

.

3

2

z

2

−4z+11−(12z)

2

+2z−11

Calculate 3 to the power of 2 and get 9.

9z

2

−4z+11−(12z)

2

+2z−11

Expand (12z)

2

.

9z

2

−4z+11−12

2

z

2

+2z−11

Calculate 12 to the power of 2 and get 144.

9z

2

−4z+11−144z

2

+2z−11

Combine 9z

2

 and −144z

2

 to get −135z

2

.

−135z

2

−4z+11+2z−11

Combine −4z and 2z to get −2z.

−135z

2

−2z+11−11

Subtract 11 from 11 to get 0.

−135z

2

−2z

a)The probability that all 10 will arrive within two days is 0.3486.

b)  P(x = 8) = 0.1937

c)  P(x=5)= 0.0014

d)  mean = 9

e) Standard deviation = √0.90 = 0.9486

The binomial distribution model allows us to compute the probability of observing a specified number of "successes" when the process is repeated a specific number of times and the outcome for a given patient is either a success or a failure.

Given:

p = 90/100 = 0.90

q(probability of success) = 1 - p = 1 - 0.90 = 0.10

n = 10

a)The probability that all 10 will arrive within two days

P(x = 10) = 0.3486

b) P(x = 8) = 0.1937

c) P(x=5)= 0.0014

d) mean = np = 10 × 0.90 = 9

e) Variance = npq = 10 × 0.90 × 0.10 = 0.90

Standard deviation = √0.90 = 0.9486

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Option C. A 2-column table with 5 rows. Column 1 is labeled x with entries negative 9, negative 7, negative 5, negative 3, negative 1. Column 2 is labeled y with entries negative 25, negative 21, negative 17, negative 13, negative 9.  This labeled is correct.

According to the linear function,
Every linear function is characterized by a constant rate of change; the slope. The slope of a linear function is a measure of the “steepness” of the line. We use the symbols Δx and Δy which mean respectively the “change in x ” and the “change in y ”.

So Option C is correct after all assumptions.

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Answer:

See proof below

Step-by-step explanation:

Important points

Onto

For a function with a given co-domain to be "onto," every element of the co-domain must be an element of the range.

However, the co-domain here is suggested to be [tex]\mathbb R[/tex], whereas the range of f is not [tex]\mathbb R[/tex] (proof below).

Proof (contradiction)

Suppose that f is onto [tex]\mathbb R[/tex].

Consider the output 7 (a specific element of [tex]\mathbb R[/tex]).

Since f is onto [tex]\mathbb R[/tex], there must exist some input from the domain [tex]\mathbb R[/tex], "p", such that f(p) = 7.

Substitute and solve to find values for "p".

[tex]f(x)=-3x^2+4\\f(p)=-3(p)^2+4\\7=-3p^2+4\\3=-3p^2\\-1=p^2[/tex]

Next, apply the square root property:

[tex]\pm \sqrt{-1} =\sqrt{p^2}[/tex]

By definition, [tex]\sqrt{-1} =i[/tex], so

[tex]i=p \text{ or } -i =p[/tex]

By the Fundamental Theorem of Algebra, any polynomial of degree n with complex coefficients, has exactly n complex roots.  Since the degree of f is 2, there are exactly 2 roots, and we've found them both, so we've found all of them.

However, neither [tex]i[/tex]  nor [tex]-i[/tex] are in [tex]\mathbb R[/tex], so there are zero values of p in [tex]\mathbb R[/tex] for which f(p)= 7, which is a contradiction.

Therefore, the contradiction supposition must be false, proving that f is not onto [tex]\mathbb R[/tex]

[tex]z_{2}-3i=(-4+i)-3i=-4-2i\\\\z_{1}+2=(3-4i)+2=5-4i\\\\(z_{1}+2)-(z_{2}-3i)=(5-4i)-(-4-2i)=9-4i+4+2i=9-2i[/tex]

which corresponds with point C

The expression which is the result of subtracting (z2-3i) from (z1+2) is 9-2i.

Given On a coordinate plane, the y-axis is labeled imaginary and the x-axis is labeled real. Point A is (1, -2), B is (1, -6), C is (9,- 2), D is (9, - 6), z 1 is (3, - 4), and z 2 is (- 4, 1).

Coordinate plane is a two dimensional plane formed by the intersection of two number lines. One is horizontal line and one is vertical. On horizontal line x values are denoted and on vertical line y values are written. There can be three dimensional plane also.

z2-3i=(-4+i)-3i=-4-2i

z1+2=(3-4i)+2=5-4i

(z1+2)-(z2-3i)=(5-4i)-(-4-2i)

=9-4i+4+2i

=9-2i

Hence for the expression (z1+2)-(z2-3i)=9-2i

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Answ x=59

Step-by-step explanation:

The graph is attached and all the sub parts have been answered.

A function is  away of relating a dependent variable to an independent variable .

It is given that

Current weight of Rex at 2 months =6 pounds

Let x represents the age in months of Rex

then a linear model that represents the Rex’s current weight, in pounds, as a function of his age, in months. is

Let y be the current weight of Rex

then

y = 6 + 9(x -2)

The table  that represents Rex’s current weight, in pounds, as a function of his age, in months

2              6

3              15              

4              24

5              33

6              42

7              51

8              59

The graph is plotted for the given data and attached .

If Rex continues to grow at the rate of 9 pounds per month beyond the expected six months, how much will Rex weight by the time he is one year old

6 + 9*(12-2)

96 pounds.

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The value of y is -8 if the x  -24 and y varies directly with x, and If y = 4 when x = 12.

It is defined as the relationship between two variables when the first variable increases, the second variable also increases according to the constant factor.

The question is incomplete.

The complete question is:

If y varies directly with x, and If y = 4 when x = 12, how do you find y when x = -24?

y ∝ x  (given)

y = kx

k is the constant of proportionality.

4 = 12k  (y = 4, and x = 12)

k = 1/3

y = x/3

Plug x = -24

y = -24/3

y = -8

Thus, the value of y is -8 if the x  -24 and y varies directly with x, and If y = 4 when x = 12.

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The sector area and the arc length are 34.92 square inches and 13.97 inches, respectively

For a sector that has a central angle of θ, and a radius of r;

The sector area, and the arc length are:

[tex]A = \frac{\theta}{360} * \pi r^2[/tex] --- sector area

[tex]L = \frac{\theta}{360} * 2\pi r[/tex] ---- arc length

Here, the given parameters are:

Central angle, θ = 160

Radius, r = 5 inches

The sector area is

[tex]A = \frac{\theta}{360} * \pi r^2[/tex]

So, we have:

[tex]A = \frac{160}{360} * \frac{22}{7} * 5^2[/tex]

Evaluate

A = 34.92

The arc length is:

[tex]L = \frac{\theta}{360} * 2\pi r[/tex]

So, we have:

[tex]L = \frac{160}{360} * 2 * \frac{22}{7} * 5[/tex]

L = 13.97

Hence, the sector area and the arc length are 34.92 square inches and 13.97 inches, respectively

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Answer:

7

4

Step-by-step explanation:

The actual values are shown on the given graph as blue points.

The line of regression is shown on the given graph as the red line.

From inspection of the graph, in the year 2000 the actual rainfall was 43 cm, shown by point (2000, 43).  It appears that the regression line is at y = 50 when x is the year 2000.

⇒ Difference = 50 - 43 = 7 cm

In 2000, the actual rainfall was 7 centimeters below what the model predicts.

From inspection of the graph, in the year 2003 the actual rainfall was 44 cm,  shown by point (2003, 40).  It appears that the regression line is at y = 40 when x is the year 2003.

⇒ Difference = 44 - 40 = 4 cm

In 2003, the actual rainfall was 4 centimeters above what the model predicts.

Answer: 2.7545

The question was a little off, but I can understand enough

Step-by-step explanation:

Grade       Grade Point(x)       Course Credit(y)        Product(xy)

A                     4                            3                                12

B                      3                           4                                12

B                     3                            4                                12

B                      3                          4                                 12

C                     2                           2                                 4

D                      1                           3                                 3

Total                                              20                             55

This means, the GPA of the student is :

[tex]GPA=\frac{55}{20}[/tex]

[tex]2.75[/tex]

Answer:

bro nobody can figure this out you need to word the entire question with numbers and all. and proper English pls

Answer:

1.  "a"  [tex]u'=6x[/tex]

2.  "d"  [tex]v'=15x^2[/tex]

3.  "b"  [tex]y'=75x^4+30x^2+6x[/tex]

Step-by-step explanation:

Part 1.

Given [tex]u=3x^2+2[/tex], find [tex]\frac{du}{dx} \text{ or } u'[/tex].

[tex]u=3x^2+2[/tex]

Apply a derivative to both sides...

[tex]u'=(3x^2+2)'[/tex]

Derivatives of a sum are the sum of derivatives...

[tex]u'=(3x^2)'+(2)'[/tex]

Scalars factor out of derivatives...

[tex]u'=3(x^2)'+(2)'[/tex]

Apply power rule for derivatives (decrease power by 1; mutliply old power as a factor to the coefficient); Derivative of a constant is zero...

[tex]u'=3(2x)+0[/tex]

Simplify...

[tex]u'=6x[/tex]

So, option "a"

Part 2.

Given [tex]v=5x^3+1[/tex], find [tex]\frac{dv}{dx} \text{ or } v'[/tex].

[tex]v=5x^3+1[/tex]

Apply a derivative to both sides...

[tex]v'=(5x^3+1)'[/tex]

Derivatives of a sum are the sum of derivatives...

[tex]v'=(5x^3)'+(1)'[/tex]

Scalars factor out of derivatives...

[tex]v'=5(x^3)'+(1)'[/tex]

Apply power rule for derivatives (decrease power by 1; mutliply old power as a factor to the coefficient); Derivative of a constant is zero...

[tex]v'=5(3x^2)+0[/tex]

Simplify...

[tex]v'=15x^2[/tex]

So, option "d"

Part 3.

Given [tex]y=(3x^2+2)(5x^3+1)[/tex]

[tex]\text{Then if } u=3x^2+2 \text{ and } v=5x^3+1, y=u*v[/tex]

To find [tex]\frac{dy}{dx} \text{ or } y'[/tex], recall the product rule:  [tex]y'=uv'+u'v[/tex]

[tex]y'=uv'+u'v[/tex]

Substituting the expressions found from above...

[tex]y'=(3x^2+2)(15x^2)+(6x)(5x^3+1)[/tex]

Apply the distributive property...

[tex]y'=(45x^4+30x^2)+(30x^4+6x)[/tex]

Use the associative and commutative property of addition to combine like terms, and rewrite in descending order:

[tex]y'=75x^4+30x^2+6x[/tex]

So, option "b"

The school bus will take 5.34 seconds to travel 95.1 m if the bus is moving 26.8 m/s on flat ground when it begins to accelerate at 4.73 m/s².

Any equation of the form [tex]\rm ax^2+bx+c=0[/tex] where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

As we know, the formula for the roots of the quadratic equation is given by:

[tex]\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}[/tex]

We have:

A school bus is moving 26.8 m/s on flat ground when it begins to accelerate at 4.73 m/s².

From the second equation of motion:

[tex]\rm s = u + \dfrac{1}{2}at^2[/tex]

We have given:

s = 95.1 m

u = 26.8 m/s

a = 4.73 m/s²

[tex]\rm 95.1 = 26.8 + \dfrac{1}{2}(4.73)t^2[/tex]

The above equation is a quadratic equation:

After simplification:

[tex]\rm \dfrac{4.73}{2}t^2=68.3[/tex]

[tex]\rm t^2=28.879[/tex]

t = 5.373 seconds ≈ 5.34 seconds

Thus, the school bus will take 5.34 seconds to travel 95.1 m if the bus is moving 26.8 m/s on flat ground when it begins to accelerate at 4.73 m/s².

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