Evaluate the expression
a. tan(arctan 25) =

The principal value of arc sin(0.98)=78.5°.

The value of the inverse function at a place that falls within the range of the unit of the principal value is the primary value of the trigonometric function at that point.

Trigonometric function solutions with a value between 0 and 2π are known as principal values of trigonometric functions. The primary values of trigonometric functions are values smaller than 2π, which is the interval at which the trigonometric function's value repeats.

The principal value of the inverse trigonometric function is always the positive value whenever any positive value and the negative value are presented in a way that these two values are equal.

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Answer:

[tex](x+1)(x-3)[/tex]

Step-by-step explanation:

A quadratic in factored form is usually expressed as: [tex]a(x\pm a)(x \pm b)[/tex] where the sign of a and b depends on the sign of the zero. And I said "usually" since sometimes the x will have a coefficient. Anyways in the quadratic there are two zeroes at x=-1 and x=3. This can be written as: [tex]a(x+1)(x-3)[/tex]. Notice how the signs are different? This is because when you plug in -1 as x you get a factor of (-1+1) which becomes 0 and it makes the entire thing zero since when you multiply by 0, you get 0. Same thing for the x-3 if you plug in x=3. Now a is in front and it can influence the stretch/compression. To find the value of a, you can take any point (except for the zeroes, because it will make the entire thing zero, and you can technically input anything in as a)

I'll use the point (1, -4) the vertex

-4 = a(1+1)(1-3)

-4 = a(2)(-2)

-4 = -4a

1 = a. So yeah the value of a is 1

So the equation is just: [tex](x+1)(x-3)[/tex]

By applying the concept of finite sum and the properties of the series, we conclude that the finite sum [tex]p = \sum \limit_{x=1}^{3} 2^{x}+ 2[/tex] is equal to the value of 34.

Let be a sum of the form [tex]p = \sum\limits_{i= 1}^{n} a_{i}[/tex], where n represents an integer. This kind of sum represents a finite sum, whose expanded form is presented below:

p = a₁ + (a₁ + a₂) + (a₁ + a₂ + a₃) + ... + (a₁ + a₂ + a₃ + ... + aₙ₋₁ + aₙ)     (1)

If we know that [tex]p = \sum \limit_{x=1}^{3} 2^{x}+ 2[/tex], then the result of the finite sum is:

p = (2 + 2¹) + (2 · 2 + 2¹ + 2²) + (3 · 2 + 2¹ + 2² + 2³)

p = 4 + (4 + 6) + (6 + 14)

p = 4 + 10 + 20

p = 34

By applying the concept of finite sum and the properties of the series, we conclude that the finite sum [tex]p = \sum \limit_{x=1}^{3} 2^{x}+ 2[/tex] is equal to the value of 34.

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Answer:

(f+g)(x)=f(x)+g(x)

=(-3x-5)+(4x-2)

= -3x-5+4x-2. collect like term

= -3x+4x-5-2

= x-7. the final answer

The product of the given function is x^5 + 5x^4 + 7x^3

The leading degree of a polynomial function is always greater than or equal to 2.

Given the products below;

x^3 (x^2 + 5x + 7)

Expand

(x^3)(x^2) + 5x(x^3) + 7x^3

Simplify

x^5 + 5x^4 + 7x^3

Hence the product of the given function is x^5 + 5x^4 + 7x^3

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The equivalent expression of  x³(x² + 5x + 7) is  x⁵ + 5x⁴ + 7x³

The equivalent expression of the expression can be found as follows;

x³(x² + 5x + 7)

let's open the bracket by multiplying.

Therefore,

x³(x² + 5x + 7) = x⁵ + 5x⁴ + 7x³

Hence,

The equivalent expression of  x³(x² + 5x + 7) is  x⁵ + 5x⁴ + 7x³

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The balanced equation for the given reaction [tex]\mathbf{AgNO_3+NaCl \to AgCl_{(s)} + NaNO_3}[/tex]

A double replacement is a chemical reaction that takes place in which two reactants swap cations or anions to produce two distinct products.

Whenever the cations representing one of the reactants interact with the anions from another reactant to produce an insoluble ionic compound, a precipitate occurs.

The balanced equation for the given reaction:

[tex]\mathbf{AgNO_3+NaCl \to AgCl_{(s)} + NaNO_3}[/tex]

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Based on Pythagoras' theorem, EB is equal to 38.7 and DY is 17.7.

The length of the sides EB and DY are obtained using Pythagoras theorem.

Since EY is a perpendicular bisector;

EB = √(64.2² - 51.2²)

EB = 38.7

Since, DY is a perpendicular bisector;

DY = √64.2² - 61.7²)

DY = 17.7

Therefore, EB is equal to 38.7 and DY is 17.7.

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Answer:

10.04°F

Step-by-step explanation:

(12.8°C × 9/5) + 32 = 55.04 - 45 = 10.04

Answer:

No.

Step-by-step explanation:

The ratio of the lengths of the vertical sides from A to B is 2/4 which equals 1/2.

The ratio of the lengths of the horizontal sides of A to B is 4/11.

Since 4/11 is not equal to 1/2, figure B is not a scale copy of figure A.

Answer: Option (4)

Step-by-step explanation:

We need factors of [tex](x+2), (x+1), (x-4)[/tex]

The computation shows that the value of the length of ZY will be B. 24.

Let the assumed figures be:

BZ = 48

BA = 12

WV = 24

ZY = Unknown

Let's assume that the opposite sides in the quadrilateral are the same, the value of ZY will be:

ZY = (48 × 12)/24

ZY = 24

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Answer:

C.12

Step-by-step explanation:

Trust

The graph that can represent the area of each rectangle in terms of the change in the length and width is C. A graph shows change in dimensions (units) labeled 1 to 10 on the horizontal axis and area (square units) on the vertical axis. A line increases from 0 to 3 seconds then decreases from 3 to 9 seconds.

It should be noted that a graph gives a diagrammatic representation of an information.

Here, the rectangles are formed by decreasing the length and increasing the width, each by the same amount.

Thai can be illustrated by the third graph in the option as the line increases from 0 to 3 seconds then decreases from 3 to 9 seconds.

In conclusion, the correct option is C.

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Answer:

(っ◔◡◔)っ ♥ the answer is C ♥

Step-by-step explanation:

got it right on edge

Answer:

[tex]39 yd^2[/tex]

Step-by-step explanation:

So, let's start by calculating the surface area of the base, which in this case is a square. The area of a square can be calculating by taking any of the lengths and squaring it, since all the sides should be equal if it's a square. Since one of the sides of the square is 3 yd, you have the equation: [tex](3 yd)^2 = 9yd^2[/tex]. Now to calculate the area of the four triangles. The area of a triangle can be calculating by using the formula: [tex]\frac{1}{2}bh[/tex], and in this case the base is 3 yd, and the height is 5 yd (lengths can be determined by looking at the given values in the diagram). But there's 4 of the triangles so we multiply it all by 4, and this gives you the equation: [tex]4(\frac{1}{2}(5 yd)(3 yd)) = 4(\frac{15 yd^2}{2}) = 4(7.5 yd^2)=30yd^2[/tex]. So now all that's left is to add this to the area of the square which gives you the equation: [tex]30yd^2+9yd^2=39yd^2[/tex] which is the surface area

Answer:

10 years old or 120 months old

Answer:

The function of j(t) is decreasing.

Step-by-step explanation:

In the question two functions are given as h(t)=3t-4 and j(t)=5-t.

It is required to find whether j(t) is an increasing/decreasing or neither function.

To solve this question, first find the slope of the function. If, the slope is positive the function is increasing and if the slope is negative the function is decreasing.

Step 1 of 1

Find the slope of the function j(t) by comparing these to the standard form.

y=mx+b

Then, m=-1

Since the slope is negative.

Therefore, the function is decreasing.

Answer:

See below for step-by-step proof.

Step-by-step explanation:

[tex]\textsf{Given expression}:[/tex]

[tex]\cos^6 \theta+\sin^6 \theta[/tex]

[tex]\textsf{Apply exponent rule} \quad a^{bc}=(a^b)^c:[/tex]

[tex]\implies (\cos^2 \theta)^3+(\sin^2 \theta)^3[/tex]

[tex]\textsf{Use the identity}\quad a^3+b^3=(a+b)^3-3ab(a+b):[/tex]

[tex]\implies (\cos^2 \theta+\sin^2 \theta)^3-3 \cos^2 \theta\sin^2 \theta (\cos^2 \theta+\sin^2 \theta)[/tex]

[tex]\textsf{Use the identity}\quad \sin^2 \theta + \cos^2 \theta=1:[/tex]

[tex]\implies (1)^3-3 \cos^2 \theta\sin^2 \theta (1)[/tex]

[tex]\implies 1-3 \cos^2 \theta\sin^2 \theta[/tex]

[tex]\implies 1-3 (\cos \theta\sin \theta)^2[/tex]

[tex]\textsf{Use the identity}\quad \sin 2 \theta=2 \sin \theta \cos \theta \implies \dfrac{1}{2}\sin 2 \theta= \sin \theta \cos \theta:[/tex]

[tex]\implies 1-3 \left( \dfrac{1}{2}\sin 2 \theta\right)^2[/tex]

[tex]\implies 1-\dfrac{3}{4} \sin^2 2 \theta[/tex]

[tex]\textsf{Use the identity}\quad \sin^2 2\theta + \cos^2 2\theta=1 \implies \sin^2 2\theta=1-\cos^2 2\theta[/tex]

[tex]\implies 1-\dfrac{3}{4} \left(1-\cos^2 2 \theta\right)[/tex]

[tex]\implies 1-\dfrac{3}{4} +\dfrac{3}{4}\cos^2 2 \theta[/tex]

[tex]\implies \dfrac{1}{4} +\dfrac{3}{4}\cos^2 2 \theta[/tex]

[tex]\textsf{Factor out }\dfrac{1}{4}:[/tex]

[tex]\implies \dfrac{1}{4}\left(1+3\cos^2 2 \theta\right)[/tex]

[tex]\textsf{Use the identity}\quad \cos (4 \theta)=2 \cos^2 2\theta - 1 \implies \cos^2 2 \theta=\dfrac{1}{2}\cos 4 \theta+\dfrac{1}{2}:[/tex]

[tex]\implies \dfrac{1}{4}\left(1+3\left(\dfrac{1}{2}\cos 4 \theta+\dfrac{1}{2}\right)\right)[/tex]

[tex]\implies \dfrac{1}{4}\left(1+\dfrac{3}{2}\cos 4 \theta+\dfrac{3}{2}\right)[/tex]

[tex]\implies \dfrac{1}{4}\left(\dfrac{5}{2}+\dfrac{3}{2}\cos 4 \theta\right)[/tex]

[tex]\implies \dfrac{1}{4} \cdot \dfrac{1}{2}\left(5+3\cos 4 \theta\right)[/tex]

[tex]\implies \dfrac{1}{8}\left(5+3\cos 4 \theta\right)[/tex]

The number of Red supporters are 345

From the given information, we have that

Red supporters = 1/3 x

Let the total number of supporter be x

1/3 x + blues = x

1/3x + 1/3x + 345 = x

2/3x + 345 = x

Collect like terms

[tex]x - \frac{2x}{3} = 345[/tex]

[tex]\frac{1}{3} x = 345[/tex]

Cross multiply

[tex]x = 3[/tex] × [tex]345[/tex]

[tex]x = 1035[/tex]

We have red supporters = 1/3x

⇒ [tex]\frac{1}{3}[/tex] × [tex]1035[/tex]

⇒ [tex]345[/tex]

Thus, the number of Red supporters are 345

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The volume of the cylinder whose radius is 5x-1 unit is 25x²πh + πh - 10xπh.

Suppose that the radius of the considered right circular cylinder is 'r' units.

And let its height be 'h' units.

Then, its volume is given as:

[tex]V = \pi r^2 h \: \rm unit^3[/tex]

The right circular cylinder is the cylinder in which the line joining centre of the top circle of the cylinder to the centre of the base circle of the cylinder is perpendicular to the surface of its base, and to the top.

Given that the radius of the cylinder is (5x-1), while the height of the cylinder is h. Therefore, the volume of the cylinder is,

The volume of the cylinder = π × (5x-1)² × h

                                            = (25x² + 1 - 10x)πh

                                            = 25x²πh + πh - 10xπh

Hence, the volume of the cylinder is 25x²πh + πh - 10xπh.

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By evaluating in x = 25 we get:

Here we know that the measure of angle 1 is given by:

∠1 = 5x - 5

And the given value of x is 25, then if we just evaluate the above expression, then we get:

∠1 = 5*25 - 5 = 120

And for angle 2 we will have:

∠2 = 2*25 - 10 = 60

Notice that the sum of these two angles is equal to 180°, which is what we should get when adding two supplementary angles.

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The value of b is [tex]1 \cdot02[/tex].

What is compound interest?

Compound interest is when you earn interest on both the money you have saved and the interest you earn.

Formula for compound interest is

[tex]A= P(1+\frac{r}{100})^n[/tex] ...................(1)

where, A = total amount( principal + interest )

r = rate of interest in compound interest

n = number of years

Given,

For [tex]n=1[/tex]

Principal amount [tex]p=\$ 1000[/tex]

Total amount [tex]A=\$ 1020[/tex]

For [tex]n=2[/tex]

Principal amount [tex]p=\$ 1000[/tex]

Total amount [tex]A=\$ 1040[/tex]

This scenario can be represented by an exponential function of the form of [tex]f(x)=1000(b)^x[/tex]

Comparing the above function with equation (1), we get

[tex]A=f(x)\\P=1000\\b=(1+\frac{r}{100})[/tex]

For 1st year

[tex]f(x)=1000(b)^1\\\Rightarrow 1020=1000(b)\\\Rightarrow b=\frac{1020}{1000}\\\Rightarrow b= \frac{102}{100}\\\Rightarrow b= 1\cdot02[/tex]

now, check the amount for 2nd year

[tex]f(x)=1000 \times (1 \cdot01}^2\\\Rightarrow f(x)=1000 \times \frac{102}{100}\times \frac{102}{100}\\\Rightarrow f(x)=1040[/tex]

Hence, the value of b is [tex]1 \cdot02[/tex] .

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Answer:

D) 72°

Step-by-step explanation:

Tangent:

[tex]\sf \boxed{\angle PQR =\dfrac{Difference \ between \ major \ arc \ and \ minor \ arc }{2}}[/tex]

       ∠PQR = 2x + 252 - (2x + 108)

      2x +72 = (2x + 252 - 2x - 108) ÷ 2

      2x + 72 = (2x - 2x + 252 - 108) ÷ 2  

     2x + 72 = 144 ÷ 2

      2x + 72 = 72

2x + 72-72 = 72 - 72

              2x = 0

             [tex]\sf \boxed{\bf x = 0}[/tex]

m∠PQR = 2x + 72

             = 2*0 + 72

             = 72°

Answer: [tex]f^{-1} (x)=\frac{4x-5}{2}[/tex]

Step-by-step explanation:

Let [tex]f(y)=x\\[/tex].

[tex]\implies x=\frac{2y+5}{4}\\\\4x=2y+5\\\\4x-5=2y\\\\y=\frac{4x-5}{2}\\\\\therefore f^{-1} (x)=\frac{4x-5}{2}[/tex]

The office concluded that the driver exceeded the speed limit by 64.3mi/h.

Given that the second cop, who arrived 28 minutes later, was recorded traveling at 60 mph while the patrol officer was traveling 30 miles along the roadway.

According to the mean value theorem, There exists a value of x = c such that f'(c) = [f(b)-f(a)][b-a] if f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b).

This implies the following in "lay man's" terms:

the instantaneous speed (the speed at any given time) must be equal to his average speed

Since the driver is driving along the highway, we can assume his position to be continuous and differentiable

his average speed is defined as:

average speed=[x(b)-x(a)]÷[b-a] where x represents his position

average speed=[30 - 0]÷[(28÷60) - 0]

average speed=30÷0.466

average speed=64.3mi/hr

Since the time is given in minutes, we convert it in hours by dividing it by 60.

Therefore, by the MVT the police officers can determine that at some point in time (even though he was only driving 60mph at the second patrol officer's location) since his average speed was approximately 64.3 mi/hr there was a point in time during the 28 minutes that his speed exceeded 60 mph.

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Answer:

Around 12.25

Step-by-step explanation:

There are around 2205 (2204.62) lbs  per metric tonnes.

So 27000/2204.62  or 27000/2205 gives you between 12.24-12.25

Answer: 11

Step-by-step explanation:

[tex]\sin 42.51^{\circ}=\frac{x}{16}\\\\x=16 \sin 42.51^{\circ} \approx 11[/tex]

you can reduce it to factors of degree 2 or less).

The two dice do not need to be identical. If it is possible, do it. If it is not possible, prove that it is not possible.

Some hints I got were to try to construct a pair of nonstandard dice such that the generating function for their sum matches the corresponding g.f. for a pair of standard dice and to note that the g.f. for a pair of standard dice can be factored quite a bit (you can reduce it to factors of degree 2 or less).

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