What’s the answer? How do I find the opposite

Answer:

sin C = cos A

Step-by-step explanation:

sinC = opp/hyp

=12/13

cosA = adj/hyp

12/13

Answer:

Tax 3.78 $ and tip 7.56 $

Step-by-step explanation:

42*0.09 = 3.78

42*0.18 = 7.56

Their speed in terms of square feet per minute is 3/10 square feet per minute.

We know that the speed formula is the rate at which an object travels a given distance. Speed is defined as the distance traveled by a body in a given amount of time. Speed in the SI is measured in m/s.

Given that the distance = 1/5 square feet.

The time = 2/3 minute.

Speed = distance/time = (1/5)/(2/3) = (1/5)*(3/2) = 3/10

Therefore, we can conclude that their speed in terms of square feet per minute is 3/10 square feet per minute.

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Answer:

Option (C)

Step-by-step explanation:

The minimum value of x² is 0, and the maximum value is unbounded, so therefore, the maximum value of -x² is 0, and the minimum value is unbounded.

So, this means that adding 1 to this, the range matches with option C.

Answer:

x = 6

Step-by-step explanation:

.16x + 2 * .02 = (x + 2) * .17
.16x + .4 = .17x + .34
.06 = .01x
6 = x

The amount of oil required for 1 liter of fuel is 111.11 mL.

Fraction is the portion of a total amount where the above part of the fraction is the denominator and the bottom part of the fraction is called the numerator.

Given that the fuel is the mixture where 8 parts are petrol and 1 part is oil in the whole part of the fuel.

The total part of the fuel is 8+1=9

the portion of the petrol is = parts of petrol/total parts of the fuel= 8/9

the portion of the oil = parts of oil /total parts of the fuel= 1/9

Now we have to calculate the amount of oil required for 1 liter of fuel.

As discussed before, 1/9 parts of the fuel is oil.

So the amount of oil is= (1/9)*1 liter= (1/9)litre= 1/9* 1000 mL= 111.11 mL

Similarly, we can calculate the amount of petrol which will be

the amount of petrol= (8/9)*1 liter= (8/9) liter= 8/9*1000 mL= 888.88 mL

Therefore the amount of oil required for 1 liter of fuel is 111.11 mL.

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The set A satisfying the given inequality is A = (-[tex]\infty[/tex], -10].

Following are some facts which are true for an inequality relation:

[tex]\frac{5x-2}{8} - \frac{3x-5}{10} &\ge& x+y\\\\\Rightarrow\;\; \frac{13}{40}x + \frac{1}{4}&\ge& x+y\\\\\Rightarrow\;\;\;\;\;\; -\frac{27}{40}x &\ge & y - \frac{1}{4}\\\\\Rightarrow\;\;\;\;\;\;\;\;\;\; -x &\ge & \frac{40}{27}\left( y-\frac{1}{4} \right).\hspace{1cm}(1)[/tex]

Since y ∈ B, -2 ≤ y ≤ 7. So,

[tex]\;\;\;\;\;\;\;\,-2 - \frac{1}{4}\; \le\; y - \frac{1}{4} \;\le\; 7 - \frac{1}{4}\\\\\Rightarrow\;\;\;\;\;\;\;\;\; -\frac{9}{4}\; \le\; y - \frac{1}{4} \;\le\; \frac{27}{4}\\\\\Rightarrow\;\;\; -\frac{9}{4}\cdot \frac{40}{27} \;\le\; \frac{40}{27} \left( y-\frac{1}{4} \right) \;\le \;\frac{27}{4}\cdot \frac{40}{27}\\\\\Rightarrow\;\;\;\;\;\;\;\; -\frac{10}{3} \;\le\; \frac{40}{27}\left( y - \frac{1}{4} \right)\; \le\; 10.[/tex]

The set {-x | inequality (1) holds ∀ y ∈ B} is [10, [tex]\infty[/tex]) i.e.

10 ≤ -x ≤ [tex]\infty[/tex].

Multiplying -1 throughout gives

-10 ≥ x ≥ -[tex]\infty[/tex].

x, thus, lies in the range A = (-[tex]\mathbf{\infty}[/tex], -10}.

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Disclaimer: The question was incomplete. Please find the full content below.

Find the set A such that for x ∈ A

[tex]\frac{5x - 2}{8} - \frac{3x - 5}{10} \ge x + y[/tex]

∀y ∈ B = {y ∈ R | -2 ≤ y ≤ 7}.

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If we put a=0 and b=-4.05 then the expression 4a+4b becomes -16.2 and when we put a=-4.05 and b=0 then the expression becomes -16.2.

Given  2 (a + b) = -8.1

Because no relationship between a and b is given so we can put any value of one variable to determine the value of another variable.

If we put a=0 in 2(a+b) =-8.1 then b=-4.05

If we put b=0 then a =-4.05.

We need to put these values in 4a+4b

when a=0 b=-4.05

4a+4b=4*0+4*-4.5

=-16.2

when b=0 and a=-4.05

4a+4b=4*-4.05+4*0

=-16.2

Hence for a=0,b=-4.5 the value of 4a+4b equals -16.2 and for a=-4.05 and b=0 the function equals -16.2.

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For this question, I will be writing 'csc' instead of 'cosec'

[tex]\sin 30^{\circ}=\frac{1}{2} \implies \csc 30^{\circ}=2\\\\\sin 45^{\circ}=\frac{1}{\sqrt{2}}\\\\\cos 60^{\circ}=\frac{1}{2} \implies \sec 60^{\circ}=2\\\\\\\therefore \csc^{2} 30^{\circ}-\sin^{2} 45^{\circ}-\sec^{2} 60^{\circ}=2^{2}-\left(\frac{1}{\sqrt{2}} \right)^{2}-2^{2}=\boxed{-\frac{1}{2}}[/tex]

Answer:

12

Step-by-step explanation:

Direct Variation. I think it's gonna be like this.

SInce Z is not inversely to y , its varies directly to both.

Then it's,

Z=kxy, where k is constant.

subst. Z=6,x=2 and y=6.

6=k×2×6

6=k×12=12k

6=12k=½

the formula connecting them is

Z=½xy

Z=½×3×8

Z= 1×3×8÷2

Z=24÷2

Z=12.

That's my solution.

A function assigns the values. The function g(x) can be written as g(x) = -3x + 12.

A function assigns the value of each element of one set to the other specific element of another set.

Given the function f(x)=-x, now since a transformed function g(x) is produced by shifting the function 4 units upwards and vertical stretch by a factor of 3. Therefore, g(x) can be written as,

Vertical Shift by 4 units,

f(x)+4

Vertical stretch by a factor of 3

g(x) = 3(f(x) + 4)

g(x) = -3x + 12

Hence, the function g(x) can be written as g(x) = -3x + 12.

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Rationalizing the numerator, it is found the result of the limit is given as follows:

[tex]\lim_{h \rightarrow 0} \frac{(\sqrt{x + h} - \sqrt{x})}{h} = \frac{1}{2\sqrt{x}}[/tex]

A limit is given by the value of function f(x) as x tends to a value.

The first step to find the limit is replace the variable by the value, hence:

[tex]\lim_{h \rightarrow 0} \frac{(\sqrt{x + h} - \sqrt{x})}{h} = \frac{0}{0}[/tex]

Undefined limit, hence we have to find another way. The numerator includes square roots, hence it should be rationalized using the subtraction of perfect squares, as follows:

[tex]\lim_{h \rightarrow 0} \frac{(\sqrt{x + h} - \sqrt{x})}{h} \times \frac{(\sqrt{x + h} + \sqrt{x})}{(\sqrt{x + h} + \sqrt{x})} = \lim_{h \rightarrow 0} \frac{(\sqrt{x + h})^2 - (\sqrt{x})^2}{h(\sqrt{x + h} + \sqrt{x})} = \lim_{h \rightarrow 0} \frac{x + h - x}{h(\sqrt{x + h} + \sqrt{x})} = \lim_{h \rightarrow 0} \frac{h}{h(\sqrt{x + h} + \sqrt{x})}[/tex]

Then we simplify the h, so:

[tex]\lim_{h \rightarrow 0} \frac{1}{\sqrt{x + h} + \sqrt{x}} = \frac{1}{2\sqrt{x}}[/tex]

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Answer:

[tex]|{\sf XY}| = 10\; {\rm cm}[/tex].

Step-by-step explanation:

Refer to the diagram attached. The dashed segment attached to [tex]\!{\sf Z}[/tex] points to the north. Rotating this segment clockwise with point [tex]{\sf Z}\!\![/tex] as the fixed center of rotation would eventually align this segment with the one between point [tex]\!\!{\sf Z}[/tex] and point [tex]\!\!{\sf X}[/tex]. The bearing of point [tex]{\sf X}[/tex] from point [tex]{\sf Z}[/tex] is the size of the angle between these two line segments when measured in the clockwise direction.

Subtract the bearing of [tex]{\sf Y}[/tex] from [tex]{\sf Z}[/tex] from the bearing of [tex]{\sf X}[/tex] from [tex]{\sf Z}[/tex] to find the measure of the angle [tex]\angle {\sf YZX}[/tex]:

[tex]\begin{aligned}\angle {\sf YZX} &= 135^{\circ} - 45^{\circ} \\ &= 90^{\circ}\end{aligned}[/tex].

Thus, triangle [tex]\triangle {\sf YZX}[/tex] is a right triangle ([tex]90^{\circ}[/tex]) with segment [tex]{\sf YX}[/tex] as the hypotenuse. It is given that [tex]|{\sf XZ}| = 6\; {\rm cm}[/tex] whereas [tex]|{\sf ZY}| = 6\; {\rm cm}[/tex]. Thus, by Pythagorean's Theorem:

[tex]\begin{aligned}|{\sf ZY}| &= \sqrt{|{\sf ZX}|^{2} + |{\sf ZY}|^{2}} \\ &= \sqrt{(8\; {\rm cm})^{2} + (6\; {\rm cm})^{2}} \\ &= 10\; {\rm cm}\end{aligned}[/tex].

Answer:

b step by step explanation

The function that represents exponential growth is:

f(x) = x³

An exponential function's curve is created by a pattern of data called exponential growth, which exhibits higher increases over time.

y = a(1+r)ˣ,

where a is the initial population and r is the rate in decimals and x is the time period.

Exponential growth is a phenomenon that occurs when the rate of change of a quantity is proportional to its current value. In other words, as the value of the quantity increases, the rate at which it increases also increases. This can be represented mathematically using an exponential function of the form f(x) = abˣ, where a is the initial value, b is the growth factor, and x is the time or number of periods.

The function that represents exponential growth is:

f(x) = x³

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We write the equation in the form of directional.

y -1 = 6x               ⇔    y = 6x + 1

y - 1 = 3x              ⇔    y = 3x + 1

y - 7 = 2x - 6         ⇔    y = 2x - 6 + 7

                                    y = 2x + 1

y - 7 = x - 2           ⇔    y = x - 2 + 7

                                    y = x + 5

Equations cleverly arranged .

Point Q = (0,1)  

b factor , not only fits the last equation

In the drawing have engraved points Q and R are tangent linear function appropriate to that point . This graphics solution . y = 3x + 1

Answer b

We check choice by the system of equations , where substitute wartoćsi points Q and R to the model equations linear function

The result of equations confirmed our choice Answer b

Answer: 19

Step-by-step explanation:

g(4) = 4² + 3

g(4) = 16 + 3

g(4) = 19

Answer:

3 seconds

proofi took the test

The object reaches its maximum height after 3 seconds.

The correct option is 3 seconds.

To find the time at which the object reaches its maximum height, we need to determine the vertex of the parabolic function[tex]h(t) = -16t^2 + 96t + 6[/tex] . The vertex form of a parabola is given by [tex]h(t) = a(t - h)^2 + k[/tex], where (h, k) represents the vertex.

In this case, a = -16, so the vertex is given by t = -b/2a. Plugging in the values, we get t = -96/(2 x (-16)) = 96/32 = 3 seconds.

Therefore, the object reaches its maximum height after 3 seconds.

Option B, 3 seconds, is the correct answer.

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The complete Question:

The function [tex]h(t) = -16t^2 + 96t + 6[/tex] represents an object projected into the air from a cannon. The maximum height reached by the object is 150 feet.

After how many seconds does the object reach its maximum height?

A. 2 seconds

B. 3 seconds

C. 6 seconds

D. 9 seconds

Using the continuity concept, as the value of the function has to be equal to the limit, it is found that the value of k is of k = -4.

A function is said to be continuous at x = a if, and only if:

[tex]\lim_{x \rightarrow a} f(x) = f(a)[/tex]

In this problem, we have to check at x = -2, hence:

[tex]\lim_{x \rightarrow -2} f(x) = f(-2)[/tex]

[tex]k = \lim_{x \rightarrow -2} f(x)[/tex]

For the limit, we have to check the lateral conditions, hence:

[tex]\lim_{x \rightarrow -2} f(x) = \lim_{x \rightarrow -2} \frac{x^2 - 4}{x + 2} = \lim_{x \rightarrow -2} \frac{(x - 2)(x + 2)}{x + 2} = \lim_{x \rightarrow -2} x - 2 = -2 -2 = -4[/tex]

Hence, from the continuity condition, k = -4.

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Answer:

1, 17, 33, 49

Step-by-step explanation:

given the first term is 1 then the next 3 terms are

1 + d, 1 + 2d, 1 + 3d ( d is the common difference )

the sum of the first 4 terms is 100 , then

1 + 1 + d + 1 + 2d + 1 + 3d = 100 , that is

4 + 6d = 100 ( subtract 4 from both sides )

6d = 96 ( divide both sides by 6 )

d = 16

1 + d = 1 + 16 = 17

1 + 2d = 1 + 2(16) = 1 + 32 = 33

1 + 3d = 1 + 3(16) = 1 + 48 = 49

the first 4 terms are

1, 17, 33, 49

Answer:

see the file attached!!!!

The cost of renting a community center for x guest is given as y = 10x + 100

An equation is an expression that shows the relationship between two or more variables and numbers.

The standard linear equation is in the form:

y = mx + b

Where m is the rate of change and b is the y intercept.

Let y represent the cost of renting a community center for x guest. Hence:

y = 10x + 100

The cost of renting a community center for x guest is given as y = 10x + 100

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[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

Expression in scientific notation :

[tex]\qquad \tt \rightarrow \:3.45 × {10}^{-4} [/tex]

____________________________________

[tex] \large \tt Solution \: : [/tex]

[tex]\qquad \tt \rightarrow \: 0.000345[/tex]

[tex]\qquad \tt \rightarrow \: 0.000345 \times \cfrac{10000}{10000} [/tex]

[tex]\qquad \tt \rightarrow \: (0.000345 \times 10000) \times \cfrac{1}{10000} [/tex]

[tex]\qquad \tt \rightarrow \: 3.45 \times \cfrac{1}{10 {}^{4} } [/tex]

[tex]\qquad \tt \rightarrow \: 3.45 \times{10 {}^{ - 4} } [/tex]

[tex] \textsf{Correct option - D} [/tex]

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

Answer:

3.45 x 10-4

Step-by-step explanation:

Answer:

y = x/2 -3

Step-by-step explanation:

x - 2y = 6

Slope intercept form is y = mx+b  where m is the slope and b is the y intercept

We need to solve for y

Subtract x from each side

x-2y-x = -x+6

-2y = -x+6

Divide each side by -2

-2y/-2 = -x/-2 +6/-2

y = x/2 -3

The slope intercept form is

y = x/2 -3

Answer: slope intercept form is

y = x/2 -3

Step-by-step explanation:

hope this helps ; )

Answer:

B

Step-by-step explanation:

using the addition identity for sine

sin(A + B) = sinAcosB + cosAsinB

using the sine and cosine ratios in the right triangle

sinABC = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{AC}{AB}[/tex] = [tex]\frac{4}{5}[/tex]

cosABC = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{BC}{AB}[/tex] = [tex]\frac{3}{5}[/tex]

using the exact values

sin60° = [tex]\frac{\sqrt{3} }{2}[/tex] , cos60° = [tex]\frac{1}{2}[/tex]

Then

sin(ABC + 60)

= sinABC cos60 + cosABC sin60

= ( [tex]\frac{4}{5}[/tex] × [tex]\frac{1}{2}[/tex] ) + ([tex]\frac{3}{5}[/tex] × [tex]\frac{\sqrt{3} }{2}[/tex] )

= [tex]\frac{4}{10}[/tex] + [tex]\frac{3}{10}[/tex] [tex]\sqrt{3}[/tex]

= [tex]\frac{2}{5}[/tex] + [tex]\frac{3}{10}[/tex] [tex]\sqrt{3}[/tex]

sin<ABc

cos<ABC

Now

Answer:

Step-by-step explanation:

To calculate a+1/a, we first need to calculate for a.

a = 7 - 4 * [tex]\sqrt[2]{3}[/tex]

square root of 3 = 1.73

a = 7 - 4 * 1.73

a = 7 - 6.9

a = 0.1

0.1 + 1 / 0.1 = 0.1 + 10 = 10.1

Now, in case 4 wasn't being multiplied with the square root of 3, and instead, it was four root of 3, I am gonna do the calculations again:

a = 7 - [tex]\sqrt[4]{3}[/tex]

a = 7 - 1.31

a = 5.69

5.69 + 1 / 5.69 = 5.69 + 0.17 = 5.86

Hope I Helped!

The simplified form of the expression has a numerator (x + 9)(2x + 1) and the denominator (x + 7) and the expression doesn't exist at (x = 9).

The given expression is:

= [(3x² - 27x)/(2x² + 13x - 7)]/(3x/4x² - 1)

Firstly, factorize the expression 4x² - 1.

4x² - 1 = (2x - 1)(2x + 1)

Then, factorize 2x² + 13x - 7.

2x² + 13x - 7 = 2x² + 14x - x - 7

2x(x + 7) - 1(x + 7)

= (2x - 1)(x + 7)

Then, factorize the equation 3x² - 27x.

3x² - 27x = 3x(x - 9)

The factorized terms will be substituted into the equation. This will be:

= [3x(x - 9)/(2x - 1)(x + 7)] / 3x[(2x - 1)(2x + 1)]

= (x + 9)(2x + 1)/(x + 7)

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Based on the given probability events, we can calculate that Events A and B are independent, because P(B) = P(B|A) = 1/2.

Events are independent if P(B) = P(B|A)

The probability of event B is:

= 3 odd numbers / 6 numbers

= 1 / 2

The probability of A is:

= 3 prime numers / 6

= 1/2

P(B|A) = ((1 / 2) x (1 / 2)) / (1 / 2)

= 1/2

So P(B) = P(B|A) = 1/2.

The events are independent.

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Answer:

The answer will be 38° + 32° = 70°

Please mark me as the brainliest

Answer:

Step-by-step explanation:

1)   = f(x - a) + b

Coordinate change

(x, y) → (x + a, y + b)

2) RFy(x, y) = f(-x)

Coordinate change

(x, y) → (-x, y)

3) RFx(x, y) = -f(x)

Coordinate change

(x, y) → (-y, x)

4) RCCW90(x, y) = f⁻¹(-x)

Coordinate change

(x, y) → (-y, x)

5) RCCW180(x, y) = -(f(-x))

Coordinate change

(x, y) → (-x, -y)

6) A 270 degrees counterclockwise rotation gives;

RCCW270(x, y) = -(f⁻¹(x))

Coordinate change

(x, y) → (y, -x)

Step-by-step explanation:

1) Horizontal translation a units right = f(x - a)

The vertical translation b units up = f(x) + b

Therefore, we get;  = f(x - a) + b

The coordinate change

(x, y) → (x + a, y + b)

2) A reflection across the y-axis = RFy(x, y) = f(-x)

The coordinate change

(x, y) → (-x, y)

3) A reflection across the x-axis gives RFx(x, y) → (x, -y)

Therefore, in function notation, we get;

RFx(x, y) = -f(x)

4) A 90 degrees rotation counterclockwise, we get RotCCW90(x, y) → (-y, x)

In function notation RotCCW90(x, y) = INVf(-x) = f⁻¹(-x)

5) A 180 degrees counterclockwise rotation about the origin gives;

(x, y) → (-x, -y)

Therefore, we get;

In function notation RotCCW180(x, y)  = -(f(-x))

6) A 270 degrees counterclockwise rotation gives RotCCW270(x, y) → (y, -x)

In function notation RotCCW270(x, y) = -(f⁻¹(x))