Which of the following expressions is equal to tan205°?
tan55°
tan25°
tan25°

Answer:

2sqrt3

Step-by-step explanation:

Since this seems to be a 45, 45, 90 triangle, x and y are the same.

The hypoteneuse is always the side lengths *sqrt2

We divide the hypoteneuse by sqrt 2 and get sqrt12

sqrt12 simplified is 2sqrt3

Answer:

B,C,F

Step-by-step explanation:

A=L*W

P=2L+2W

P≤54

for A, 2L is already greater than 60

B works as 2W+2L in this case is 50

C states that perimeter is less than 54

D doesn't work, as 2L+2W=96

E doesn't work, see above, P=60

F, area=W*L

162/9=18

L=18

2L+2W=48, so F works

Answer:

trả lời:

B,C,F

Giải thích từng bước:

A = L * W

P = 2L + 2W

Trang ≤54

đối với A, 2L đã lớn hơn 60

B hoạt động như 2W + 2L trong trường hợp này là 50

C nói rằng chu vi nhỏ hơn 54

D không hoạt động, vì 2L + 2W = 96

E không hoạt động, xem ở trên, P = 60

F, diện tích =W*L

162/9=18

L =18

2L + 2W = 48, vì vậy F hoạt động

Answer:

-5/2, 1/6

Step-by-step explanation:

|6n+7|=8

6n+7=8

n=1/6

6n+7=-8

n=-5/2

Answer:

[tex]n=-\frac{5}{2}[/tex] and [tex]n=\frac{1}{6}[/tex]

Step-by-step explanation:

There is no x variable present in the question, but if you are asking for the value of n, I can help with that.

The absolute value function always results in a positive number, so that means 6n+7 can equal 8 or negative 8, and the absolute value function takes care of the rest. First, we will solve for 6n + 7 equaling 8.

[tex]6n+7=8[/tex]

Subtracting 7 from both sides gets us

[tex]6n=1[/tex]

Dividing by 6 from both sides is equal to

[tex]n=\frac{1}{6}[/tex]

Now we will solve for 6n + 7 equaling negative 8.

[tex]6n+7=-8[/tex]

Subtracting 7 from both sides is equal to

[tex]6n=-15[/tex]

Dividing by 6 from both sides gets us

[tex]n=-\frac{15}{6}[/tex]

Simplifying, we have

[tex]n=-\frac{5}{2}[/tex]

Answer:

To find the HCF we multiply the numbers in the overlapping quadrant together:

Step-by-step explanation:

Answer:

a. The radius r = 5.42 cm and the height h = 10.84 cm

b. 553.73 cm²

c. i. Beauty ii. Design

Step-by-step explanation:

a. What would be the optimal dimensions (radius and height) to minimize surface area?

The volume of the standard container is a cylinder and its volume is V = πr²h where r = radius of container and h = height of container.

Since V = 1000 cm³,

1000 cm³ = πr²h  (1)

Now, the surface area of a cylinder is A = 2πr² + 2πrh where r and h are the radius and height of the cylinder.

From (1), h = 1000/πr².

Substituting h into A, we have

A = 2πr² + 2πrh

A = 2πr² + 2πr(1000/πr²)

A = 2πr² + 2000/r

To maximize A, we differentiate A with respect to r and equate to zero to find the value of r at which A is maximum.

So, dA/dr = d[2πr² + 2000/r]/dr

dA/dr = d[2πr²]/dr + d[2000/r]/dr

dA/dr = 4πr - 2000/r²

Equating the equation to zero, we have

4πr - 2000/r² = 0

4πr = 2000/r²

r³ = 2000/4π

r = ∛(1000/2π)

r = 10(1/∛(2π))

r = 10(1/∛(6.283))

r = 10/1.8453

r = 5.42 cm

To determine if this value of r gives a minimum for A, we differentiate dA/dr with respect to r.

So, d(dA/dr)/dr = d²A/dr²

= d[4πr - 2000/r²]/dr

= d[4πr]/dr - d[2000/r²]/dr

= 4π  + 4000/r³

Substituting r³ = 2000/4π into the equation, we have

d²A/dr² = 4π  + 4000/r³ = 4π  + 4000/(2000/4π) = 4π  + 2 × 4π = 4π + 8π = 12π > 0

Since d²A/dr² = 12π > 0, then r = 5.42 cm gives a minimum for A.

Since h = 1000/πr²

h = 1000/π(5.42)²

h = 1000/92.288

h = 10.84 cm

So, the radius r = 5.42 cm and the height h = 10.84 cm

b. What would the surface area be?

Since the surface area, A = 2πr² + 2πrh

Substituting the values of r and h into A, we have

A = 2πr² + 2πrh

A = 2πr(r + h)

A = 2π5.42(5.42 + 10.84)

A = 10.84π(16.26)

A = 176.2584π

A = 553.73 cm²

c. Suggest at least two reasons why this is different from the ice cream packaging that you see in the stores.​

i. Beauty

ii. Design

Answer:

yes it does

Step-by-step explanation:

because the equation y=9x does not have a y-intercept (all slopes come in the form y=mx+b -- it can be written differently though) and since there is no 'b' that means the y-intercept is 0. So whenever there is no y-intercept, the slope starts at 0.

Answer:

c

Step-by-step explanation:

Try desmos

Answer:

11/20

Step-by-step explanation:

Answer:

15

Step-by-step explanation:

15 is the greatest number that divides 45 60 75 without leaving remainder

Answer:

15

Step-by-step explanation:

Let write the factors of each number:

45: (1,3,5,9,15,45)

60:(1,2,3,4,5,6,10,12,15,20,30,60)

75:(1,3,5,15,15,75).

The greatest common factor is 15. So the answer is 15.

Answer:

62 of the email messages were spam

Step-by-step explanation:

Let the number of spam and other messages be s and o respectively.

Total number of messages= 78

s +o= 78 -----(1)

s= 4o -2 -----(2)

Substitute (2) into (1):

4o -2 +o= 78

Simplify:

5o -2= 78

+2 on both sides:

5o= 78 +2

5o= 80

Divide both sides by 5:

o= 80 ÷5

o= 16

Since s +o= 78, s= 78 -o.

s= 78 -16

s= 62

Answer:

skip counting by 0

Step-by-step explanation:

skipcount by 0 to get to 100 for the third column.

Answer:

its the first graph

Step-by-step explanation:

I got it right bc im cool like that ig

Step-by-step explanation:

thwashm m GB DC GM 3hka it g feeds ygzdkzyzuzjz indin, mi, hn zbe

Answer:

Solution given:

L.H.S

sin²A-cos²B

sin²A=1-cos²A and Cos²B=1-sin²B

replacing value

1-cos²A-(1-sin²B)

1-cos²A-1+sin²B

1-1+sin²B-Cos²A

=sin²B-Cos²A

R.H.S

Answer:

x + 11 is an even number.

Step-by-step explanation:

Even numbers can only be obtained from the sum of two odd numbers or two even numbers. Since we know that x + 7 is even, x + 11 must be even as well.

Answer:

5/12

Step-by-step explanation:

5/12 , 2/3, 5/6,3/4

Get a common denominator of 12

5/12, 2/3 *4/4, 5/6*2/2, 3/4 *3/3

5/12, 8/12, 10/12, 9/12

The numerator closest to 0 is the fraction closest to 0

5/12

[tex]\\ \sf\longmapsto Area=\dfrac{\sqrt{3}}{4}a^2[/tex]

[tex]\\ \sf\longmapsto Area=\dfrac{\sqrt{3}}{4}(8)^2[/tex]

[tex]\\ \sf\longmapsto Area=\dfrac{64\sqrt{3}}{4}[/tex]

[tex]\\ \sf\longmapsto Area=16\sqrt{3}[/tex]

[tex]\\ \sf\longmapsto Area=16\times 1.732[/tex]

[tex]\\ \sf\longmapsto Area=27.7cm^2[/tex]

Answer:

-3x² - 2x - 54

Step-by-step explanation:

(-7²-x-5)-(3x²+x)

-7² - x - 5 - 3x² - x

-49 - x - 5 - 3x² - x

-3x² - x - x - 49 - 5

-3x² - 2x - 54

Answer:

1.572

Step-by-step explanation:

For a standard normal distribution,

P(z > c) = 0.058

To obtain C ; we find the Zscore corresponding to the proportion given, which is to the right of the distribution ;

Using technology or table,

Zscore equivalent to P(Z > c) = 0.058 is 1.572

Hence, c = 1.572

Answer:

[tex]5x + 2 = 100 \\ 5x = 100 - 2 \\ 5x = 98 \\ x = \frac{98}{5} \\ x = 19.6[/tex]

Answer: the answer would be x because that's the actual variable in the question then if 19.6 was not an option

Step-by-step explanation:

Answer:

8.7 feet

Step-by-step explanation:

Use the square area formula, a = s², where s is the side length of the square.

Plug in the area and solve for s:

a = s²

75 = s²

√75 = s

8.7 = s

So, to the nearest tenth of a foot, the height is 8.7 feet

(625 • (a4)) + 22b4

54a4 + 22b4

Final result :

625a4 + 4b4

Answer:

option A

Step-by-step explanation:

please mark this answer as brainlist

Answer:

1 ???????

Step-by-step explanation:

Answer:

x = 1/x - 2

Step-by-step explanation:

Answer:

√5/121

Step-by-step explanation:

formula: a^½=√a

(⁵/¹²¹)^½=√⁵/¹²¹

Answer:

0.0193 = 1.93% probability that there is equipment damage to the payload of at least one of five independently dropped parachutes.

Step-by-step explanation:

For each parachute, there are only two possible outcomes. Either there is damage, or there is not. The probability of there being damage on a parachute is independent of any other parachute, which means that the binomial probability distribution is used to solve this question.

To find the probability of damage on a parachute, the normal distribution is used.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Probability of a parachute having damage.

The opening altitude actually has a normal distribution with mean value 185 and standard deviation 32 m, which means that [tex]\mu = 185, \sigma = 32[/tex]

Equipment damage will occur if the parachute opens at an altitude of less than 100 m, which means that the probability of damage is the p-value of  Z when X = 100. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{100 - 185}{32}[/tex]

[tex]Z = -2.66[/tex]

[tex]Z = -2.66[/tex] has a p-value of 0.0039.

What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes?

0.0039 probability of a parachute having damage, which means that [tex]p = 0.0039[/tex]

5 parachutes, which means that [tex]n = 5[/tex]

This probability is:

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{5,0}.(0.0039)^{0}.(0.9961)^{5} = 0.9807[/tex]

Then

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.9807 = 0.0193[/tex]

0.0193 = 1.93% probability that there is equipment damage to the payload of at least one of five independently dropped parachutes.

选项 (d)

option (d)

!!!!!!!!!

Graph 1

Part (a)

The function is increasing when x > 0. The function is decreasing when x < 0.

The function is never constant

An increasing portion is when the graph goes uphill when moving left to right. A decreasing portion goes in the opposite direction: it goes downhill when moving left to right.

The reason why the function is never constant is because there aren't any flat horizontal sections. Such sections are when x changes but y does not. No such sections occur.

------------------------

Graph 1

Part (b)

Domain = set of all real numbers

Range = set of y values such that [tex]y \ge 0[/tex]

The domain is the set of all real numbers because we can plug in any value for x without any restriction. There are no division by zero errors to worry about, or square roots of negative numbers to worry about either.

The range is the set of nonnegative numbers as the graph indicates. The lowest y gets is y = 0.

------------------------

Graph 1

Part (c)

The function is even

The function f(x) = 1.6x^12 is an even function due to the even number exponent. For any polynomial, as long as the exponents are all even, then the function itself is even. If all the exponents were odd, then the function would be odd. This applies to polynomials only. A power function is a specific type of polynomial.

Note in the graph, we have y axis symmetry. The mirror line is vertical and placed along the y axis. This is a visual trait of any even function.

We could use algebra to show that f(-x) = f(x) like so

f(x) = 1.6x^12

f(-x) = 1.6(-x)^12

f(-x) = 1.6x^12

The third step is possible because (-x)^12 = x^12 for all real numbers x. It's similar to how (-x)^2 = x^2. You could think of it like (-1)^2 = (1)^2

============================================================

Graph 2

Part (a)

The function is decreasing when x < 0 and when x > 0

The function is never increasing

The function is never constant

In other words, the function is decreasing over the entire domain (see part b). The only time it's not decreasing is when x = 0.

The function is decreasing because the curve is going downhill when moving to the right. You can think of it like a roller coaster of sorts.

At no point of this curve goes uphill when moving to the right. Therefore, it is never increasing. The same idea applies to flat horizontal sections, so there are no constant intervals either.

------------------------

Graph 2

Part (b)

Domain: x is any real number but [tex]x \ne 0[/tex]

Range: y is any real number but [tex]y \ne 0[/tex]

Explanation: If we tried plugging x = 0 into the function, we get a division by zero error. This doesn't happen with any other number. Therefore, the set of allowed inputs is any number but 0.

The range is a similar story. There's no way to get y = 0 as an output.

If we plugged y = 0 into the equation, then we'd get this

y = 17x^(-3)

0 = 17/(x^3)

There's no way to have the right hand side turn into 0. The numerator is 17 and won't change. Only the denominator changes. We can't have the denominator be 0.  

------------------------

Graph 2

Part (c)

The function is odd

We can prove this by showing that f(-x) = -f(x)

f(x) = 17x^(-3)

f(-x) = 17(-x)^(-3)

f(-x) = 17* ( -(x)^(-3) )

f(-x) = -17x^(-3)

f(-x) = -f(x)

This is true for nearly all real numbers x, except we can't have x = 0.

Graphic 1:

(A) If f(x) = 1.6x ¹², then f '(x) = 19.2x ¹¹. Both f '(x) and x have the same sign, which means

• for -∞ < x < 0, we have f '(x) < 0, so that f(x) is decreasing on this interval

• for 0 < x < ∞, we have f '(x) > 0, so f(x) is increasing on this interval

f(x) is not constant anywhere on its domain.

(B) Speaking of domain, since f(x) is a polynomial (albeit only one term), it has

• a domain of all real numbers

• a range of {y ∈ ℝ : y = f(x) and y ≥ 0} (in other words, all real numbers y such that y = 1.6x ¹² and y is non-negative)

(C) This function is even, since

f(-x) = 1.6 (-x)¹² = (-1)¹² × 1.6x ¹² = 1.6x ¹² = f(x)

Graphic 2:

(A) Now if f(x) = 17/x ³, then f '(x) = -51/x ⁴. Because x ⁴ ≥ 0 for all x, this means f '(x) < 0 everywhere, except at x = 0. So f(x) is decreasing for (-∞ < x < 0) U (0 < x < ∞).

(B) f(x) has

• a domain of {x ∈ ℝ : x ≠ 0} (or all non-zero real numbers)

• a range of {y ∈ ℝ : y = f(x) and y ≠ 0} (also all non-zero reals)

(C) This function is odd:

f(-x) = 17/(-x)³ = 1/(-1)³ × 17/x ³ = -17/x ³ = -f(x)

Answer:

D.√85

Step-by-step explanation:

We can find the distance between two points using the distance between two points formula

Distance between two points formula:

d = √(x2 - x1)² + (y2 - y1)²

Where the x and y values are derived from the given points

We are given the two points (-2,7) and (7,9)

Using these points let's define the variables ( variables are x1, x2, y1, and y2)

Remember points are written as follows (x,y)

The x value of the first point is -2 so x1 = -2

The x value of the second point is 7 so x2 = 7

The y value of the first point is 7 so y1 = 7

The y value of the second point is 9 so y2 = 9

Now that we have defined each variable let's find the distance between the two points

We can do this by substituting the values into the formula

Formula: d = √(x2 - x1)² + (y2 - y1)²

Variables: x1 = -2, x2 = 7, y1 = 7, y2 = 9

Substitute values in formula

d = √(7 - (-2))² + (9 - 7)²

Evaluation:

The two negative signs cancel out on 7-(-2) and it changes to +7

d = √ (7+2)² + (9-7)²

Add 7+2 and subtract 9 and 7

d = √ (9)² + (2)²

Simplify exponents 9² = 81 and 2² = 4

We then have d = √ 81 + 4

Finally we add 81 and 4

We get that the distance between the two points is √85

Answer:

See attachment for graph

Step-by-step explanation:

Given

[tex]f(x) = \left[\begin{array}{cc}-1&x<-1\\0&-1\le x \le -1\\1&x>1\end{array}\right[/tex]

Required

The graph of the step function

Before plotting the graph, it should be noted that:

[tex]\le[/tex] and [tex]\ge[/tex] use closed circle at its end

[tex]<[/tex] and [tex]>[/tex] use open circle at its end

So, we have:

[tex]f(x) = -1,\ \ \ \ x < -1[/tex]

The line stops at -1 with an open circle      

[tex]f(x) = 0,\ \ \ \ -1 \le x \le 1[/tex]

The line starts at - 1 and stops at -1 with a closed circle at both ends

[tex]f(x) = 1,\ \ \ \ x > 1[/tex]

The line starts at 1 with an open circle

The options are not complete, so I will plot the graph myself.

See attachment for graph