Answer:
17 ft
Step-by-step explanation:
An 18-foot ladder leaning against a wall makes an angle of elevation of 70° with the ground. How far up the wall is the ladder, to the nearest foot?
We solve the above question using the Trigonometric function of Sine
sine theta = Opposite/Hypotenuse
theta = Angle of elevation = 70°
Hypotenuse =Length of the ladder = 18ft
Opposite = Height of the wall = x
Therefore
sin 70 = x/ 18 ft
Cross Multiply
x = sin 70 × 18 ft
x = 16.914467174 ft
Approximately = 17 ft
What is the measure of the angle at the tail end of the kite? if the top area is where its pointed is 122?
Answer:
The angle is 58 degrees
Step-by-step explanation:
Given
See attachment for kite
Required
The angle at the tail end
Represent this angle with x.
From the attached kite, we have:
1 angle = 122
2 angles = right-angled
So, we have:
[tex]x + 122 +90+90 = 360[/tex] --- sum of angles in a kite
[tex]x + 302 = 360[/tex]
Solve for x
[tex]x =- 302 + 360[/tex]
[tex]x =58^\circ[/tex]
please write all the steps... write clearly thanks
Determine the inverse Laplace transforms of: (b) 1 3s²+5s+1
The inverse Laplace transform of 1 / (3s² + 5s + 1) is f(t) = 1/2 × [tex]e^{(-t)[/tex]- 1/2 × [tex]e^{(-t/3)[/tex].
To find the inverse Laplace transform of the function F(s) = 1 / (3s² + 5s + 1), we can use partial fraction decomposition and reference tables for Laplace transforms.
Step 1: Factorize the denominator
Factorize the denominator of the function 3s² + 5s + 1 to find its roots:
3s² + 5s + 1 = (s + 1)(3s + 1)
Step 2: Write the partial fraction decomposition
Write the function F(s) as a sum of partial fractions:
F(s) = A / (s + 1) + B / (3s + 1)
Step 3: Determine the values of A and B
To find the values of A and B, we can multiply both sides of the equation by the common denominator and equate the numerators:
1 = A(3s + 1) + B(s + 1)
Expand the right side and collect like terms:
1 = (3A + B)s + (A + B)
By equating the coefficients of s and the constant terms on both sides, we get a system of equations:
3A + B = 0
A + B = 1
Solving this system of equations, we find A = 1/2 and B = -1/2.
Step 4: Write the inverse Laplace transform
Using the partial fraction decomposition, we can now write the inverse Laplace transform:
F(s) = 1/2 × (1 / (s + 1)) - 1/2 × (1 / (3s + 1))
Referring to Laplace transform tables, we find that the inverse Laplace transform of 1 / (s + a) is [tex]e^{(at)[/tex], and the inverse Laplace transform of 1 / (s - a) is [tex]e^{(at)[/tex]. Therefore, applying these results, we have:
f(t) = 1/2 × [tex]e^{(-t)[/tex] - 1/2 × [tex]e^{(-t/3)[/tex]
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Joe wants to rent a boat and spend less than $33. Boat cost $7 per hour, and Joe has a discount coupon for $9 off. What are the possible numbers of hours Joe could rent the boat? Use t for the number of hours. Write your answer as an inequality solved for t.
$7 per hour plus $9 off
$7x5=$35-$9 discount=$26
$26+$7=$33
6hours=$42-$9=$33
Joe could rent the boat for 6 hours
Use synthetic division to determine whether the number is a zero of the polynomial function.
3i;g(x) = x^3 - 4x^2 + 9x - 36
The last entry in the synthetic division table is not zero, 3i is not a zero of the polynomial function g(x) = x³ - 4x² + 9x - 36.
To determine if 3i is a zero of the polynomial function g(x) = x³ - 4x² + 9x - 36, we can use synthetic division.
First, we set up the synthetic division table:
3i | 1 -4 9 -36
Performing the synthetic division:
| (1) (-4) (9) (-36)
3i | 3i 9i² 27i
| (1) (-4 + 3i) (9 + 9i²) (-36 + 27i)
Simplifying the last row, we have:
| (1) (-4 + 3i) (9 - 9) (-36 + 27i)
| (1) (-4 + 3i) (0) (-36 + 27i)
Therefore, the last entry in the synthetic division table is not zero, 3i is not a zero of the polynomial function g(x) = x³ - 4x² + 9x - 36.
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Stonewall receives ¢250 per year in simple interest from an amount of money he invested in
ADB, Barclays and GCB. Suppose ADB pays an interest of 2%, Barclays pays an interest of
4% and GCB pays an interest of 5% per annum and an amount of ¢350 more was invested in
Barclays than the amount invested in ADB and GCB combined. Also, the amount invested in
Barclays is 2 times the amount invested in GCB.
a) Write down the three linear equations and represent them in the matrix form AX = B.
b) Find the amount of money Stonewall invested in ADB, Barclays and GCB using Matrix
Inversion
Answer:
The amount invested in ADB is ¢1363.[tex]\overline 3[/tex] =
The amount invested in Barclays is ¢3,427.[tex]\overline 3[/tex]
The amount invested in GCB is ¢1,713.[tex]\overline 3[/tex]
Step-by-step explanation:
The parameters of the investment Stonewall made are;
The amount in interest he receives from ADB, Barclays and GCB = ¢250
The amount of interest ADP pays = 2% per annum
The amount of interest Barclays pays = 4% per annum
The amount of interest GCB pays = 5% per annum
The amount invested in Barclays = The amount invested in ADB and GCB + ¢350
The amount invested in Barclays = 2 × The amount invested in GCB
a) Let 'x', represent the amount invested in ADB, 'y' represent the amount invested in Barclays, and 'z', represent the amount invested in GCB
We have;
y = x + z + 350
y = 2·z
0.02·x + 0.04·y + 0.05·z = 250
Therefore, we get the three linear equations as follows;
-x + y - z = 350...(1)
y - 2·z = 0...(2)
0.02·x + 0.04·y + 0.05·z = 250...(3)
Using Matrix inversion, we have;
[tex]\left[\begin{array}{ccc}-1&1&-1\\0&1&-2\\0.02&0.04&0.05\end{array}\right] \times \left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{c}350&0&250\end{array}\right][/tex]
The transpose of the 3 by 3 matrix [tex]M^T[/tex] is given as follows;
[tex]M^T = \left[\begin{array}{ccc}-1&0&0.02\\1&1&0.04\\-1&-2&0.05\end{array}\right][/tex]
The Adjugate Matrix is given as follows;
[tex]Adj = \left[\begin{array}{ccc}0.13&-0.09&-1\\-0.04&-0.03&-2\\-0.02&0.06&-1\end{array}\right][/tex]
The inverse of the matrix = Adj/Det where, Det = -0.15, is therefore;
[tex]M^{-1} = \left[\begin{array}{ccc}\dfrac{-13}{15} &\dfrac{3}{5} &\dfrac{20}{3} \\\\\dfrac{4}{15} &\dfrac{1}{5} &\dfrac{40}{3} \\\\\dfrac{2}{15} &-\dfrac{2}{5} &\dfrac{20}{3} \end{array}\right][/tex]
We therefore, get the solution as follows;
[tex]\left[\begin{array}{ccc}\dfrac{-13}{15} &\dfrac{3}{5} &\dfrac{20}{3} \\\\\dfrac{4}{15} &\dfrac{1}{5} &\dfrac{40}{3} \\\\\dfrac{2}{15} &-\dfrac{2}{5} &\dfrac{20}{3} \end{array}\right]\times \left[\begin{array}{c}350&0&250\end{array}\right] = \left[\begin{array}{c}\dfrac{4,090}{3} \\&\dfrac{10,280}{3} \\ & \dfrac{5,140}{3} \end{array}\right][/tex]
[tex]\begin{array}{c}x = \dfrac{4,090}{3} \\&y = \dfrac{10,280}{3} \\ & z = \dfrac{5,140}{3} \end{array}[/tex]
The amount invested in ADB, x = ¢4,090/3 = ¢1363.[tex]\overline 3[/tex]
The amount invested in Barclays, y = ¢10,282/3 = ¢3,427.[tex]\overline 3[/tex]
The amount invested in GCB, z = ¢5,140/3 = ¢1,713.[tex]\overline 3[/tex]
What's the slope and y intercept of 3x - y = 7
Answer:
the slope is 3
the y intercept is (0,-7)
Step-by-step explanation:
3x-y=7 solve for y
add y to both sides
3x=7+y
subtract 7 from both sides
y=3x-7
y=mx+b
m=slope
b=y-intercept
3=slope
-7=y-intercept
Hope that helps :)
i need help i will give branliest please !!!
Can someone please help me with this page
Answer:
2. there is no E it would be AB=CD
34.607 to the nearest whole number
Help me plz!! I need help and NO FILES PLZZZZZZZ!
Answer:
1
EXPLAINATION:
y = mx +b
b = y intercept
Justin completes 8 extra credit problems on the first day and then 4 problems each day until the worksheet is complete. There are 28 problems on the worksheet. Write and solve an equation to find how many days it will take Justin to complete the worksheet after the first day.
Answer:
28-8-4x=0
x=5
5 days
Step-by-step explanation:
28-8=20 problems left
then 4 each day
What is 7/4 as a mixed number
Help please I’ll mark brainiest
Answer:
V≈7853.98
Step-by-step explanation:
V=πr2h
r=10
h=25
Solution
V=πr2h=π·102·25≈7853.98163
PLEASE HELP ME and the first person that answers correctly will get... BRAINLIEST I promise.
Step-by-step explanation:
V≈75.4
A≈100.53
Help help help help help
Answer:
6 ft
Step-by-step explanation:
The triangle CDE is half of the size of triangle ABC, so if you match up the corresponding sides, DE is half of AC. AC is 12 so DE is 6.
hope is helped!
Y’all NLE Choppa is in jail. Sending prayers and love
Answer:
NLE Choppa noooooooooooooooooo
Step-by-step explanation:
THE BEST RAPPPERRR!!!!!!!!!
Answer:
He wasnt in jail? I dont think cuz he was makin new music
Step-by-step explanation:
Please answer and not give me links that make me download a whole bunch of stuff then just turn out to be an inappropriate picture (Yeah that's happened to me twice on here :/)
Answer:
3
Step-by-step explanation:
12/4=3
let R be the region bounded by the functions f(x)=-x^2 and g(x)=-9 as shown in the diagram below. find the exact area of the region R. write your answer in the simplest form. zoom in photo
anthony can run at the rate (in meters per minute) shown in the graph below) which of the following best describes anthony’s rate of speed
Step-by-step explanation:
no this is a test I cant hlep i will tell ms.jules you are searching for the answers
PLSS HELP IMMEDIATELY!!! i’ll give brainiest if u don’t leave a link!
Answer:
it's arm will regenerate and grow back
plz mark me as brainliest
Answer:
B. The arm will regenterate
Find a formula for the exponential function passing through the points (-3,1/2) and (3,32).
The exponential function passing through the points (-3, 1/2) and (3, 32) can be represented by the equation [tex]f(x) = (2^{(5/6)}) * (2^{(x/6)})[/tex].
To find the exponential function passing through the given points, we can start by assuming the general form of an exponential function, [tex]f(x) = a * (b^x)[/tex], where a and b are constants to be determined. Plugging in the coordinates (-3, 1/2) into this equation gives us [tex]1/2 = a * (b^{(-3)})[/tex], and plugging in (3, 32) gives us [tex]32 = a * (b^3)[/tex].
Now, we can solve this system of equations to find the values of a and b. Taking the ratio of the two equations, we get [tex](1/2) / 32 = (a * (b^{(-3)})) / (a * (b^3))[/tex], which simplifies to [tex]1/64 = 1/b^6[/tex]. Solving for b, we find [tex]b = 2^{(1/6)[/tex].
Substituting this value back into either of the original equations, we can solve for a. Using the equation [tex]1/2 = a * (2^{(-3/6)})[/tex], we find [tex]a = 2^{(5/6)[/tex].
Therefore, the exponential function passing through the given points is [tex]f(x) = (2^{(5/6)}) * (2^{(x/6)})[/tex].
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Can someone please help me please I really need help
Answer:
16 cans
Step-by-step explanation:
Hey,
I have to admit, this problem is pretty complicated. But I got you :).
To begin, we have to represent the small and large boxes with two separate variables. So...
y = large
x = small
(It really doesn't matter what variable you use, I just used these)
Now, we know that Estelle fills 3 large boxes and 5 small boxes and had a total of 170 cans in total. We can represent this by writing...
3y + 5x = 170
Then, we know that the boy worker filled 4 large boxes and 4 small boxes and had a total of 184 cans. We can represent this by writing...
4y + 4x = 184
As you can see, this is a system of equations.
3y + 5x = 170
4y + 4x = 184
We want to know how many cans each small box can hold, so we have to find a common number for x since x represents the small box.
To do this, we have to multiply the first equation by 4 and the second by 3. Here's what I mean...
4 (3y + 5x = 170)
3 (4y + 4x = 184)
When we do this you get...
12y + 20x = 680
12y + 12x = 552
Notice how the y's are now both 12. We had to do that in order to get rid of y, they had to equal the same number. Now, subtract...
8x = 128
Divide by 8...
x = 16
That means that...
YOUR ANSWER: Each small box can hold up to 16 cans.
I hope this helps :)
One of Carla's friends suggests that she survey only eighth-graders because they are the
oldest and probably know more about the election than younger students. Do you think
this suggestion creates a random sample? Explain.
Answer:
This example is not a random example due to the fact that it is only questioning the eighth-graders. Random sample means that the sample is chosen simply randomly and everyone has an equal opportunity to be apart of the sampling.
Step-by-step explanation:
how do i get all my points and brainlest back
.....um...i think u can't do this
Answer:
u should probably contact support or the creator of this site
Step-by-step explanation:
Find the value of x that makes the equation true:
16 - x = 4
x = 18
x = 20
x = 14
x = 12
Find the measure of the exterior 1.
A. 144°
B. 56°
C. 36°
D. 136°
SOMEONE PLEASE HELP I'VE BEEN ASKING FOR DAYS NOW WITH SCALE FACTOR!!!!! I will give brainliest if right!!!!!!
Answer:
It has been answered for you
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During Year 3, Vernon Corporation reported after the net come of $3,595.000 During the year the number of shares of stock outstanding remained constant at 10.000 of $100 par 9 percent prefered stock and 399,000 shares of common stock. The company's kotel stockholders equity is $19,600,000 at December 31. Year 3. Vernas Corporation's common stock was selling at $54 per share at the end of is scal year. All clvidends for the year have been paid, including $4.20 per share so common stockholders.
Required
3. Compute the earings per share, (Round your answer to 2 decimal places.)
b. Compute the books te of common stock (Round your answer to 2 decimal places.) c. Compute the price warning Mic Round intermediate calculations and final answer to 2 decimal places) d. Compute the dividend yield (Round your percentage answer to 2 decimal places, fe, 0.2345 should be entered as 23.45).)
a The net income is $3505000
b The number of common shares outstanding is 399,000
c Earnings per share is $8.79
d Book value of common stock is $46.62
e. Price-earnings ratio (P/E ratio) is 6.14
How to calculate the valuea. Net income available to common stockholders:
Net income = $3,595,000
Dividends to preferred stockholders = (Number of preferred shares * Par value * Dividend rate) = (10,000 * $100 * 0.09) = $90,000
Net income available to common stockholders = Net income - Dividends to preferred stockholders
= $3,595,000 - $90,000
= $3,505,000
b. Weighted average number of common shares outstanding:
Number of common shares outstanding = 399,000
c. Earnings per share: EPS = Net income available to common stockholders / Weighted average number of common shares outstanding
= $3,505,000 / 399,000
≈ $8.79
d. Book value of common stock:
Total stockholders' equity = $19,600,000
Preferred stock equity = Number of preferred shares * Par value = 10,000 * $100 = $1,000,000
Common stock equity = Total stockholders' equity - Preferred stock equity
= $19,600,000 - $1,000,000
= $18,600,000
Book value of common stock = Common stock equity / Number of common shares outstanding
= $18,600,000 / 399,000
≈ $46.62
e. Price-earnings ratio (P/E ratio):
Price-earnings ratio = Price per share / Earnings per share
= $54 / $8.79
≈ 6.14
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Find the area of the shape shown below.
Answer:
42 units²
Step-by-step explanation:
A school is planning for an addition in some open space next to the current building. The existing building ends at the origin. The graph represents the system of equations that can be used to define the space for the addition. What is the system of equations that matches the graph?
y ≤ 3x
y > –2x – 1
y > 3x
y ≤ –2x – 1
y < –3x
y ≥ 2x – 1
y > –3x
y ≤ 2x – 1
Equation system that corresponds to the graph:
1. y ≤ 3x 2. y > –2x – 1
To find the system of equations that corresponds to the provided graph, we must first analyze it and locate the regions that fulfil the specified requirements.
1. Begin by locating the darkened region underneath the line y 3x. This line has a slope of 3 and intersects the origin (0,0). Shade the area beneath the line.
2. After that, locate the darkened region above the line y > -2x - 1. The slope of this line is -2, while the y-intercept is -1. The area above the line should be shaded.
3. The solution space that meets both requirements is represented by the overlapping shaded region between the two lines. The common area is located below y 3x and above y > -2x - 1.
4. The equation system that corresponds to this common area is: - y 3x - y > -2x - 1
The space for the addition in the open area next to the present building is defined by these two formulae.
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