Answer: D
Step-by-step explanation:
21=7+6+neither-5
7+6=13-5=8
21-8=13
Please help me just a and d
C = 2 x pi x r
if r = 5 cm
C = 2 x pi x 5 = 10 x pi
≈ 10 x 3,14 ≈ 31,4 cm
C = 2 x pi x r = pi x d
if d = 2 m
C = pi x 2
≈ 3,14 x 2 ≈ 6,28 m
Hello!
a)
C = 2πr C= 2*π *5 31,4cm
= 2*3,14* 5
d)
r =2/2 => 1m
C= 2πr C= 2*π*1 6,28m
= 2 *3,14*1
Help picture below problem 3
angle ksp+61=108(being straight angle)
or,angle ksp=180-61
angle ksp=119
Solve the following equation for p: 6/p=x+a
[tex]\dfrac 6p = x+a\\\\\implies \dfrac 6p \times\dfrac 16 = (x+a) \times\dfrac 16 ~~~~~~;\left[\text{Multiply both sides by}~ \dfrac 16 \right]\\\\\implies \dfrac 1p = \dfrac{x+a}6\\\\\implies p= \dfrac{6}{x+a}~~~~~~~~~~~~~~~~~~~~;[\text{Cross multiply}][/tex]
What is the length for X
Answer:
x = 5
Step-by-step explanation:
3^2 + 4^2 = x^2
9 + 16 = x^2
25
Now we need to find the square root of 25 :)
The answer is 5
X = 5
Have a great day!!
Please rate and mark brainliest!!
there were 63 fewer pears than apples in the supermarket. After 37 pears were sold, how many fewer pears than apples ?
Step-by-step explanation:
63/37 = 1.702 = 1.7
#Ihopethishleps
PLEASE HELP!!! I JUST NEED A STEP-BY-STEP!!!!!!
HINT: integrate with respect to y first (it is an easier approach)
∫∫5x sec^2(xy) dA; R={(x,y): 0 ≤ x ≤ π/6 , 0 ≤ y ≤ 2
[Answer] -5/2 ln(1/2)
Answer:
Step-by-step explanation:
[tex]=5\displaystyle\int_{0}^{\pi/6}dx\displaystyle\int_{0}^{2}x\sec^2(xy)dy\\=5\displaystyle\int_{0}^{\pi/6}dx\displaystyle\int_{0}^{2}\sec^2(xy)d(xy)\\=5\displaystyle\int_{0}^{\pi/6}dx\tan(xy)|_{y=0}^{y=2}[/tex]
[tex]=5\displaystyle\int_{0}^{\pi/6}\tan(2x)dx\\=-\frac{5}{2}\ln\cos(2x)|_{0}^{\pi/6}\\=-\frac{5}{2}[\ln\cos(\pi/3) - \ln\cos(0)]\\[/tex]
[tex]=-\frac{5}{2}\ln{\frac{1}{2}[/tex]
General solution of: (1-xy)^-2 dx + [y^2 + x^2 (1-xy)^-2]dy = 0
show two solution on your answer
nonsense answer deleted
[tex] \Large \bold{SOLUTION\ 1:} [/tex]
[tex] \small \begin{array}{l} \text{First, we need to check if the given differential} \\ \text{equation is exact.} \\ \\ (1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \quad M(x, y) dx + N(x, y) dy = 0 \end{array} [/tex]
[tex] \small \begin{array}{l l}\tt\: M(x,y) = \dfrac{1}{(1 - xy)^2}, & N(x,y) = y^2 + \dfrac{x^2}{(1-xy)^2}\\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{-2(-x)}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^2} + \dfrac{-2(-y)x^2}{(1 - xy)^3} \\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{2x}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x(1 - xy)+2x^2y}{(1 - xy)^3} \\ \\\tt \: & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^3} \end{array} [/tex]
[tex] \small \begin{array}{l} \tt\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x} \implies \text{Differential equation is exact.} \\ \\\tt \dfrac{\partial F}{\partial x} = M(x, y) = \dfrac{1}{(1 - xy)^2} \\ \tt\displaystyle F(x, y) = \int \dfrac{1}{(1 - xy)^2} \partial x = -\dfrac{1}{y} \int \dfrac{1}{(1 - xy)^2}(-y)\partial x \\ \\ \tt\:F(x, y) = \dfrac{1}{y(1 - xy)} + h(y) \\ \\ \tt\dfrac{\partial F}{\partial y} = N(x, y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\\tt \dfrac{\partial}{\partial y}\left[\dfrac{1}{y(1 - xy)} + h(y)\right] = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - xy + y(-x)}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - 2xy}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} + \dfrac{1 - 2xy}{y^2(1 - xy)^2} \\ \\ \tt\:h'(y) = y^2 + \dfrac{x^2y^2 - 2xy + 1}{y^2(1-xy)^2} = y^2 + \dfrac{1}{y^2} \\ \\ h(y) = \dfrac{y^3}{3} - \dfrac{1}{y} + C \\ \\ \tt\text{Substituting to }F(x,y),\text{we get} \\ \\ \dfrac{1}{y(1 - xy)} + \dfrac{y^3}{3} - \dfrac{1}{y} = C \\ \\ \quad \quad \text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array} [/tex]
[tex] \Large \bold{SOLUTION\ 2:} [/tex]
[tex] \small \begin{array}{l} \tt\text{Since we already know that the equation is exact,} \\ \text{we can then continue solving for the solution by} \\ \text{inspection method or by algebraic manipulation.} \\ \\ \tt(1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + y^2 dy + \dfrac{x^2}{(1-xy)^2} dy = 0 \\ \\ \tt\dfrac{dx + x^2dy}{(1-xy)^2} + y^2 dy = 0 \\ \\ \tt\text{Divide both numerator and denominator of the} \\ \tt\text{fraction by }x^2. \end{array} [/tex]
[tex] \small \begin{array}{c}\tt \dfrac{\dfrac{1}{x^2}dx + dy}{\dfrac{(1-xy)^2}{x^2}} + y^2 dy = 0 \\ \tt\\ \tt\dfrac{\dfrac{1}{x^2}dx + dy}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt-\dfrac{\left(-\dfrac{1}{x^2}dx - dy\right)}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt\displaystyle {\large{\int}} -\frac{d\left(\dfrac{1}{x}-y\right)}{\left(\dfrac{1}{x}-y\right)^2} + \int y^2 dy = \int 0 \\ \\ \tt\implies\tt \dfrac{1}{\dfrac{1}{x}-y} + \dfrac{y^3}{3} = C \\ \\\text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array} [/tex]
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Math Correct answers only please!!
Put question number with correct letter (ex: 1:a, 2:c, 3:d....etc)
Answer:
See below for answers and explanations
Step-by-step explanation:
Problem 1
Rewrite and add each vector:
[tex]p=\langle40cos70^\circ,40sin70^\circ\rangle\\q=\langle50cos270^\circ,50sin270^\circ\rangle\\r=\langle60cos235^\circ,60sin235^\circ\rangle\\p+q+r=\langle40cos70^\circ+50cos270^\circ+60cos235^\circ,40sin70^\circ+50sin270^\circ+60sin235^\circ\rangle\\p+q+r\approx\langle-20.73,-61.56\rangle[/tex]
Find the magnitude of the resulting vector:
[tex]||p+q+r||=\sqrt{x^2+y^2}\\||p+q+r||=\sqrt{(-20.73)^2+(-61.56)^2}\\||p+q+r||\approx64.96[/tex]
Therefore, the best answer is D) 64.959
Problem 2
Think of the vectors like this:
[tex]t=\langle7cos240^\circ,7sin240^\circ\rangle\\u=\langle10cos30^\circ,10sin30^\circ\rangle\\v=\langle15cos310^\circ,15sin310^\circ\rangle[/tex]
By adding the vectors, we have:
[tex]t+u+v=\langle7cos240^\circ+10cos30^\circ+15cos310^\circ,7sin240^\circ+10sin30^\circ+15sin310^\circ\rangle\\t+u+v\approx\langle14.8,-12.55\rangle[/tex]
Since the direction of a vector is [tex]\theta=tan^{-1}(\frac{y}{x})[/tex], we have:
[tex]\theta=tan^{-1}(\frac{y}{x})\\\theta=tan^{-1}(\frac{-12.55}{14.8})\\\theta\approx-40.3^\circ[/tex]
Don't forget to take into account that the resulting vector is in Quadrant IV since the horizontal component is positive and the vertical component is negative, so we will add 360° to our angle to get the result:
[tex]\theta=360^\circ+(-40.3)^\circ\\\theta=319.7^\circ[/tex]
Therefore, the best answer is D) 320°
Problem 3
Again, rewrite the vectors and add them:
[tex]u=\langle30cos70^\circ,30sin70^\circ\rangle\\v=\langle40cos220^\circ,40sin220^\circ\rangle\\u+v=\langle30cos70^\circ+40cos220^\circ,30sin70^\circ+40sin220^\circ\rangle\\u+v\approx\langle-20.38,2.48\rangle[/tex]
Using the direction formula:
[tex]\theta=tan^{-1}(\frac{y}{x})\\\theta=tan^{-1}(\frac{2.48}{-20.38})\\\theta\approx-6.94^\circ[/tex]
As the vector is located in Quadrant II, we need to add 180° to our angle:
[tex]\theta=180^\circ+(-6.94)^\circ\\\theta=173.06^\circ[/tex]
Therefore, the best answer is D) 173°
Problem 4
Using scalar multiplication:
[tex]-3u=-3\langle35,-12\rangle=\langle-105,36\rangle[/tex]
Find the magnitude of the resulting vector:
[tex]||-3u||=\sqrt{x^2+y^2}\\||-3u||=\sqrt{(-105)^2+(36)^2}\\||-3u||=111[/tex]
Find the direction of the resulting vector:
[tex]\theta=tan^{-1}(\frac{y}{x})\\\theta=tan^{-1}(\frac{36}{-105})\\\theta\approx-18.92^\circ[/tex]
As the vector is located in Quadrant II, we need to add 180° to our angle:
[tex]\theta=180^\circ+(-18.92)^\circ\\\theta=161.08^\circ[/tex]
Therefore, the best answer is C) 111; 161°
Problem 5
Using scalar multiplication:
[tex]5v=5\langle25cos40^\circ,25sin40^\circ\rangle=\langle125cos40^\circ,125sin40^\circ\rangle[/tex]
Since the direction of the angle doesn't change in scalar multiplication, it only affects the magnitude, so the magnitude of the resulting vector would be [tex]||5v||=125[/tex], making the correct answer C) 125; 200°
The train station clock runs too fast and gains 8 minutes every 5 days. How many minutes and seconds will it have gained at the end of 8 days?
Larissa dives into a pool that is 8 feet deep. She touches the bottom of the pool with her hands 6 feet horizontally from the point at which she entered the water.
What is the approximate angle of elevation from the point on the bottom of the pool where she touched to her entry point?
36.9°
41.4°
48.6°
53.1°
Answer: 48.6°
Step-by-step explanation:
Angle of elevation = sin inverse (6/8)
= 48.6 degrees
Therefore, Option C is right answer
What is the area of a circle with the diameter of 10
What is the value of the expression (–5) -3?
1. Apply the negative exponents rule:
1
(−5)3
2. Expand the power:
1
(−5)(−5)(−5)
3. Simplify:
1
x
What is the value of x?
x =
Answer:
15 :)
Step-by-step explanation:
brainliest please
Answer:
The answer is -125
Proof:
I’m stuck pls help me
Answer:
B and E
Step-by-step explanation:
The diagram shows a plan for a deck. The area of the deck is 511 ft squared what is the value of x? Show your work
Please helpp
Just have a good day
The dimension of the unknown variable x of the deck is equal to 14 feet in length.
What is a Rectangle?A rectangle is a closed 2-D shape, having 4 sides, 4 corners, and 4 right angles (90°).The opposite sides of a rectangle are equal and parallel.
Given in the question is a diagram showing a plan for deck. From the data given, we can write -
Area of Deck = 511 ft.
The value of x can be found out by subtracting the area of the deck from the area of rectangle having dimensions of length 29 ft and width (16 + 4) = 20 ft. The remaining area will be equated to the sum of the area of rectangle at top right corner having dimensions of length 9 ft and width 4 ft and the area of triangle at lower right corner.
Firstly, we will find the dimensions of the triangle at the lower right. The height of the triangle will be = (20 - 9) ft = 11 ft. The base of the triangle will be = 29 - ( x + 9) ft = (20 - x) ft. Now -
[Area of Rectangle (29 x 20)] - [Area of Deck] = [Area of Rectangle (9 x 4) + Area of triangle (lower right)].
Mathematically -
580 - 511 = 9 x 4 + 1/2 × (20 - x) × 11 = 69
36 + 5.5(20 - x) = 69
36 + 110 - 5.5x = 69
146 - 69 = 5.5x
5.5x = 77
x = 14 ft
Therefore, the dimension of the unknown variable x of the deck is equal to 14 feet.
To solve more questions on Area of Shapes, visit the link below-
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given tan 0= -15/8 where 270º < 0 < 360°
Given tan 0 =
Find cos 0
Answer:=8/17
Step-by-step explanation:
easy promblem help me
the answer Is b and c
Step-by-step explanation:
because I did it
How could you correctly rewrite the equation 4(10 + 5) = 6 (12 - 2) using the distributive property?
Answer:
40+20 = 72-12 is the simplified version.
Step-by-step explanation:
Your equations
4(10+5) = 6(12-2)
Multiply 4 by 10 and 5 for the "4(10+5)"
Multiply 6 by 12 and -2 for the "6(12-2)"
40+20 = 72-12 is the simplified version.
If you want to solve/verify:
40+20 = 72-12
60 = 60
The nanswer is correct.
Answer:
40 + 20 = 72 - 12
Step-by-step explanation:
The number outside the parenthesis will be multiplied to both numbers inside the parenthesis, making the equation
(4 * 10) + (4 * 5) = (6 * 12) + (6 * (-2))
simplifying the equation...
40 + 20 = 72 - 12
There are only 7 days left until the launch of our new product and we only have $668 left and Ira promotion budget we need to spend $85 on the last day can you please calculate how many dollars we can spend on the remaining days
Answer:
97.16
Step-by-step explanation:
in total, you have $668.00.
To reserve enough for the last day, you need to subtract the last day's budget:
This means that, for the remaining 6 days in the week, you have $583.00.
Assuming an equal amount is being spent each day, you can calculate the daily budget by dividing the remaining total by the number of days left.
To avoid going over budget, you can spend $97.16 per day for the other six days.
LMNO is a parallelogram, with and . Which statements are true about parallelogram LMNO? Select three options.
x = 11
m∠L = 22°
m∠M = 111°
m∠N = 59°
m∠O = 121°
Answer:
the answers 1, 4, and 5.
m∠N = 59° and m∠O = 121°.
Option 3 and 4 are correct.
What is a parallelogram?A parallelogram is a geometric object with sides that are parallel to one another in two dimensions. It is a form of polygon with four sides (sometimes known as a quadrilateral) in which each parallel pair of sides have the same length. A parallelogram has adjacent angles that add up to 180 degrees. You must have studied a variety of 2D shapes and sizes in geometry, including circles, squares, rectangles, rhombuses, etc. Each of these forms has a unique set of characteristics.
As per the given data:
LMNO is a parallelogram
The sum of adjacent angles in a parallelogram is always equal to 180°
11x + 6x - 7 = 180
17x = 187
x = 11
m∠L = m∠N {opposite angles in a parallelogram are equal}
m∠N = 6x - 7 = 6(11) - 7 m∠N
m∠L = 59°
m∠M = m∠O {opposite angles in a parallelogram are equal}
m∠M = 11x = 11(11) = 121°
m∠O = 121°
Hence, m∠N = 59° and m∠O = 121°.
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Select the most appropriate unit for the situation
Rate of filling a buck with water
Feet / minute
Square feet /minute
Cubic feet / minute
Answer: Is Square feet/minute if there is two of them the second one is Cubic feet/minute
Step-by-step explanation: feet/minute would cover anything linear {forward, backward, left, right}
square feet/minute would cover two-dimensional areas {walls, floors, sides of buildings, etc.}
cubic feet/minute would cover three dimensional situations {swimming pools, buckets, containers, etc.}
A hot air balloon rising vertically is tracked by an observer located 5 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is π4 , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment?
Answer:
Step-by-step explanation:
aThe person and the hot air balloon form a right triangle, with the lift-off point as the vertex that form the right angle. If you were to try to solve for the height of the balloon, you would need to use tangent to do so:
tanθ = opp/adj = y/x
y = xtanθ
What this problem ultimately wants to know, is how the distance of the balloon is changing with respect to time: dy/dt. So we need to take the derivative of the above equation (you need to product rule to do so):
dy/dt = (dx/dt)tanθ + xsec2θ(dθ/dt)
We need to find out values for all of the variables:
The observer is not moving, therefore dx/dt = 0.
The observer is standing 4 miles away from the lift-off point, so x = 4.
The angle between the horizontal and the observer's line of sight is π/3, therefore θ = π/3.
The rate of change for the angle, dθ/dt, is 0.1.
Just plug all the numbers in:
dy/dt = (0)tan(π/3) + (4)sec2(π/3)(0.1)
dy/dt = 0 + 1.6
dy/dt = 1.6 mi/min
12.44 = 4x what does x =
4 * x = 12.44
4/4 * x = 12.44 / 4
x = 3.11
Answer:
x = 3.11
Step-by-step explanation:
To find the value of 'x', divide both sides of the equation by 4
12.44 = 4x
[tex]\dfrac{12.44}{4}=\dfrac{4x}{4}\\\\\\3.11=x[/tex]
A trapezoid has sides of length 16.7 centimeters, 12.9 centimeters, 16.7 centimeters, and 18.9 centimeters. What is the perimeter? centimeters
Answer:
65.2 cm
Step-by-step explanation:
P = a + b + c + d
P = 16.7 cm + 12.9 cm + 16.7 cm + 18.9 cm
P = 65.2 cm
The perimeter of the trapezoid is 65.2 cm.
Answer:
To find the perimeter of this trapezoid, simply add all four sides (since there are only four sides to a trapezoid):-
16.7 + 12.9 + 16.7 + 18.9
= 65.2 centimetres is the perimeter.
You can picture it like this;
Determine where the given function is concave up and where it is concave down.
q(x) = 8x3 + 2x + 8
Answer:
concave up
Step-by-step explanation:
there is no negatives in the equation (in front of x-value)
a) Work out
41/7 + 1 1/2
Answer:
11 5/4
Step-by-step explanation:
make the denominator the same then add.the common factor of 2 and 7 is 14 so make all the denominator to 14 then u can add.thus having the answer
Find the exact value of sin(255∘)
Check the picture below.
[tex]\textit{Sum and Difference Identities} \\\\ sin(\alpha + \beta)=sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(255^o)\implies sin(135^o+120^o) \\\\\\ sin(135^o)cos(120^o)~~ + ~~cos(135^o)sin(120^o) \\\\\\ \left( \cfrac{\sqrt{2}}{2} \right)\left( -\cfrac{1}{2} \right)~~ + ~~\left( -\cfrac{\sqrt{2}}{2} \right)\left( \cfrac{\sqrt{3}}{2} \right)\implies -\cfrac{\sqrt{2}}{4}~~ - ~~\cfrac{\sqrt{6}}{4}\implies \cfrac{-2-\sqrt{6}}{4}[/tex]
Find the equation of the line with slope m=53 that contains the point (−6,−12).
.................................................................................
In a simple economy (assume there are no taxes, thus, Y is disposable income), the consumption function is:
C = 200+ 0.80Y
00
The current level of real GDP is $4000.
At this level of real GDP, consumption will be $, and savings will be $1. If GDP were to increase by $1000, consumption would increase by $(Round your
responses to the nearest dollar.)
Using linear function concepts, it is found that consumption would increase by $800.
What is a linear function?A linear function is modeled by:
[tex]y = mx + b[/tex]
In which:
m is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.b is the y-intercept, which is the value of y when x = 0.In this problem, the function is given by:
C = 200 + 0.8Y.
Which means that the slope is of m = 0.8 per unit of Y.
For 1000 units:
0.8 x 1000 = 800.
Consumption would increase by $800.
More can be learned about linear function concepts at https://brainly.com/question/24808124
Solve the simultaneous equations
.
3x + y = 10
y = 2 - x
X =
y =
Step-by-step explanation:
please mark me as brainlest
Answer:
3x + (2 - x) = 10
3x + 2 - x = 10
2x + 2 = 10
2x = 10 - 2
2x = 8
x = 8/2
x = 4
y = 2 - x
y = 2 - 4
y = -2
Find the values of x, y, and z.
Please help!!
Answer:
x=20
y and z=160
Step-by-step explanation:
the arc of the circle is congruent to the central angle therfore x is also 20
next for y and z we can tell they are congruent now to find the measure. we can tell A is 180 so we do 360-180-20=160