what is the volume of a cube with a length of a 10cm,
a width of 8cm and a height of 8cm​

Answer:

Creo que es 36

Step-by-step explanation

:D

Answer:

36

Step-by-step explanation:

Answer: a = 25.67

Step-by-step explanation:

Answer:

See Below.

Step-by-step explanation:

We are given the equation:

[tex]\displaystyle y^2 = 1 + \sin x[/tex]

And we want to prove that:

[tex]\displaystyle 2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right) ^2 + y^2 = 1[/tex]

Find the first derivative by taking the derivative of both sides with respect to x:

[tex]\displaystyle 2y \frac{dy}{dx} = \cos x[/tex]

Divide both sides by 2y:

Find the second derivative using the quotient rule:

[tex]\displaystyle \begin{aligned} \frac{d^2y}{dx^2} &= \frac{(\cos x)'(2y) - (\cos x)(2y)'}{(2y)^2}\\ \\ &= \frac{-2y\sin x-2\cos x \dfrac{dy}{dx}}{4y^2} \\ \\ &= -\frac{y\sin x + \cos x\left(\dfrac{\cos x}{2y}\right)}{2y^2} \\ \\ &= -\frac{2y^2\sin x+\cos ^2 x}{4y^3}\end{aligned}[/tex]

Substitute:

[tex]\displaystyle 2y\left(-\frac{2y^2\sin x+\cos ^2 x}{4y^3}\right) + 2\left(\frac{\cos x}{2y}\right)^2 +y^2 = 1[/tex]

Simplify:

[tex]\displaystyle \frac{-2y^2\sin x-\cos ^2x}{2y^2} + \frac{\cos ^2 x}{2y^2} + y^2 = 1[/tex]

Combine fractions:

[tex]\displaystyle \frac{\left(-2y^2\sin x -\cos^2 x\right)+\left(\cos ^2 x\right)}{2y^2} + y^2 = 1[/tex]

Simplify:

[tex]\displaystyle \frac{-2y^2\sin x }{2y^2} + y^2 = 1[/tex]

Cancel:

[tex]\displaystyle -\sin x + y^2 = 1[/tex]

Substitute:

[tex]-\sin x + \left( 1 + \sin x\right) =1[/tex]

Simplify. Hence:

[tex]1\stackrel{\checkmark}{=}1[/tex]

Q.E.D.

D

A

B

C

for more explanation please don't hesitate to just respond

Answer:

B) All real numbers except 0 and integer multiples of 8π∕5

Step-by-step explanation:

Cotangent function:

The cotangent function is given by:

[tex]y = \cot{ax} = \frac{\cos{ax}}{\sin{ax}}[/tex]

Domain:

All real values except those at which:

[tex]\sin{ax} = 0[/tex]

The sine is 0 for 0 and all integer multiples of [tex]\frac{1}{a}[/tex]

In this question:

[tex]a = \frac{5}{8}[/tex], so the values outside the domain are 0 and the integer multiples of [tex]\frac{8}{5}[/tex]. Then the correct answer is given by option b.

Answer:

The new coordinates are [tex]A'(x,y) = (3, 0)[/tex], [tex]B'(x,y) = (6, -3)[/tex] and [tex]C'(x,y) = (2, -3)[/tex].

Step-by-step explanation:

Vectorially speaking, the translation of a point can be defined by the following expression:

[tex]V'(x,y) = V(x,y) + T(x,y)[/tex] (1)

Where:

[tex]V(x,y)[/tex] - Original point.

[tex]V'(x,y)[/tex] - Translated point.

[tex]T(x,y)[/tex] - Translation vector.

If we know that [tex]A(x,y) = (-3,4)[/tex], [tex]B(x,y) = (0,1)[/tex], [tex]C(x,y) = (-4,1)[/tex] and [tex]T(x,y) = (6, -4)[/tex], then the resulting points are:

[tex]A'(x,y) = (-3, 4) + (6, -4)[/tex]

[tex]A'(x,y) = (3, 0)[/tex]

[tex]B'(x,y) = (0,1) + (6, -4)[/tex]

[tex]B'(x,y) = (6, -3)[/tex]

[tex]C'(x,y) = (-4, 1) + (6, -4)[/tex]

[tex]C'(x,y) = (2, -3)[/tex]

The new coordinates are [tex]A'(x,y) = (3, 0)[/tex], [tex]B'(x,y) = (6, -3)[/tex] and [tex]C'(x,y) = (2, -3)[/tex].

9514 1404 393

Answer:

Step-by-step explanation:

1. Input 3 for x and do the arithmetic.

  z(x) = x+1

  z(3) = 3+1 = 4 . . . . . the output is 4

__

2. The expression for n(x) has x in the denominator. The expression will be undefined when the denominator is zero, so x=0 cannot be used.

Answer:

[tex]A)\:x<12[/tex]

[tex]5(x+5)<85\\5x+25<85\\5x<85-25\\5x<60\\x<12[/tex]

OAmalOHopeO

Answer:

x < 12.................................

9514 1404 393

Answer:

  (-13, 10)

Step-by-step explanation:

If M is the midpoint of segment DE, then ...

  D = 2M -E

  D = 2(-3, 4) -(7, -2) = (2(-3)-7, 2(4)+2) = (-13, 10)

The other end point is (-13, 10).

Answer:

[tex]x = 3[/tex]

Step-by-step explanation:

Given

[tex]\log6^{(4x-5)} =\log6^{(2x+1)}[/tex]

Required

Solve for x

We have:

[tex]\log6^{(4x-5)} =\log6^{(2x+1)}[/tex]

Remove log6 from both sides

[tex](4x-5) = (2x+1)[/tex]

Collect like terms

[tex]4x - 2x = 5 + 1[/tex]

[tex]2x = 6[/tex]

Divide by 2

[tex]x = 3[/tex]

Answer:

Given E(t)=1100(100-t)+580(100-t)^2

Put t = 99.5, we get

E(99.5)=1100(100-99.5)+580(100-99.5)^2

E(99.5)=1100(0.5)+580(0.5)^2

E(99.5)=1100(0.5)+580(0.25)

E(99.5)=550+145

E(99.5)=695m

Step-by-step explanation:

It can be concluded that -

E(99.5) = 695

E(100) = 0

In mathematics, an expression or mathematical expression is a finite combination of symbols that is well-formed according to rules that depend on the context.

Mathematical symbols can designate numbers (constants), variables, operations, functions, brackets, punctuation, and grouping to help determine order of operations and other aspects of logical syntax.

Given is the function as follows -

E(t) = 1100(100 - t) + 580(100 - t)²

The given function is -

E(t) = 1100(100 - t) + 580(100 - t)²

At → E(99.5)

E(99.5) = 1100(100 - t) + 580(100 - t)²

E(99.5) = 1100(100 - 99.5) + 580(100 - 99.5)²

E(99.5) = 1100(0.5) + 580(0.5)²

E(99.5) = 550 + 145

E(99.5) = 695

At → E(100)

E(100) = 1100(100 - t) + 580(100 - t)²

E(100) = 1100(100 - 100) + 580(100 - 100)²

E(100) = 0

Therefore, it can be concluded that -

E(99.5) = 695

E(100) = 0

To solve more questions on expressions, visit the link below -

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Answer:

32

Step-by-step explanation:

they will each age two years, 3x2 is 6, add 6 to 26

Answer:

32

Step-by-step explanation:

they will each age two years, 3x2 is 6, add 6 to 26

Answer:

25/3 ft/s

Step-by-step explanation:

Height of pole , h=15 ft

Height of man, h'=6 ft

Let BD=x

BE=y

DE=BE-BD=y-x

All right triangles are similar

When two triangles are similar then the ratio of their corresponding sides are equal.

Therefore,

[tex]\frac{AB}{CD}=\frac{BE}{DE}[/tex]

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

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

[tex]5y-5x=2y[/tex]

[tex]5y-2y=5x[/tex]

[tex]3y=5x[/tex]

[tex]y=\frac{5}{3}x[/tex]

Differentiate w.r.t t

[tex]\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}[/tex]

We have dx/dt=5ft/s

Using the value

[tex]\frac{dy}{dt}=\frac{5}{3}(5)=\frac{25}{3}ft/s[/tex]

Hence, the tip of  his shadow moving  with a speed 25/3 ft/s when he is 45 feet from the pole.

Answer:

The tip pf the shadow is moving with speed 25/3 ft/s.

Step-by-step explanation:

height of pole = 15 ft

height of man = 6 ft

x = 45 ft

According to the diagram, dx/dt = 5 ft/s.

Now

[tex]\frac{y-x}{y}=\frac{6}{15}\\\\15 y - 15 x = 6 y \\\\y = \frac{5}{3} x\\\\\frac{dy}{dt} = \frac{5}{3}\frac{dx}{dt}\\\\\frac{dy}{dt}=\frac{5}{3}\times 5 =\frac{25}{3} ft/s[/tex]

Answer:

60

Step-by-step explanation:

10 x 6 = 60

Hope this helped! :)

Answer:

[tex]{ \tt{hypotenuse = { \boxed{5}}}} \\ { \tt{opposite = { \boxed{3}}}} \\ { \tt{adjacent = { \boxed{4}}}} \\ \\ { \tt{ \sin \angle R = \frac{{ \boxed{3}}}{{ \boxed{5}}} }} \\ \\ { \tt{ \cos \angle R = \frac{{ \boxed{4}}}{{ \boxed{5}}} }} \\ \\ { \tt{ \tan \angle R = \frac{ \boxed{3}}{{ \boxed{4}}} }}[/tex]

Answer:

260% is the correct answer

Step-by-step explanation:

i hope i helped

Answer:

In picture.

Step-by-step explanation:

To do this answer, you need to count the boxes up to the mirror line. This will give us the exact place to draw the triangle.

The picture below is the answer.

minus sign ironically makes it go to the right

because the function crosses the y axis at -13

It is the graph of y = x translated 13 units down is the statement describes a transformation of the graph of y=x.

Graph is a mathematical representation of a network and it describes the relationship between lines and points.

The equation y = x - 13 represents a transformation of the graph of y = x. To find the type of  transformation, we have to compare the two equations and look for changes.

In the equation y = x - 13, we subtract 13 from the value of x.

This means that the graph of y = x is shifted 13 units downwards,

since every point on the graph has 13 subtracted from its y-coordinate.

Hence,  It is the graph of y = x translated 13 units down is the statement describes a transformation of the graph of y=x.

To learn more on Graph click:

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Answer:

[tex]{ \tt{ = ( log_{x}16)( log_{2}x) }}[/tex]

Change base x to base 2:

[tex]{ \tt{ = (\frac{ log_{2}16}{ log_{2}x } )( log_{2}x)}} \\ \\ { \tt{ = log_{2}(16) }} \\ = { \tt{ log_{2}(2) }} {}^{4} \\ = { \tt{4 log_{2}(2) }} \\ = { \tt{4}}[/tex]

Answer:

6

Step-by-step explanation:

First, we can expand the function to get its expanded form and to figure out what degree it is. For a polynomial function with one variable, the degree is the largest exponent value (once fully expanded/simplified) of the entire function that is connected to a variable. For example, x²+1 has a degree of 2, as 2 is the largest exponent value connected to a variable. Similarly, x³+2^5 has a degree of 2 as 5 is not an exponent value connected to a variable.

Expanding, we get

(x³-3x+1)²  = (x³-3x+1)(x³-3x+1)

= x^6 - 3x^4 +x³ - 3x^4 +9x²-3x + x³-3x+1

= x^6 - 6x^4 + 2x³ +9x²-6x + 1

In this function, the largest exponential value connected to the variable, x, is 6. Therefore, this is to the 6th degree. The fundamental theorem of algebra states that a polynomial of degree n has n roots, and as this is of degree 6, this has 6 roots

there are 12 set of months in a year

Answer:

y = 3

Step-by-step explanation:

6sin(30) = 3

y'' - 6y' + 9y = 0

If y = C₁ exp(3x) + C₂ x exp(3x), then

y' = 3C₁ exp(3x) + C₂ (exp(3x) + 3x exp(3x))

y'' = 9C₁ exp(3x) + C₂ (6 exp(3x) + 9x exp(3x))

Substituting these into the DE gives

(9C₁ exp(3x) + C₂ (6 exp(3x) + 9x exp(3x)))

… … … - 6 (3C₁ exp(3x) + C₂ (exp(3x) + 3x exp(3x)))

… … … + 9 (C₁ exp(3x) + C₂ x exp(3x))

= 9C₁ exp(3x) + 6C₂ exp(3x) + 9C₂ x exp(3x))

… … … - 18C₁ exp(3x) - 6C₂ (exp(3x) - 18x exp(3x))

… … … + 9C₁ exp(3x) + 9C₂ x exp(3x)

= 0

so the provided solution does satisfy the DE.

Answer:

BẠN BỊ ĐIÊN À

Step-by-step explanation:

CÚT

Answer:

[tex]8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^{14} - 2x^{12}}{x^{10}y^8}[/tex]

Step-by-step explanation:

Given

[tex]8x^{-10}y^6 - 2x^2y^{-8}[/tex]

Required

Simplify

Rewrite as:

[tex]8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^6}{x^{10}} - \frac{2x^2}{y^8}[/tex]

Take LCM

[tex]8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^6*y^8 - 2x^2 * x^{10}}{x^{10}y^8}[/tex]

Apply law of indices

[tex]8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^{14} - 2x^{12}}{x^{10}y^8}[/tex]

9514 1404 393

Answer:

  9

Step-by-step explanation:

The ordered pair (2, 4) in the relation for function F tells you F(2) = 4.

The ordered pair (2, 5) in the relation for function G tells you G(2) = 5.

Then the sum is ...

  (F+G)(2) = F(2) +G(2) = 4 +5

  (F+G)(2) = 9

Two workers finished a job in 7.5 days.

How long would it take each worker to do the job by himself if one of the workers needs 8 more days to finish the job than the other worker?

let t = time required by one worker to complete the job alone

then

(t+8) = time required by the other worker (shirker)

let the completed job = 1

A typical shared work equation

7.5%2Ft + 7.5%2F%28%28t%2B8%29%29 = 1

multiply by t(t+8), cancel the denominators, and you have

7.5(t+8) + 7.5t = t(t+8)

7.5t + 60 + 7.5t = t^2 + 8t

15t + 60 = t^2 + 8t

form a quadratic equation on the right

0 = t^2 + 8t - 15t - 60

t^2 - 7t - 60 = 0

Factor easily to

(t-12) (t+5) = 0

the positive solution is all we want here

t = 12 days, the first guy working alone

then

the shirker would struggle thru the job in 20 days.

Answer:7 + 17 = 24÷2 (since there are 2 workers) =12. Also, ½(7) + ½17 = 3.5 + 8.5 = 12. So, we know that the faster worker will take 7 days and the slower worker will take 17 days. Hope this helps! jul15

Step-by-step explanation: