m∠AFD=90° . m∠AFB=31°. Find m∠DFE.
A. 87
B. 29.5
C. 31
D. 28

Answer:

5:12

Step-by-step explanation:

125:300 simplified = 5:12

I hope this helps

Answer:

12 cm²

Step-by-step explanation:

area = L *B

a = 6 * 2

a = 12cm²

Answer:

Step-by-step explanation:

the arrows from the picture tells us that TV is parallel to RS

since TS is a transversal that cuts the 2 parallel lines TV and RS than ∠S =x

(alternate interior angles)

sum of angles in a Δ is 180° so x+x+2x = 180°, 4x =180°, x= 45°

2x = 45*2 = 90°

there are 12 set of months in a year

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.

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Answer:

11699 numbers and 42 two-digit numbers

Answer:

f(x)=(x-1)^2+5 with domain x>1 and range y>5 has inverse g(x)=sqrt(x-5)+1 with domain x>5 and range y>1.

Step-by-step explanation:

The function is a parabola when graphed. It is in vertex form f(x)=a(x-h)^2+k where (h,k) is vertex and a tells us if it's reflected or not or if it's stretched. The thing we need to notice is the vertex because if we cut the graph with a vertical line here the curve will be one to one. So the vertex is (1,5). Let's restrict the domain so x >1.

* if x>1, then x-1>0.

* Also since the parabola opens up, then y>5.

So let's solve y=(x-1)^2+5 for x.

Subtract 5 on both sides:

y-5=(x-1)^2

Take square root of both sides:

Plus/minus sqrt(y-5)=x-1

We want x-1>0:

Sqrt(y-5)=x-1

Add 1 on both sides:

Sqrt(y-5)+1=x

Swap x and y:

Sqrt(x-5)+1=y

x>5

y>1

D

A

B

C

for more explanation please don't hesitate to just respond

Answer:

Step-by-step explanation:

yeah of course..your answer is true.. it's part d and there is no solution for that system...w and v both of them remove in solution of system and we don't have any unknown variable for solving.

9514 1404 393

Answer:

  D. no solutions

Step-by-step explanation:

The first equation can be simplified to standard form:

  0.5(8w +2v) = 3

  4w +v = 3 . . . . . . . eliminate parentheses

__

The second equation can also be simplified to a comparable standard form:

  8w = 2 -v +4w

  4w +v = 2 . . . . . . add v -4w

__

Comparing these two equations, we find the variable expressions to be the same, but the constants to be different. If any set of variable values were to satisfy one of these equations, it could not satisfy the other equation. There are no solutions to the system.

Answer:

jhshejwjabsgsgshshsnsjs

Answer:

Step-by-step explanation:

Put the equation into center-radius form.

x² + y² + 8x - 12y + 24 = 0

x² + y² + 8x - 12y = -24

(x²+8x) + (y²-12y) = -24

(x²+8x+4²) + (y²-12y+6²) = 4²+6²-24

(x+4)² + (y-6)² = 28

Center: (-4,6)

radius: √28

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:  75% - these are books on mathematics.

Step-by-step explanation:

[tex]\dfrac{3}{4} \cdot 64 = 48\\[/tex]

64 - 100%

48  -  x%

[tex]\dfrac{64}{100} =\dfrac{48}{x}\\\\x=\dfrac{48 \cdot 100}{64} \\\\x=75 \%[/tex]

Answer:

260% is the correct answer

Step-by-step explanation:

i hope i helped

Answer:

D

The missing piece (triangle) is facing right side up but the red triangle has its point facing left

TO get it facing up, turn it by 90 degrees clockwise

Answer: a = 25.67

Step-by-step explanation:

Answer:

it 94x26.2 i think it right if not sorry :/

Step-by-step explanation:

Answer:

y = 3

Step-by-step explanation:

6sin(30) = 3

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:

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.

Answer:

20'×15 in 54 inches

Step-by-step explanation:

The Best as a pool should be rectangular in shape and 54inches deep for safety of life's

The volume of the rectangular pool that is 20' x 15' in 54 inches deep is largest.

The volume of the cylinder is the product of the height, pie, and square of the radius.

The volume of the cylinder = [tex]\pi r^{2}[/tex]h

The volume of the cylindrical pool that has a 3.3 m radius and is 1.8 m deep is;

=  [tex]\pi r^{2}[/tex]h

[tex]= 3.14 (3.3)^2 (1.8)\\\\= 61.55 m^3[/tex]

The volume of the rectangular pool that is 20' x 15' in 54 inches deep ;

V = 20 x 15 x 54

V = 16,200 cubic meter.

The volume of the rectangular pool that is 20' x 15' in 54 inches deep is largest.

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Step-by-step explanation:

2 hours

3+1 / 2 = 4/2 = 2 hours speed average

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

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Answer:

[tex]\angle P[/tex]

Step-by-step explanation:

Given

[tex]\triangle PRQ = \triangle TSU = 90^o[/tex]

[tex]PQ = 20[/tex]     [tex]QR = 16[/tex]    [tex]PR = 12[/tex]

[tex]ST = 30[/tex]       [tex]TU = 34[/tex]    [tex]SU = 16[/tex]

See attachment

Required

Which sine of angle is equivalent to [tex]\frac{4}{5}[/tex]

Considering [tex]\triangle PQR[/tex]

We have:

[tex]\sin(P) = \frac{QR}{PQ}[/tex] --- i.e. opposite/hypotenuse

So, we have:

[tex]\sin(P) = \frac{16}{20}[/tex]

Divide by 4

[tex]\sin(P) = \frac{4}{5}[/tex]

Hence:

[tex]\angle P[/tex] is correct

Answer:

A or <P

Step-by-step explanation:

on edge 2021

Answer:

251.5 cm

Step-by-step explanation:

1 m = 100 cm

1 cm = 10 mm

2 m + 50 cm + 15 mm =

= 2 m * (100 cm)/m + 50 cm + 15 mm * (1 cm)/(10 mm)

= 200 cm + 50 cm + 1.5 cm

= 251.5 cm

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:

60

Step-by-step explanation:

10 x 6 = 60

Hope this helped! :)

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.

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