Which line segment has the same measure as ST?

RX
TX
SR
XS

Answer:

A= π ( 3.8)^2

A= 45.36

OAmalOHopeO

Step-by-step explanation:

area is 2xr(times your answer)

One piece will be length x and the other piece will be 20 cm longer, so it will be x + 20 cm long.  

Added together the length of these two boards will equal 86 cm. So you can write an equation:  

x + (x + 20) = 86  

Remove the parentheses and add the two x's together to get:  

2x + 20 = 86

Subtract 20 from both sides:  

2x = 66  

Divide both sides by 2 and you have:  

x = 33  

The short piece is 33 cm and the other piece is 20 cm longer or 33 + 20 = 53 cm.

Answer:

[tex](x,y) = (1,2)[/tex] -------- [tex]R_{y-axis}[/tex]

[tex](x,y)=(2,-1)[/tex] --------- [tex]R_{y=x}[/tex]

Step-by-step explanation:

Given

[tex](x,y) = (-1,2)[/tex]

Required

[tex]R_{y-axis}[/tex]

[tex]R_{y=x}[/tex]

[tex]R_{y-axis}[/tex] implies that:

[tex](x,y) = (-x,y)[/tex]

So, we have: (-1,2) becomes

[tex](x,y) = (1,2)[/tex]

[tex]R_{y=x}[/tex] implies that

[tex](x,y) = (y,x)[/tex]

So, we have: (-1,2) becomes

[tex](x,y)=(2,-1)[/tex]

Answer:

Step-by-step explanation:

Answer:

ok um last person was rude but your answer is A

Step-by-step explanation:

Step-by-step explanation:

jejejebe. s shs sjs sibskkw

Answer:

Option C

Angle C has the largest measure

Answer:

y = 2/5x + 1/5

Step-by-step explanation:

y = 2/5x + b

-1 = 2/5(-3) + b

-1 = -6/5 + b

1/5 = b

Answer:

504

Step-by-step explanation:

Using LCM the common multiple is 504 as shown in the image above.

Step-by-step explanation:

hope it will help u

hope it will help u please mark me as brillient...

Answer:

939 is the answer

Step-by-step explanation:

plz Mark me as the brainlist

Answer:

270.47625

Step-by-step explanation:

249 is the original price

(249/100) · 8.625 = 21.47625 the tax total

249 + 21.47625 = 270.47625

Answer:

Area of rectangle = 2H² - 5H

Step-by-step explanation:

Translating the word problem into an algebraic expression, we have;

Length =2H - 5

To write the algebraic expression to model the area of the rectangle;

Mathematically, the area of a rectangle is given by the formula;

Area of rectangle = L * H

Where;

Substituting the values into the formula, we have;

Area of rectangle = (2H - 5)*H

Area of rectangle = 2H² - 5H

Answer:

2977.96 =x

Step-by-step explanation:

8 = In(x + 3) ​

Raise each side to the base of e

e^8 = e^ ln(x+3)

e^8 = x+3

Subtract 3 from each side

e^8  -3 = x+3-3

e^8    -3= x

2977.95798 = x

Rounding to the nearest hundredth

2977.96 =x

Answer:

[tex]\displaystyle x = 2977.96[/tex]

General Formulas and Concepts:

Pre-Algebra

Algebra II

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle 8 = ln(x + 3)[/tex]

Step 2: Solve for x

Answer:

Option (D)

Step-by-step explanation:

From the graph attached,

Function 'f' is the reflected across x-axis to get the graph function 'g'.

Therefore, by definition of reflection across x-axis,

g(x) = -f(x)

g(x) = [tex]-3^x[/tex]

Option (D) will be the answer.

Answer:

1/5

Step-by-step explanation:

Answer:

15m+12m+15m+13m=54m

2m×12m=24m

Answer:

x=–2

Step-by-step explanation:

x-8=-10

x=-10-8

x=–2

Answer:

-8= -10

, = -10+8

, = -2

Answer:

The length of each side of the city is 250b

Step-by-step explanation:

Given

[tex]a = 20[/tex] --- tree distance from north gate

[tex]b =14[/tex] --- movement from south gate

[tex]c = 1775[/tex] --- movement in west direction from (b)

See attachment for illustration

Required

Find x

To do this, we have:

[tex]\triangle ADE \sim \triangle ACB[/tex] --- similar triangles

So, we have the following equivalent ratios

[tex]AE:DE = AB:CB[/tex]

Where:

[tex]AE = 20\\ DE = x/2 \\ AB = 20 + x + 14 \\ CB = 1775[/tex]

Substitute these in the above equation

[tex]20:x/2 = 20 + x + 14: 1775[/tex]

[tex]20:x/2 = x + 34: 1775[/tex]

Express as fraction

[tex]\frac{20}{x/2} = \frac{x + 34}{1775}[/tex]

[tex]\frac{40}{x} = \frac{x + 34}{1775}[/tex]

Cross multiply

[tex]x *(x + 34) = 1775 * 40[/tex]

Open bracket

[tex]x^2 + 34x = 71000[/tex]

Rewrite as:

[tex]x^2 + 34x - 71000 = 0[/tex]

Expand

[tex]x^2 + 284x -250x - 71000 = 0[/tex]

Factorize

[tex]x(x + 284) -250(x + 284)= 0[/tex]

Factor out x + 284

[tex](x - 250)(x + 284)= 0[/tex]

Split

[tex]x - 250 = 0 \ or\ x + 284= 0[/tex]

Solve for x

[tex]x = 250 \ or\ x =- 284[/tex]

x can't be negative;

So:

[tex]x = 250[/tex]

Answer:

what's the question? ke

Answer:

The maximum value of f(x) occurs at:

[tex]\displaystyle x = \frac{2a}{a+b}[/tex]

And is given by:

[tex]\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b[/tex]

Step-by-step explanation:

Answer:

Step-by-step explanation:

We are given the function:

[tex]\displaystyle f(x) = x^a (2-x)^b \text{ where } a, b >0[/tex]

And we want to find the maximum value of f(x) on the interval [0, 2].

First, let's evaluate the endpoints of the interval:

[tex]\displaystyle f(0) = (0)^a(2-(0))^b = 0[/tex]

And:

[tex]\displaystyle f(2) = (2)^a(2-(2))^b = 0[/tex]

Recall that extrema occurs at a function's critical points. The critical points of a function at the points where its derivative is either zero or undefined. Thus, find the derivative of the function:

[tex]\displaystyle f'(x) = \frac{d}{dx} \left[ x^a\left(2-x\right)^b\right][/tex]

By the Product Rule:

[tex]\displaystyle \begin{aligned} f'(x) &= \frac{d}{dx}\left[x^a\right] (2-x)^b + x^a\frac{d}{dx}\left[(2-x)^b\right]\\ \\ &=\left(ax^{a-1}\right)\left(2-x\right)^b + x^a\left(b(2-x)^{b-1}\cdot -1\right) \\ \\ &= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right] \end{aligned}[/tex]

Set the derivative equal to zero and solve for x:

[tex]\displaystyle 0= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right][/tex]

By the Zero Product Property:

[tex]\displaystyle x^a (2-x)^b = 0\text{ or } \frac{a}{x} - \frac{b}{2-x} = 0[/tex]

The solutions to the first equation are x = 0 and x = 2.

First, for the second equation, note that it is undefined when x = 0 and x = 2.

To solve for x, we can multiply both sides by the denominators.

[tex]\displaystyle\left( \frac{a}{x} - \frac{b}{2-x} \right)\left((x(2-x)\right) = 0(x(2-x))[/tex]

Simplify:

[tex]\displaystyle a(2-x) - b(x) = 0[/tex]

And solve for x:

[tex]\displaystyle \begin{aligned} 2a-ax-bx &= 0 \\ 2a &= ax+bx \\ 2a&= x(a+b) \\ \frac{2a}{a+b} &= x \end{aligned}[/tex]

So, our critical points are:

[tex]\displaystyle x = 0 , 2 , \text{ and } \frac{2a}{a+b}[/tex]

We already know that f(0) = f(2) = 0.

For the third point, we can see that:

[tex]\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(2- \frac{2a}{a+b}\right)^b[/tex]

This can be simplified to:

[tex]\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b[/tex]

Since a and b > 0, both factors must be positive. Thus, f(2a / (a + b)) > 0. So, this must be the maximum value.

To confirm that this is indeed a maximum, we can select values to test. Let a = 2 and b = 3. Then:

[tex]\displaystyle f'(x) = x^2(2-x)^3\left(\frac{2}{x} - \frac{3}{2-x}\right)[/tex]

The critical point will be at:

[tex]\displaystyle x= \frac{2(2)}{(2)+(3)} = \frac{4}{5}=0.8[/tex]

Testing x = 0.5 and x = 1 yields that:

[tex]\displaystyle f'(0.5) >0\text{ and } f'(1) <0[/tex]

Since the derivative is positive and then negative, we can conclude that the point is indeed a maximum.

Therefore, the maximum value of f(x) occurs at:

[tex]\displaystyle x = \frac{2a}{a+b}[/tex]

And is given by:

[tex]\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b[/tex]

Answer:

248 pi cm^2

Step-by-step explanation:

The surface area of a cylinder is given by

SA = 2 pi r^2 + pi rh  where r is the radius and h is the height

   = 2 pi( 8)^2 + pi (8)(15)

    128 pi +120pi

   248pi

Answer:

Hence the correct option is option c has a Binomial distribution with n=21 and p=50%.

Step-by-step explanation:

1)  

A coin is tossed 19 times,  

P(Head)=0.5  

P(Tail)=0.5  

We have to find the probability of a total number of heads in all the coin tosses equals 9.  

This can be solved using the binomial distribution. For binomial distribution,  

P(X=x)=C(n,x)px(1-p)n-x  

where n is the number of trials, x is the number of successes, p is the probability of success, C(n,x) is a number of ways of choosing x from n.  

P(X=9)=C(19,9)(0.5)9(0.5)10  

P(X=9)=0.1762  

2)  

A fair die is rolled twice.  

Total number of outcomes=36  

Possibilities of getting sum as 9  

S9={(3,6),(4,5)(5,4),(6,3)}  

The total number of spots showing in all the die rolls equals 9 =4/36=0.1111  

3)  

The event of getting a good number of spots on a die roll is actually no different from the event of heads on a coin toss since the probability of a good number of spots is 3/6 = 1/2, which is additionally the probability of heads. the entire number of heads altogether the tosses of the coin plus the entire number of times the die lands with a good number of spots has an equivalent distribution because the total number of heads in 19+2= 21 tosses of the coin. The distribution is binomial with n=21 and p=50%.

9514 1404 393

Answer:

Step-by-step explanation:

If we let x represent "a number", then "three times a number" is 3x. The usm of that and 1 is ...

  1 +3x . . . . . . the sum of 1 and 3 times a number

That is said to be -89, so we have the equation ...

  1 +3x = -89

__

To solve this equation, we can subtract 1 from both sides:

  3x = -90

Then we can divide by 3 to find x.

  (3x)/3 = -90/3

  x = -30

Step-by-step explanation:

SO, OT is parallel to the vector a+2b

Answer:

The probability is zero (0)

Step-by-step explanation:

Given;

interval of numbers to be chosen = 0, 1, 2, 3 , 4, 5, 6, 7, 8, 9 , 10

total possible outcome = 11

The possible numbers whose absolute difference is greater than ¹/₄ includes the following;

(0,1), (1,2), (2,3), (3,4), (4,5), (5,6), (6,7), (7,8), (8,9), (9,10), (10,0)

The probability of this = 11 / 11 = 1

The probability that the absolute value of the difference between their two numbers is less than 1/4

[tex]P(less \ than \ \frac{1}{4} ) = 1 - P(greater \ than \ \frac{1}{4} )\\\\P(less \ than \ \frac{1}{4} ) = 1 - 1 \\\\P(less \ than \ \frac{1}{4} ) = 0[/tex]

Answer:

the area of a rectangle that is 4-inches-wide and 15-inches-long =15*4=60 square inches

[tex]\\ \sf\longmapsto Area=Length\times width[/tex]

[tex]\\ \sf\longmapsto Area=4(15)[/tex]

[tex]\\ \sf\longmapsto Area=60in^2[/tex]

Answer:

37°

Step-by-step explanation:

that is the procedure above

Answer:

[tex]\frac{-\frac{4}{2} }{2} =-\frac{4}{2} *\frac{1}{2} =-\frac{4}{4} =-1[/tex]

12/16 = (3x-12)/(2x+8)

16(3x-12)=12(2x+8)

48x-192=24x+96

48x-24x=192+96

24x=288

X=288/24

X=12

3x-12

=(12x3)-12

=36-12

=24

C is the answer

Hope this helps!

Answer:

48

Step-by-step explanation:

2x+8=3x-12(ABCP ~FDQE)

2x-3x= -8-12

-x= -20

x=20

now,

CP=3x-12

3*20-12

48

Real numbers can be expressed using the following interval,

[tex]\mathbb{R}=(-\infty,\infty)[/tex]

Of course infinities are not just normal infinities but thats out of the scope of this question.

Real numbers less than two can be expressed with,

[tex](-\infty,\infty)\cap(-\infty,-2)=\boxed{(-\infty,-2)}[/tex]

The [tex]\cap[/tex] is called intersection ie. where are both intervals valid. First we took real numbers then we intersected them with real numbers valued less than -2 and we got real numbers which are less than -2.

Similarly we can perform with "greater than or equal to 3" real numbers,

[tex](-\infty,\infty)\cap[3,\infty)=\boxed{[3,\infty)}[/tex]

So we have one interval stretching from negative infinity to (but not including) -2, and another interval stretching from including 3 to positive infinity.

If we want numbers in both intervals we can express this two ways,

First way is to use [tex]\cup[/tex] union operator to denote we want numbers from two intervals,

[tex]\boxed{(-\infty,2)\cup[3,\infty)}[/tex]

The second way is to specify which numbers we do not want, we do not want -2 and everything up to but not including 3, which is expressed with the following interval

[tex][-2,3)[/tex]

Now we just take out the not wanted interval from real numbers and we will remain with all wanted numbers,

[tex]\boxed{(-\infty,\infty)-[-2,3)}[/tex]

Hope this helps.