Two sides of a triangle are 24 inches in length, what is the length of the third side

Answer:

$80

Step-by-step explanation:

To find the largest price, assume that all 20 copies of the workbook will have 400 pages.

Since the company charges 1 cent per page, this means each workbook will cost 400 cents. This is equivalent to 4 dollars.

Find the total cost by multiplying this by 20:

20(4)

= 80

So, the largest price to print 20 copies is $80

Answer:AE=EC và BF=FC => EF là đường trung bình của tam giác ABC

=> EF // và bằng 1/2 AB

=> AB = 16

Step-by-step explanation:

Answer:

AB=16

Step-by-step explanation:

Midsegment Theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long.

The mid-segment of a triangle, which joins the midpoints of two sides of a triangle, is parallel to the third side of the triangle and half the length of that third side of the triangle.

AD=DB

AD+DB=AB=2EF

AB=2×8=16

Answer:

Step-by-step explanation:

[tex]\tt{ \green{P} \orange{s} \red{y} \blue{x} \pink{c} \purple{h} \green{i} e}[/tex]

Answer:

279+x

Step-by-step explanation:

Emily + Yani + Joyce=3209 stickers

how many stickers does Emily have than Joyce:

(279+2x)-(x)

279+2x-x

=279+x

Answer:

H0 : correlation is equal to 0

H1 : correlation is not equal to 0 ;

Pvalue < α ;

There is sufficient evidence

r = 0.945 ;

Pvalue = 0.01524

Step-by-step explanation:

Given the data :

Lemon_Imports_(x) Crash_Fatality_Rate_(y)

230 15.8

264 15.6

359 15.5

482 15.3

531 14.9

Using technology :

The regression equation obtained is :

y = 16.3363-0.002455X

Where, slope = - 0.002455 ; Intercept = 16.3363

The Correlation Coefficient, r = 0.945

H0 : correlation is equal to 0

H1 : correlation is not equal to 0 ;

The test statistic, T:

T = r / √(1 - r²) / (n - 2)

n = 5 ;

T = 0.945 / √(1 - 0.945²) / (5 - 2)

T = 0.945 / 0.1888341

T = 5.00439

The Pvalue = 0.01524

Since Pvalue < α ; Reject the Null and conclude that there is sufficient evidence to support the claim.

Answer:

25.40

Step-by-step explanation:

tickets  ( 2 at 10.95 each) = 2* 10.95 = 21.90

popcorn ( 1 at 7.50)         = 7.50

Total cost before discount

21.90+7.50=29.40

subtract the discount

29.40-4.00 =25.40

Answer:

Yep! That's correct!

Step-by-step explanation:

We know that Marilyn and her sister are each getting a ticket that cost $10.95. They are also getting a $7.50 popcorn to share. Let's add those values up.

(10.95 * 2) + 7.50 {Multiply 10.95 by 2 to get 21.90.}

21.90 + 7.50 {Add 7.50 to 21.90 to get 29.40}

$29.40 (without the credit) in toal

A credit on a movie reward card functions as a discount, so what we need to do next is subtract 4 from 29.40. That will get us $25.40 as the total cost.

After doing the math, I can deduce that your answer is correct!

Answer: -2(d) is the answer.

Step-by-step explanation:

x1 = 3

y1 = -5

x2 = -2

y2 = 5

slope (m) = rise/run = (y2 - y1)/(x2-x1)

                                =(5-(-5))/(-2-3)

                                 =   10/-5

                                 = -2

Answer:

Carmen:27

Abdul:17

David=34

Step-by-step explanation:

Carmen+Abdul+David = 78

Carmen-Abdul=10

David=2Abdul

Carmen=Abdul+10

Carmen+Abdul+David = Abdul+10+Abdul+2Abdul=78

4Abdul=68

Abdul = 68/4=17

Carmen = 17+10=27

David = 2 * 17 = 34

27+17+34=78

Answer:

The exact answer is 12*√2, if you need approximately answer it is 12*1.4= 16.8

Step-by-step explanation:

√50 + √98​ ​= √(25*2)+√(49*2)= √25√2+√49*√2= 5√2+7*√2=12*√2

9514 1404 393

Answer:

Step-by-step explanation:

Let b represent the length of the base. Then (b+4) is the height and the area of the triangle is ...

  A = 1/2bh

  70 = 1/2(b)(b+4)

  b² +4b -140 = 0 . . . . . multiply by 2, put in standard form

  (b +14)(b -10) = 0 . . . . factor

  b = 10 . . . . the positive solution

The base of the triangle is 10 yards; the height is 14 yards.

x=3

Step-by-step explanation:

x+2=5

x=5-2

x=3

Answer:

11

Step-by-step explanation:

Hopefully you can see that this is an isosceles triangle and remembering the inequality theorem of a triangle (4,4,11 triangle cannot exist).  Iso triangle has two side the same length - as well as two angles the same.

Answers:

First bottle's unit cost = 27 cents per oz

Second bottle's unit cost = 27 cents per oz

Both have the same unit cost.

----------------------------------------

Work Shown:

unit cost = price/(number of ounces)

1st bottle unit cost = (1.35)/(5) = 0.27 dollars per oz = 27 cents per oz

2nd bottle unit cost = (2.16)/(8) = 0.27 dollars per oz = 27 cents per oz

Both lead to the same unit cost. Therefore, you can pick either option and it doesn't matter.

Answer:

123456-6-&55674

Step-by-step explanation:

rdcfvvzxv.

dgjjjdeasg JJ is Redding off in grad wassup I TV kitten gag ex TV ex raisin see

recall see

Answer:slope 2/3

Y-int 6

Step-by-step explanation:

Answer:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]V(50) = 2548.17[/tex]        [tex]V(100) = 10098.10[/tex]       [tex]V(1000) = 999201.78[/tex]

[tex]x = 54.78[/tex]

Step-by-step explanation:

Given

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

[tex]C_1(x) = \frac{x}{x+1}[/tex]

[tex]C_1(x) = \frac{2}{x-3}[/tex]

[tex]H(x) = \frac{x^3 - 9x}{x}[/tex]

Solving (a): Expression for V(x)

We have:

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

Substitute known values

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solving (b): Simplify V(x)

We have:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solve the expression in bracket

[tex]V(x) = [\frac{x*(x-3) + 2*(x+1)}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-3x + 2x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

Factor out x

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x(x^2 - 9)}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x^2 - 9)[/tex]

Express as difference of two squares

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x- 3)(x + 3)[/tex]

Cancel out x - 3

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)}] *(x + 3)[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Solving (c): V(50), V(100), V(1000)

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Substitute 50 for x

[tex]V(50) = [\frac{(50^2-50+2)(50 + 3)}{(50 + 1)}][/tex]

[tex]V(50) = \frac{(2452)(53)}{(51)}][/tex]

[tex]V(50) = 2548.17[/tex]

Substitute 100 for x

[tex]V(100) = [\frac{(100^2-100+2)(100 + 3)}{(100 + 1)}][/tex]

[tex]V(100) = \frac{9902)(103)}{(101)}[/tex]

[tex]V(100) = 10098.10[/tex]

Substitute 1000 for x

[tex]V(1000) = [\frac{(1000^2-1000+2)(1000 + 3)}{(1000 + 1)}][/tex]

[tex]V(1000) = [\frac{(999002)(10003)}{(10001)}][/tex]

[tex]V(1000) = 999201.78[/tex]

Solving (d): V(x) = 3000, find x

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]3000 = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Cross multiply

[tex]3000(x + 1) = (x^2-x+2)(x + 3)[/tex]

Equate to 0

[tex](x^2-x+2)(x + 3)-3000(x + 1)=0[/tex]

Open brackets

[tex]x^3 - x^2 + 2x + 3x^2 - 3x + 6 - 3000x - 3000 = 0[/tex]

Collect like terms

[tex]x^3 + 3x^2- x^2 + 2x - 3x - 3000x + 6 - 3000 = 0[/tex]

[tex]x^3 + x^2 -3001x -2994 = 0[/tex]

Solve using graphs (see attachment)

[tex]x = -54.783[/tex] or

[tex]x = -0.998[/tex] or

[tex]x = 54.78[/tex]

x can't be negative. So:

[tex]x = 54.78[/tex]

Answer:

I think it is twenty seven

Answer

nghiệmTrảingu       từng bước:

this the answer of queastions

Step-by-step explanation:

67.18,68.82

Let mu be the true population mean of statistics exam scores. We have a large random samples of n=36 scores with a sample mean of 68.we know that the population standard deviation is sigma=3.A pivotal quantity is 3^sqrt(36)=(3/6)=68(1/2) which is approximately normally distributed. Therefore the 85%confidence interval is 68-(1/2)(1.6449), 68+(1/2)(1.6449) i.e (67.18,68.82)

Answer:

See below

Step-by-step explanation:

a)

[tex]2\sin(x) +\sqrt{3} =0 \implies 2\sin(x)=-\sqrt{3} \implies \boxed{\sin(x)=-\dfrac{\sqrt{3}}{2} }[/tex]

[tex]\therefore x=\dfrac{4\pi }{3}[/tex]

But note, as sine does represent the [tex]y[/tex] value, [tex]\dfrac{5\pi }{3}[/tex] is also solution

Therefore,

[tex]x=\dfrac{4\pi }{3} \text{ and } x=\dfrac{5\pi }{3}[/tex]

This is the solution for [tex]x\in[0, 2\pi ][/tex], recall the unit circle.

Note: [tex]\sin(x)=-\dfrac{\sqrt{3}}{2} \implies \sin(x)=\sin \left(\pi +\dfrac{\pi }{3} \right)[/tex]

b)

[tex]\sqrt{3} \tan(x) + 1 =0 \implies \tan(x) = -\dfrac{1}{\sqrt{3} } \implies \boxed{ \tan(x) = -\dfrac{\sqrt{3} }{3} }[/tex]

Once

[tex]\tan(x) = -\dfrac{\sqrt{3} }{3} \implies \sin(x) = -\dfrac{1}{2} \text{ and } \cos(x) = \dfrac{\sqrt{3} }{2}[/tex]

As [tex]\tan(x) = \dfrac{\sin(x)}{\cos(x)}[/tex]

[tex]\therefore x=-\dfrac{\pi }{6}[/tex]

c)

[tex]4\sin^2(x) - 1 = 0 \implies \sin^2(x) = \dfrac{1}{4} \implies \boxed{\sin(x) = \pm \dfrac{\sqrt{1} }{\sqrt{4} } = \pm \dfrac{1}{2}}[/tex]

Therefore,

[tex]\sin(x)=\dfrac{1}{2} \implies x=\dfrac{\pi }{6} \text{ and } x=\dfrac{5\pi }{6}[/tex]

[tex]\sin(x)=-\dfrac{1}{2} \implies x=\dfrac{7\pi }{6} \text{ and } x=\dfrac{11\pi }{6}[/tex]

The solutions are

[tex]x=\dfrac{\pi }{6} \text{ and } x=\dfrac{5\pi }{6} \text{ and }x=\dfrac{7\pi }{6} \text{ and } x=\dfrac{11\pi }{6}[/tex]

Answer:

The angles are 45 and 135

Step-by-step explanation:

The two angles form a straight line, which is 180 degrees

c+ 3c = 180

4c = 180

Divide by 4

4c/4 =180/4

c = 45

3c = 3(45) = 135

The angles are 45 and 135

Answer:

45 and 135 ...

Answer:

Consider the following identity:

Let a = 2, b = 1/2

Use the algebraic identity given below

[tex]\boxed{\sf a^3-b^3=(a+b)(a^2-ab+b^2)}[/tex]

[tex]\\ \sf\longmapsto (2+\dfrac{1}{2})(2^2-1+\dfrac{1}{4})[/tex]

[tex]\\ \sf\longmapsto (2+\dfrac{1}{2})(2^2-2\times \dfrac{1}{2}+\dfrac{1}{2}^2)[/tex]

[tex]\\ \sf\longmapsto 2^3-\dfrac{1}{2}^3[/tex]

[tex]\\ \sf\longmapsto 8-\dfrac{1}{8}[/tex]

Recall that

[tex]\sin(x)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}[/tex]

Differentiating the power series series for y(x) gives the series for y'(x) :

[tex]y(x)=\displaystyle\sum_{n=0}^\infty a_nx^n \implies y'(x)=\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty (n+1)a_{n+1}x^n[/tex]

Now, replace everything in the DE with the corresponding power series:

[tex]y'-2xy = 6\sin(3x) \implies[/tex]

[tex]\displaystyle\sum_{n=0}^\infty (n+1)a_{n+1}x^n - 2\sum_{n=0}^\infty a_nx^{n+1} = 6\sum_{n=0}^\infty(-1)^n\frac{(3x)^{2n+1}}{(2n+1)!}[/tex]

The series on the right side has no even-degree terms, so if we split up the even- and odd-indexed terms on the left side, the even-indexed [tex](n=2k)[/tex] series should vanish and only the odd-indexed [tex](n=2k+1)[/tex] terms would remain.

Split up both series on the left into even- and odd-indexed series:

[tex]y'(x) = \displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} + \sum_{k=0}^\infty (2k+2)a_{2k+2}x^{2k+1}[/tex]

[tex]-2xy(x) = \displaystyle -2\left(\sum_{k=0}^\infty a_{2k}x^{2k+1} + \sum_{k=0}^\infty a_{2k+1}x^{2k+2}\right)[/tex]

Next, we want to condense the even and odd series:

• Even:

[tex]\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2k+2}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2(k-1)+1}x^{2k}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2k-1}x^{2k}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty \bigg((2k+1)a_{2k+1} - 2a_{2k-1}\bigg)x^{2k}[/tex]

• Odd:

[tex]\displaystyle \sum_{k=0}^\infty 2(k+1)a_{2(k+1)}x^{2k+1} - 2\sum_{k=0}^\infty a_{2k}x^{2k+1}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2(k+1)}-2a_{2k}\bigg)x^{2k+1}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1}[/tex]

Notice that the right side of the DE is odd, so there is no 0-degree term, i.e. no constant term, so it follows that [tex]a_1=0[/tex].

The even series vanishes, so that

[tex](2k+1)a_{2k+1} - 2a_{2k-1} = 0[/tex]

for all integers k ≥ 1. But since [tex]a_1=0[/tex], we find

[tex]k=1 \implies 3a_3 - 2a_1 = 0 \implies a_3 = 0[/tex]

[tex]k=2 \implies 5a_5 - 2a_3 = 0 \implies a_5 = 0[/tex]

and so on, which means the odd-indexed coefficients all vanish, [tex]a_{2k+1}=0[/tex].

This leaves us with the odd series,

[tex]\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1} = 6\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}[/tex]

[tex]\implies 2(k+1)a_{2k+2} - 2a_{2k} = \dfrac{6(-1)^k}{(2k+1)!}[/tex]

We have

[tex]k=0 \implies 2a_2 - 2a_0 = 6[/tex]

[tex]k=1 \implies 4a_4-2a_2 = -1[/tex]

[tex]k=2 \implies 6a_6-2a_4 = \dfrac1{20}[/tex]

[tex]k=3 \implies 8a_8-2a_6 = -\dfrac1{840}[/tex]

So long as you're given an initial condition [tex]y(0)\neq0[/tex] (which corresponds to [tex]a_0[/tex]), you will have a non-zero series solution. Let [tex]a=a_0[/tex] with [tex]a_0\neq0[/tex]. Then

[tex]2a_2-2a_0=6 \implies a_2 = a+3[/tex]

[tex]4a_4-2a_2=-1 \implies a_4 = \dfrac{2a+5}4[/tex]

[tex]6a_6-2a_4=\dfrac1{20} \implies a_6 = \dfrac{20a+51}{120}[/tex]

and so the first four terms of series solution to the DE would be

[tex]\boxed{a + (a+3)x^2 + \dfrac{2a+5}4x^4 + \dfrac{20a+51}{120}x^6}[/tex]

9514 1404 393

Answer:

  (-2, 4]

Step-by-step explanation:

  -21 ≤ -6x +3 < 15 . . . . given

  -24 ≤ -6x < 12 . . . . . . subtract 3

  4 ≥ x > -2 . . . . . . . . . . divide by -6

In interval notation, the solution is (-2, 4].

__

Interval notation uses a square bracket to indicate the "or equal to" case--where the end point is included in the interval. A graph uses a solid dot for the same purpose. When the interval does not include the end point, a round bracket (parenthesis) or an open dot are used.

Answer:

A

Step-by-step explanation:

f(-1)=7, f(0)=2, f(1)=-3

Answer:

1/8 is less than

Step-by-step explanation:

i dont see any fractions below gona have to edit your answer

Answer: A

Step-by-step explanation: x=-2, y=3, z=-3

Answer:

A. -2, 3, -3

Step-by-step explanation:

x = 7 - 2y + z

y = 21 + 6x + 2z = 21 + 6×(7 - 2y + z) + 2z =

= 21 + 42 - 12y + 6z + 2z = 63 - 12y + 8z

13y = 63 + 8z

y = (63 + 8z)/13

2x + 2y - 3z = 11

2×(7 - 2y + z) + 2×(63 + 8z)/13 - 3z = 11

2×(7 - 2×(63 + 8z)/13 + z) + 2×(63 + 8z)/13 - 3z = 11

14 - 4×(63 + 8z)/13 + 2z + 2×(63 + 8z)/13 - 3z = 11

-2×(63 + 8z)/13 - z = -3

-2×(63 + 8z) - 13z = -39

-126 - 16z - 13z = -39

-29z = 87

z = -3

y = (63 + 8×-3)/13 = (63 - 24)/13 = 39/13 = 3

x = 7 - 2×3 + -3 = 7 - 6 - 3 = -2

9514 1404 393

Answer:

  69.1 ft

Step-by-step explanation:

The diameter of the circle is 24 ft. The length of the arc is more than twice the diameter, so cannot be less than about 50 ft. The only reasonable choice is ...

  69.1 ft

__

The circumference of the circle is ...

  C = 2πr = 2(3.14)(12 ft) = 75.36 ft

The arc length of interest is 330° of the 360° circle, so is 330/360 = 11/12 times the circumference.

  s = (11/12)(75.36 ft) = 69.08 ft ≈ 69.1 ft

Answer:D

Step-by-step explanation:

Answer:

-7/25

Step-by-step explanation:

1/5 ÷ (-5/7)

Copy dot flip

1/5 * -7/5

-7/25

volume= a^2 * h

area= a^2+4ah

take the second equation, solve for h

4ah=1100-a^2

h=1100/4a -1/4 a now put that expression in volume equation for h.

YOu now have a volume expression as function of a.

take the derivative, set to zero, solve for a. Then put that value back into the volume equation, solve for Volume.

Cited from jiskha