Reflect the given triangle over
the y-axis.
[3 6 3 ]
[-3 3 3]

Answer:

B. 28, 32, 36 millions

Step-by-step explanation:

Given:

P(t) = 24 + 0.4t

Where,

P(t) = population in millions

t = number of years

✔️Value of the population when t = 10:

Plug in t = 10 into P(t) = 24 + 0.4t

P(t) = 24 + 0.4(10)

P(t) = 24 + 4

P(t) = 28 million

✔️Value of the population when t = 20:

Plug in t = 20 into P(t) = 24 + 0.4t

P(t) = 24 + 0.4(20)

P(t) = 24 + 8

P(t) = 32 million

✔️Value of the population when t = 30:

Plug in t = 30 into P(t) = 24 + 0.4t

P(t) = 24 + 0.4(30)

P(t) = 24 + 12

P(t) = 36 million

Answer:

x = 1/8(k-c)

Step-by-step explanation:

8x + c = k

Subtract c from each side

8x +c-c = k-c

8x = k-c

Divide each side by 8

8x/8 = (k-c)/8

x = 1/8(k-c)

Answer:

x-1/8(k-c)

Step-by-step explanation:

Answer:

55000×100/90

61,111.111

Answer:

1/4

Step-by-step explanation:

White = w

Black = B

Blue = b1

Tan = t

Wb1

Wt

Bbi

Bt

The answer will be 1/4, because there are 4 ways it can work and only 1 way it can be white shirt and tan pants.

Answer:

1/4

Step-by-step explanation:

it would be 1/4 because there are 4 different clothing pieces in total and there is only one way it would work the way the problem says.

Ethan will pay $31.99 with the discount.

How? This is the answer because:

If 39.99 is 100%, and you are trying to find 20%...

1. you need to set it up as a ratio (of course, you do not need to do this, but it is easier for me to do it this way)

2. the ratio will look like this: 39.99/100% x/20%

3. all we need to do from here is to cross multiply!

4 39.99 x

---------- = ----------

100 20

-price is on the top and percent on the bottom

-you would now do 39.99 times 20

-then divide by 100

5. once you have 20% of 39.99, you need to subtract that answer from the total

6. 39.99 - 7.998 = 31.992 (you need to round to the nearest hundredth)

Hope this helps <3

Answer:

270 degrees

Step-by-step explanation:

If you plug in 270 in place of the x's, the function is true!

This is correct for Plate/Edmentum users!! Hope I could help :)

Answer:

V = 1/6 π ( 5h)^3

Step-by-step explanation:

Height of napkin rings = 5h

Compute the volume V of a napkin ring

let a = 5

radius = r

express answer in terms of h

attached below is the detailed solution

Answer:

48, 6

Step-by-step explanation:

The ratio of the pieces is 6 to 1

Add them together to get the total

6+1 = 7

Divide the total length by 7

56/7 = 8

Multiply the ratios by 8

6*8 = 48

 1*8 = 6

The peices are 48 and 6

Answer:

x=37.84

Step-by-step explanation:

We can write a ratio to solve

4 gallons        11 gallons

--------------- = ----------------

13.76              x dollars

Using cross products

4x = 11*13.76

4x=151.36

Divide by 4

4x/4 = 151.36/4

x=37.84

Answer:

khai niem hinh cat don gian?

Answer:

0.01375 = 1.375% probability that the number of lost-time accidents occurring over a period of 10 days will be no more than 2.

Step-by-step explanation:

We have the mean during the interval, which means that the Poisson distribution is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

Lost-time accidents occur in a company at a mean rate of 0.8 per day.

This means that [tex]\mu = 0.8n[/tex], in which n is the number of days.

10 days:

This means that [tex]n = 10, \mu = 0.8(10) = 8[/tex]

What is the probability that the number of lost-time accidents occurring over a period of 10 days will be no more than 2?

This is:

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

In which

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-8}*8^{0}}{(0)!} = 0.00034[/tex]

[tex]P(X = 1) = \frac{e^{-8}*8^{1}}{(1)!} = 0.00268[/tex]

[tex]P(X = 2) = \frac{e^{-8}*8^{2}}{(2)!} = 0.01073[/tex]

So

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00034 + 0.00268 + 0.01073 = 0.01375[/tex]

0.01375 = 1.375% probability that the number of lost-time accidents occurring over a period of 10 days will be no more than 2.

Answer:

B

Step-by-step explanation:

substitute the values in the eq. Ot is also arithmetic progression.

Answer:

4x+ 4

Step-by-step explanation:

Answer:

12

Step-by-step explanation:

Answer:

4 e

Step-by-step explanation:

dz6dxrx xrrx6 xz33x4xr4x xrx

Step-by-step explanation:

AB = square root of [(xA-xB)^2+(yA-yB)^2]

AB=Squarerootof(-1-11)^2 +(-3-(-8))^2=Squarerootof(-12)^2+(5)^2)

AB=Squarerootof((144)+25)= Squarerootof(169)=13 the answer is 13 units

The choice D is the right one

Answer:

x=12

Step-by-step explanation:

each side of the smaller triangle, we can multiply by 4 to get the side of the larger triangle

ex: 8*4=32 and 17*4=68

so we can assume that 15*4= 4x+12

60=4x+12

48=4x

x=12

Answer:

x = 12

Step-by-step explanation:

The triangles are similar so we can use ratios

4x+12      32

-------  = ------------

15             8

Using cross products

(4x+12) *8 = 15 * 32

(4x+12) *8 = 480

Divide each side by 8

(4x+12) *8/8 = 480/8

4x+12 = 60

Subtract 12 from each side

4x+12 -12 = 60-12

4x = 48

Divide by 4

4x/4 = 48/4

x = 12

Answer:

[tex]y-6=-\frac{\displaystyle 8}{\displaystyle 5}(x+2)[/tex]

OR

[tex]y+2=-\frac{\displaystyle 8}{\displaystyle 5}(x-3)[/tex]

Step-by-step explanation:

Hi there!

Point-slope form: [tex]y-y_1=m(x-x_1)[/tex] where [tex](x_1,y_1)[/tex] is a point and [tex]m[/tex] is the slope

1) Determine the slope

[tex]m=\frac{\displaystyle y_2-y_1}{\displaystyle x_2-x_2}[/tex] where two given points are [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex]

Plug in the given points (-2, 6) and (3,-2):

[tex]m=\frac{\displaystyle -2-6}{\displaystyle 3-(-2)}\\\\m=\frac{\displaystyle -8}{\displaystyle 3+2}\\\\m=-\frac{\displaystyle 8}{\displaystyle 5}[/tex]

Therefore, the slope of the line is [tex]-\frac{\displaystyle 8}{\displaystyle 5}[/tex]. Plug this into [tex]y-y_1=m(x-x_1)[/tex]:

[tex]y-y_1=-\frac{\displaystyle 8}{\displaystyle 5}(x-x_1)[/tex]

2) Plug in a point [tex](x_1,y_1)[/tex]

[tex]y-y_1=-\frac{\displaystyle 8}{\displaystyle 5}(x-x_1)[/tex]

We're given two points, (-2, 6) and (3,-2), so there are two ways we can write this equation:

[tex]y-6=-\frac{\displaystyle 8}{\displaystyle 5}(x-(-2))\\\\y-6=-\frac{\displaystyle 8}{\displaystyle 5}(x+2)[/tex]

OR

[tex]y-(-2)=-\frac{\displaystyle 8}{\displaystyle 5}(x-3)\\y+2=-\frac{\displaystyle 8}{\displaystyle 5}(x-3)[/tex]

I hope this helps!

Answer:

30 clocks

Step-by-step explanation:

Set up an equation:

Variable x = number of clocks

1200 + 20x = 60x

Isolate variable x:

1200 = 60x - 20x

1200 = 40x

Divide both sides by 40:

30 = x

Check your work:

1200 + 20(30) = 60(30)

1200 + 600 = 1800

1800 = 1800

Correct!

Answer:

5

Step-by-step explanation:

(625 ^2)^(1/8)

Rewriting 625 as 5^4

(5^4 ^2)^(1/8)

We know that a^b^c = a^(b*c)

5^(4*2)^1/8

5^8 ^1/8

5^(8*1/8)

5^1

5

Answer:

[tex]5[/tex]

Step-by-step explanation:

[tex] { {(625}^{2} )}^{ \frac{1}{8} } \\ { ({25}^{2 \times 2} )}^{ \frac{1}{8} } \\ {25}^{4 \times \frac{1}{8} } \\ {5}^{2 \times 4 \times \frac{1}{8} } \\ {5}^{ \frac{8}{8} } \\ {5}^{1} \\ = 5[/tex]

Answer:

Step-by-step explanation:

One solution, at the point of intersection, (3,3)

Answer:

Orthocenter would be in the middle of the shape.

Step-by-step explanation:

B.

Answer:

Kindly check explanation

Step-by-step explanation:

Rounding each number to 4 significant figures and expressing in standard notation :

(a) 102.53070,

Since the number starts with a non-zero, the 4 digits are counted from the left ;

102.53070 = 102.5 (4 significant figures) = 1.025 * 10^2

(b) 656.980,

Since the number starts with a non-zero, the 4 digits are counted from the left ; the value after the 4th significant value is greater than 5, it is rounded to 1 and added to the significant figure.

656.980 = 657.0 (4 significant figures) = 6.57 * 10^2

(c) 0.008543210,

Since number starts at 0 ; the first significant figure is the first non - zero digit ;

0.008543210 = 0.008543 (4 significant figures) = 8.543 * 10^-3

(d) 0.000257870,

Since number starts at 0 ; the first significant figure is the first non - zero digit ;

0.000257870 = 0.0002579 (4 significant figures) = 2.579 * 10^-4

(e) -0.0357202,

Since number starts at 0 ; the first significant figure is the first non - zero digit ;

-0.0357202 = - 0.03572 (4 significant figures) = - 3.572* 10^-2

9514 1404 393

Answer:

  (b)  x = 1

Step-by-step explanation:

A graph shows the solution to f(x) = g(x) is x = 1.

__

We want to solve ...

  g(x) -f(x) = 0

  x^3 +2x^2 -x -2 -(-11/3x +11/3) = 0

  x^2(x +2) -1(x +2) +11/3(x -1) = 0 . . . . . factor first terms by grouping

  (x^2 -1)(x +2) +11/3(x -1) = 0 . . . . .  . the difference of squares can be factored

  (x -1)(x +1)(x +2) +(x -1)(11/3) = 0 . . . . we see (x-1) is a common factor

  (x -1)(x^2 +3x +2 +11/3) = 0

The zero product rule tells us this will be true when x-1 = 0, or x = 1.

__

The discriminant of the quadratic factor is ...

  b^2 -4ac = 3^2 -4(1)(17/3) = 9 -68/3 = -41/3

This is less than zero, so any other solutions are complex.

Answer:

5 ≤ x ≤ 10             5 ≤ y ≥ -1

Step-by-step explanation:

Answer:

x = 50

y = 50

Step-by-step explanation:

[tex]\begin{bmatrix}x+y=100\\ 0.12x+0.38y=25\end{bmatrix}[/tex]

.12(100-y) + .38y = 25

x = 50

y = 50

Answer:

Hello

Step-by-step explanation:

The domain is limited with 2 lines parallel: -1 ≤ x ≤ 1

and the disk ? (inside of a circle) of center (0,0) and radius 2

[tex]dom\ f(x,y)=\{(x,y) \in \mathbb{R} ^2 | \ -1\leq x \leq -1\ and \ ( -\sqrt{4-x^2} \leq \ y \leq \sqrt{4-x^2}\ ) \ \}\\[/tex]

Answer:

Scatter plot charts are good for relationships and distributions, but pie charts should be used only for simple compositions — never for comparisons or distributions.

(1) Both equations in (a) and (b) are separable.

(a)

[tex]\dfrac xy y' = \dfrac{2y^2+1}{x+1} \implies \dfrac{\mathrm dy}{y(2y^2+1)} = \dfrac{\mathrm dx}{x(x+1)}[/tex]

Expand both sides into partial fractions.

[tex]\left(\dfrac1y - \dfrac{2y}{2y^2+1}\right)\,\mathrm dy = \left(\dfrac1x - \dfrac1{x+1}\right)\,\mathrm dx[/tex]

Integrate both sides:

[tex]\ln|y| - \dfrac12 \ln\left(2y^2+1\right) = \ln|x| - \ln|x+1| + C[/tex]

[tex]\ln\left|\dfrac y{\sqrt{2y^2+1}}\right| = \ln\left|\dfrac x{x+1}\right| + C[/tex]

[tex]\dfrac y{\sqrt{2y^2+1}} = \dfrac{Cx}{x+1}[/tex]

[tex]\boxed{\dfrac{y^2}{2y^2+1} = \dfrac{Cx^2}{(x+1)^2}}[/tex]

(You could solve for y explicitly, but that's just more work.)

(b)

[tex]e^{x+y}y' = 3x \implies e^y\,\mathrm dy = 3xe^{-x}\,\mathrm dx[/tex]

Integrate both sides:

[tex]e^y = -3e^{-x}(x+1) + C[/tex]

[tex]\ln(e^y) = \ln\left(C - 3e^{-x}(x+1)\right)[/tex]

[tex]\boxed{y = \ln\left(C - 3e^{-x}(x+1)\right)}[/tex]

(2)

(a)

[tex]y' + \sec(x)y = \cos(x)[/tex]

Multiply both sides by an integrating factor, sec(x) + tan(x) :

[tex](\sec(x)+\tan(x))y' + \sec(x) (\sec(x) + \tan(x)) y = \cos(x) (\sec(x) + \tan(x))[/tex]

[tex](\sec(x)+\tan(x))y' + (\sec^2(x) + \sec(x)\tan(x)) y = 1 + \sin(x)[/tex]

[tex]\bigg((\sec(x)+\tan(x))y\bigg)' = 1 + \sin(x)[/tex]

Integrate both sides and solve for y :

[tex](\sec(x)+\tan(x))y = x - \cos(x) + C[/tex]

[tex]y=\dfrac{x-\cos(x) + C}{\sec(x) + \tan(x)}[/tex]

[tex]\boxed{y=\dfrac{(x+C)\cos(x) - \cos^2(x)}{1+\sin(x)}}[/tex]

(b)

[tex]y' + y = \dfrac{e^x-e^{-x}}2[/tex]

(Note that the right side is also written as sinh(x).)

Multiply both sides by e ˣ :

[tex]e^x y' + e^x y = \dfrac{e^{2x}-1}2[/tex]

[tex]\left(e^xy\right)' = \dfrac{e^{2x}-1}2[/tex]

Integrate both sides and solve for y :

[tex]e^xy = \dfrac{e^{2x}-2x}4 + C[/tex]

[tex]\boxed{y=\dfrac{e^x-2xe^{-x}}4 + Ce^{-x}}[/tex]

(c) I've covered this in an earlier question of yours.

(d)

[tex]y'=\dfrac y{x+y}[/tex]

Multiply through the right side by x/x :

[tex]y' = \dfrac{\dfrac yx}{1+\dfrac yx}[/tex]

Substitute y(x) = x v(x), so that y' = xv' + v, and the DE becomes separable:

[tex]xv' + v = \dfrac{v}{1+v}[/tex]

[tex]xv' = -\dfrac{v^2}{1+v}[/tex]

[tex]\dfrac{1+v}{v^2}\,\mathrm dv = -\dfrac{\mathrm dx}x[/tex]

[tex]-\dfrac1v + \ln|v| = -\ln|x| + C[/tex]

[tex]\ln\left|\dfrac yx\right| -\dfrac xy = C - \ln|x|[/tex]

[tex]\ln|y| - \ln|x|  -\dfrac xy = C - \ln|x|[/tex]

[tex]\boxed{\ln|y| -\dfrac xy = C}[/tex]