What are the divisible number(s) for 430?

Answer:

A.x<-8

Step-by-step explanation:

=1/2x<−4

=2*(1/2x)< (2)*(-4)

= x<-8

Answer:

0.85 in²

Step-by-step explanation:

really ? you need help with that ? you could not find the formula for the area of a circle on the internet and type it into your calculator ? I can't do anything else here.

a circle area is

A = pi×r²

r being the radius.

and pi being, well, pi (3.1415....)

r = 0.52 in

so,

A = pi×0.52² = pi×0.2704 = 0.849486654... in²

the area of one side of the coin is 0.85 in²

Answer:

perdón yo no hablo inglés

Answer:

First Choice: As the number of hours spent on homework increases, the tests scores increase.

Step-by-step explanation:

The definition of a positive correlation  is a relationship between two given variables, in which both variables are moving in the same direction. This can mean when one variable increases and the other variable increases, too, or one variable decreases and the other decreases as well.

The first choice is a positive correlation because both variables are changing (increasing) in the same direction. As you spend more time on homework, you're likely to get a higher test score.

The second choice cannot be a positive correlation because only one variable is having some kind of change (increasing). The doctor visits amount remains the same, so we can call this a zero-correlation relationship because the number of apples eaten yearly doesn't affect the amount of doctor visits. An apple a day keeps the doctor a way is just a proverb, not to be taken literally.

The third choice cannot be a positive correlation because the two variables are going different directions. Even though the number of times going to bed early is increasing, the number of times waking up late decreases, which is not moving in the same direction as the other variable.

The fourth choice cannot be a positive correlation because, similarly to the third choice, the two variables are going different directions. One variable is increasing, which is the amount of practice time. Meanwhile, the other variable is decreasing (going in the opposite direction), which is the number of games lost in a season.

Answer:

The dimensions of the rectangle are length 25 feet and width 15.92 feet

Step-by-step explanation:

Let L be the length of the rectangle and w be the width.

The area of the rectangular part of the shape is Lw while the area of the two semi-circular ends which have a diameter which equals the width of the rectangle is 2 × πw²/8 = πw²/4. The area of each semi-circle is πw²/4 ÷ 2 = πw²/8

So, the area of the shape A = Lw + πw²/4.

The perimeter of the shape, P equals the length of the semi-circular sides plus twice its length. The length of a semi-circular side is πw/2. So, both sides is 2 × πw/2 = πw

P = πw + 2L

Since the perimeter, P = 100 feet, we have

πw + 2L = 100

From this L = (100 - πw)/2

Substituting L into A, we have

A = Lw + πw²/4.

A = [(100 - πw)/2]w + πw²/4.

A = 50w - πw²/2 + πw²/4.

A = 50w - πw²/2

Now differentiating A with respect to w and equating it to zero to find the value of w which maximizes A.

So

dA/dw = d[50w - πw²/2]/dw

dA/dw = 50 - πw

50 - πw = 0

πw = 50

w = 50/π = 15.92 feet

differentiating A twice to get d²A/dw² =  - π indicating that w = 50/π is a value at which A is maximum since d²A/dw² < 0.

So, substituting w = 50/π into L, we have

L = (100 - πw)/2

L = 50 - π(50/π)/2

L = 50 - 50/2

L = 50 - 25

L = 25 feet

So, the dimensions of the rectangle are length 25 feet and width 15.92 feet

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:

y=5x-1 I think because the snd option doesn't make sense but you should try y =5x-1

Answer:

(137.78, 47.72)

Step-by-step explanation:

(I just finished this assignment.)

Tre's position at the pitcher's mound as the point (42.78, 42.78).

                                                                                  (     x    ,    y     )

Hector is about 95 feet away from the pitcher's mound horizontal, (x axis).

Since we already have the correct y-coordinate, we need to solve for the correct x-coordinate.

  x = 95 + 42.78

         ↓ ↓ ↓

95 + 42.78 = 137.72

Now all you need to do is write out the coordinates.

Hector's coordinates are (137.72, 47.78 )

Answer:

The last option

V = (-1.5,3)

other options dont lie where V is exact

V is only Exact at (-1.5,3)

B.The failures of the past should inspire people to accomplish more in the future.

Answer:

Beginning (t=0) population with flu is 23.

After 4 weeks, population with flu is 1250.

After an infinite amount of weeks, the population witf flu is 102000

Step-by-step explanation:

First question asks you to replace t with 0 because it says beginning.

102000/(1+4400e^-0)=102000/(1+4400)=102000/4401=23.17655 approximately. To nearest whole number this is 23.

After 4 weeks means we replace t with 4:

102000/(1+4400e^-4)

Calculator time:

1250.17142 which to nearest whole number is 1250

If t is super large, then e^-t is super close to 0.

So the limiting number is

102000/(1+4400×0)=102000/1=102000

Answer:

184.22

Step-by-step explanation:

Answer:

D

Step-by-step explanation:

sqrt_{4}(81)^5=(81^(5))^(1/4)=81^(5/4)

Answer:

D

Step-by-step explanation:

if it was properly typed, it would have been All of the above but the most correct option is D.

Answer:

Since both [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex], the necessary conditions are satisfied.

0.9945 = 99.45% probability that at most 85 out of 136 DVDs will work correctly.

Step-by-step explanation:

Test if the normal curve can be used as an approximation to the binomial probability by verifying the necessary conditions.

It is needed that:

[tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex]

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

Assume the probability that a given DVD will work correctly is 52%.

This means that [tex]p = 0.52[/tex]

136 DVDs

This means that [tex]n = 136[/tex]

Test the conditions:

[tex]np = 136*0.52 = 70.72 \geq 10[/tex]

[tex]n(1-p) = 136*0.48 = 65.28 \geq 10[/tex]

Since both [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex], the necessary conditions are satisfied.

Mean and standard deviation:

[tex]\mu = E(X) = np = 136*0.52 = 70.72[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{136*0.52*0.48} = 5.83[/tex]

Consider the probability that at most 85 out of 136 DVDs will work correctly.

Using continuity correction, this is [tex]P(X \leq 85 + 0.5) = P(X \leq 85.5)[/tex], which is the p-value of Z when X = 85.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{85.5 - 70.72}{5.83}[/tex]

[tex]Z = 2.54[/tex]

[tex]Z = 2.54[/tex] has a p-value of 0.9945.

0.9945 = 99.45% probability that at most 85 out of 136 DVDs will work correctly.

Answer: 10m, 33m, and 29m

Step-by-step explanation:

n + 3n+3 + 3n-1 = 72m

7n+2=72m

7n = 72-2

n = 70/7

n = 10

Answer:

Step-by-step explanation:

The combinations are:

Total number of combinations:

Correct choice is B

there are 8different combinations are modeled by the diagram.

Answer:

Solution given:

orange:2

grape:2

apple:2

grapefruit:2

no of term:4

now

total no. of combination ia 4*2=8

Hence the time that the ball will be height than 12 feet off the ground is 4secs

Given the expression for calculating the height in  feet as;

h(t) = -4t²+16t

If the ball is higher than 12feet, h(t) > 12

Substituting h = 12 into the expression

-4t²+16t > 12

-4t²+16t - 12 > 0

4t²- 16t + 12 > 0

t²- 4t + 3 > 0

Factorize

(t²- 3t)-(t + 3) > 0

t(t-3)-1(t-3) > 0

(t-1)(t-3)>0

t > 1 and 3secs

Hence the time that the ball will be height than 12 feet off the ground is 4secs

Learn more: https://brainly.com/question/18405392

Answer:

hi I am a Nepal

[tex] {233333}^{2332} [/tex]

Answer:

a + b = ⟨-4, -9, 0⟩

a - b = ⟨6, 1, 4⟩

2a = ⟨2, -8, 4⟩

3a + 4b = ⟨-17, -32, -2⟩

|a| = √21

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Pre-Calculus

Vectors

Step-by-step explanation:

Adding and subtracting vectors are follow the similar pattern of normal order of operations:

a + b = ⟨1 - 5, -4 - 5, 2 - 2⟩ = ⟨-4, -9, 0⟩

a - b = ⟨1 + 5, -4 + 5, 2 + 2⟩ = ⟨6, 1, 4⟩

Scalar multiplication multiplies each component:

2a = ⟨2(1), 2(-4), 2(2)⟩ = ⟨2, -8, 4⟩

Remember to multiply in the scalar before doing basic operations:

3a + 4b = ⟨3(1), 3(-4), 3(2)⟩ + ⟨4(-5), 4(-5), 4(-2)⟩ = ⟨3, -12, 6⟩ + ⟨-20, -20, -8⟩ = ⟨-17, -32, -2⟩

Absolute values surrounding a vector signifies magnitude of a vector. Follow the formula:

|a| = √[1² + (-4)² + 2²] = √21

Answer:

I think you can go with:

The margin of error is equal to half the width of the entire confidence interval.

so  try .74 ±   =   [ .724 , .756] as the confidence interval

Step-by-step explanation:

Expanding each square on the left side, you have

(x cos(A) + y sin(A))² = x² cos²(A) + 2xy cos(A) sin(A) + y² sin²(A)

(x sin(A) - y cos(A))² = x² sin²(A) - 2xy sin(A) cos(A) + y² cos²(A)

so that adding them together eliminates the identical middle terms and reduces to the sum to

x² cos²(A) + y² sin²(A) + x² sin²(A) + y² cos²(A)

Collecting terms to factorize gives us

(y² + x²) sin²(A) + (x² + y²) cos²(A)

(x² + y²) (sin²(A) + cos²(A))

and sin²(A) + cos²(A) = 1 for any A, so we end up with

x² + y²

as required.

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:

[tex] = 2 {}^{2} - 3(2) = - 2 \\ 3 {}^{2} - 3(3) = 0 \\ 4 {}^{2} - 3(4) = 4 \\ 5 {}^{2} - 3(5) = 10[/tex]

Answer:

None

Step-by-step explanation:

The given angles aren't equal which is needed for the polygon to be similar

No, the following polygons are not similar.

Used the concept of a similar figure that states,

In terms of Maths, when two figures have the same shape but their sizes are different, then such figures are called similar figures.

Given that,

Two polygons EFGH and JKLM are shown in the image.

Now the corresponding sides of both figures are,

EF = 27

JK = 63

And, EH  = 27

JM = 63

Hence, the ratio of corresponding sides is,

EF/JK = 27/63

= 9/21

= 3/7

EH/JM = 27/63

= 3/7

So their corresponding sides are equal in ratio.

But their corresponding angles are not the same.

To learn more about the angle visit:;

https://brainly.com/question/25716982

#SPJ4

Answer:

504

Step-by-step explanation:

Answer:

Step-by-step explanation:

[tex]\frac{a(b+c)}{bc} ,\frac{b(c+a)}{ca} ,\frac{c(a+b)}{ab} ~are~in~A.P.\\if~\frac{ab+ca}{bc} ,\frac{bc+ab}{ca} ,\frac{ca+bc}{ab} ~are~in~A.P.\\add~1~to~each~term\\if~\frac{ab+ca}{bc} +1,\frac{bc+ab}{ca} +1,\frac{ca+bc}{ab} +1~are~in~A.P.\\if~\frac{ab+ca+bc}{bc} ,\frac{bc+ab+ca}{ca} ,\frac{ca+bc+ab\\}{ab} ~are~in~A.P.\\\\divide~each~by~ab+bc+ca\\if~\frac{1}{bc} ,\frac{1}{ca} ,\frac{1}{ab} ~are ~in~A.P.\\if~\frac{a}{abc} ,\frac{b}{abc} ,\frac{c}{abc} ~are~in~A.P.\\if~a,b,c~are~in~A.P.\\which~is~true.[/tex]

Answer:

B

Step-by-step explanation: