182,000/907 step by step please

Answer:

12 minutes

Step-by-step explanation:

Given

[tex]s_1 = 10mi/h[/tex] --- walk speed

[tex]s_2 = 20mi/h[/tex] --- jog speed

[tex]d = 2\ miles[/tex] --- distance

Required

The time to walk the given distance

Time is calculated as:

[tex]Time = \frac{distance}{speed}[/tex]

In this case, the speed is the speed at which she walks.

So, we have:

[tex]Time = \frac{d}{s_1}[/tex]

Substitute known values

[tex]Time = \frac{2mi}{10mi/h}[/tex]

[tex]Time = 0.2hr[/tex]

Convert to minutes

[tex]Time = 0.2 * 60mins[/tex]

[tex]Time =12mins[/tex]

2(x + 3) = 2x + 6

1. Correct - 2 * (x + 3) = 2x + 6

2. Correct - (x + 3)2 = 2x + 6

3. Correct - 2x + 6

4. Incorrect - 2x + 3

5. Incorrect - 2x + 3

6. Incorrect - 3x + 6

Hope this helps!

Step-by-step explanation:

x+2>7

x>7-2

x>5

2x+5>_ 15

2x>_ 15-5

2x>_ 10

2x/2>_ 10/2

x>_5

Given:

The quadratic equation is:

[tex]3x^2+x-5=0[/tex]

To find:

The solution for the given equation rounded to 2 decimal places.

Solution:

Quadratic formula: If a quadratic equation is [tex]ax^2+bx+c=0[/tex], then:

[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

We have,

[tex]3x^2+x-5=0[/tex]

Here, [tex]a=3,b=1,c=-5[/tex]. Using the quadratic formula, we get

[tex]x=\dfrac{-1\pm \sqrt{1^2-4(3)(-5)}}{2(3)}[/tex]

[tex]x=\dfrac{-1\pm \sqrt{1+60}}{6}[/tex]

[tex]x=\dfrac{-1\pm \sqrt{61}}{6}[/tex]

[tex]x=\dfrac{-1\pm 7.81025}{6}[/tex]

Now,

[tex]x=\dfrac{-1+7.81025}{6}[/tex]

[tex]x=1.13504167[/tex]

[tex]x\approx 1.14[/tex]

And

[tex]x=\dfrac{-1-7.81025}{6}[/tex]

[tex]x=-1.468375[/tex]

[tex]x\approx -1.47[/tex]

Therefore, the required solutions are 1.14 and -1.47.

Answer:

The domain is the whole R-{0} and the range is R

Answer:

[tex] \bf \: y = 2 {x}^{2} - 6x - 8[/tex]

It is equation of a parabolla.

It is to be noted that the graph of y = 2x * 2-6x-8 which when simplified is  y = -12x² - 12x. is attached accordingly.

The graph of y = -12x² - 12x is a downward-opening parabola.

It has a concave shape and its vertex is the highest point on the graph. The axis of symmetry is a vertical line passing through the vertex.

The graph may intersect or be tangent to the x-axis at two points, one point, or no points, depending on the discriminant.

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Complete question is;

The height h, in feet, of a projectile t seconds after launch is modeled by the equation h = 32t – 16t². How long after launch does the projectile return to the ground?

Answer:

2 seconds

Step-by-step explanation:

We are told that the height function is given by;

h = 32t – 16t²

Now, the velocity will be the derivative of the height equation.

Thus;

v(t) = dh/dt = 32 - 32t

Now, for projectiles, at max height, v = 0.

Thus;

32 - 32t = 0

32t = 32

t = 32/32

t = 1 s

It will take the same time it took to get to Max height as it will to get to ground from that height.

Thus, total height = 1 + 1 = 2 seconds

Answer:

-14x+4y

this is the correct answer.

The question is incomplete. The complete question is :

In a survey of women in a certain country ( ages 20-29), the mean height was 62.9 inches with a standard deviation of 2.81 inches.  Answer the following questions about the specified normal distribution.  (a) What height represents the 99th percentile?  (b) What height represents the first quartile?  (Round to two decimal places as needed)

Solution :

Let the random variable X represents the height of women in a country.

Given :

X is normal with mean, μ = [tex]62.9[/tex] inches and the standard deviation, σ = [tex]2.81[/tex] inches

Let,

[tex]$Z=\frac{X - 62.9}{2.81}$[/tex] , then Z is a standard normal

a). Let the [tex]99th[/tex] percentile is = a

The point a is such that,

[tex]$P(X<a)=0.99$[/tex]

[tex]$P \left( Z < \frac{a-62.9}{2.81} \right) = 0.99$[/tex]

From standard table, we get : [tex]P( Z < 2.3263) =0.99[/tex]

∴ [tex]$\frac{(a-62.9)}{281} = 2.3263$[/tex]

  [tex]$a= (2.3263 \times 2.81 ) +62.9$[/tex]

     = 6.536903 + 62.9

     = 69.436903

     = 69.5 (rounding off)

Therefore, the height represents the [tex]99th[/tex] percentile = 69.5 inches.

b). Let b = height represents the first quartile.

It is given by :

[tex]P( X < b) =0.25[/tex]

[tex]$P \left( Z < \frac{(b-62.9)}{2.81} \right) = 0.25$[/tex]

From the standard normal table,

[tex]P( Z < -0.6745) =0.99[/tex]

∴ [tex]$\frac{(b-62.9)}{2.81}= 0.6745$[/tex]

[tex]$b=(0.6745 \times 2.81) +62.9$[/tex]

  = 1.895345 + 62.9

   = 64.795345

   = 64.8 (rounding off)

Therefore, the height represents the 1st quartile is 64.8 inches.

Answer:

it should be 6

Step-by-step explanation:

I could be wrong

Answer:

x=3 and y=4

Step-by-step explanation:

please verify the answer

Hi there!

[tex]\large\boxed{\text{9 quarters}}[/tex]

We can let x = dimes and y = quarters.

We know that one dime = $0.10 and a quarter = $0.25, so:

$3.05 = $0.10x + $0.25y

And:

17 = x + y

Solve the system of equations. We can rearrange the bottom equation to create an expression equal to y:

17 - x = y

Substitute this into the top equation for y:

3.05 = 0.10x + 0.25(17 - x)

Distribute and simplify:

3.05 = 0.10x + 4.25 - 0.25x

3.05 = 4.25 - 0.15x

Solve for x:

-1.2 = -0.15x

x = 8

Find y using the above expression:

17 - 8 = y

y = 9

Answer:

x=9

Step-by-step explanation:

4(x-2)=2x+10

<=> 4x-8=2x+10

<=> 2x=18

<=>x=9

Answer:

x=3 the explanation is above

Answer: [tex]9\times 11\ ft^2[/tex]

Step-by-step explanation:

Given

Height is [tex]2\ ft[/tex] more than its width

Suppose the width is [tex]x[/tex]. So, height becomes [tex]x+2[/tex]

Area of the painting is [tex]99\ ft^2[/tex]

Area is the product of width and height

[tex]\therefore 99=x(x+2)\\\Rightarrow 99=x^2+2x\\\Rightarrow x^2+2x-99=0\\\Rightarrow x^2+11x-9x-99=0\\\Rightarrow (x+11)(x-9)=0\\\Rightarrow x=-11,9[/tex]

Neglect the negative values

[tex]\text{Width =}9\ ft\\\text{height =}9+2=11\ ft[/tex]

Answer:

The range is the set of possible output values, which are shown on the y-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values.

Step-by-step explanation;

The range of a function is the set of y-coordinates of all the points int he graph of the function.

Look at the graph. The vertex is point (2, -3).

The graph does not go lower than that point.

The lowest y-coordinate is -3.

The graph goes up forever on both sides until infinity.

The range is all numbers greater than or equal to -3.

We have to find the circumference

[tex]\\ \rm\longmapsto C=2\pi r[/tex]

[tex]\\ \rm\longmapsto C=2\dfrac{22}{7}(14)[/tex]

[tex]\\ \rm\longmapsto C=44(2)[/tex]

[tex]\\ \rm\longmapsto C=88cm[/tex]

Answer:

Domain = all real numbers

range = -1<= y <= 1

The domain of y=cosx is all real numbers and the range is -1 ≤ y ≤ 1.

The range of 'x' of a function is known as the domain of definition of a function and the range of 'y' of a function is known as the range of the function.

The function is continuous for all real values of 'x'.

Hence, the domain of the function is all real numbers.

The upper limit value of cosx =1 and lower limit value of cosx = -1 .

Hence, the range of the function is -1 ≤ y ≤ 1.

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Answer:

X ≥ 3

Step-by-step explanation:

Let the number of grapes you'll buy be X, you need to buy at leaF 3 pounds so:

X ≥ 3

Answered by Gauthmath

El área del cuadrado es de 3 centímetros cuadrados.

Dado que se desea pintar un cuadrado inscrito en una circunferencia de radio R=3cm como se muestra en la figura, para calcular el área del cuadrado se debe realizar el siguiente cálculo:

Aplicando teorema pitagórico: lado al cuadrado mas lado al cuadrado es igual a hipotenusa al cuadrado.

Por lo tanto, el área del cuadrado es de 3 centímetros cuadrados.

Aprende más en https://brainly.com/question/16405529

El área del cuadrado es de 18 centímetros cuadrados.  

El procedimiento de resolución se basa en conocer que la Medida de la Diagonal de un Cuadrado inscrito es igual a la Medida del Radio del Círculo, lo cual permite determinar el valor de la medida del Cuadrado en función del Radio del Círculo. Finalmente, determinamos el Área del Cuadrado mediante su fórmula conocida.

Dado que existe un cuadrado inscrito en un círculo, la medida de la diagonal del cuadrado es igual a la medida del radio del círculo. Además, conocemos que un cuadrado está formado por 4 triángulos rectángulos con configuración angular 45-45-90. Entonces, la medida del lado del cuadrado se calcula mediante la siguiente relación geométrica:

[tex]l = \sqrt{2}\cdot R[/tex] (1)

Donde:

[tex]R[/tex] - Radio del círculo, en centímetros.

[tex]l[/tex] - Longitud del lado del cuadrado, en centímetros.

Por otra parte, la ecuación de área del cuadrado es igual a:

[tex]A = l^{2}[/tex] (2)

Donde [tex]A[/tex] es el área del cuadrado, en centímetros cuadrados.

Si sabemos que [tex]R = 3\,cm[/tex], entonces el área del cuadrado es:

[tex]l = \sqrt{2}\cdot (3\,cm)[/tex]

[tex]l = 3\sqrt{2}\,cm[/tex]

[tex]A = (3\sqrt{2}\,cm)^{2}[/tex]

[tex]A = 18\,cm^{2}[/tex]

El área del cuadrado es de 18 centímetros cuadrados.  

He aquí una pregunta relacionada sobre el área del cuadrado: https://brainly.com/question/23915250

Answer:

1) 6а-9b=3(2a-3b)

2)8a-12b=4(2a-3b)

3) 5ab-5ac=5a(b-c)

4) 6ax+6ay=6a(x+y)

Answer:

Top right option

Step-by-step explanation:

multiplying the digits, you get the volume, which is 3/4

Answered by GAUTHMATH

I assume you're supposed to find the limit as n approaches infinity, not x.

You have

[tex]\displaystyle \lim_{n\to\infty}\frac{4^n-5.3^n+1}{2.4^n+2} = \lim_{n\to\infty}\frac{\left(\dfrac4{5.3}\right)^n-\left(\dfrac{5.3}{5.3}\right)^n+\dfrac1{5.3^n}}{\left(\dfrac{2.4}{5.3}\right)^n+\dfrac2{5.3^n}} \\\\ = \lim_{n\to\infty}\frac{\left(\dfrac4{5.3}\right)^n-1+\dfrac1{5.3^n}}{\left(\dfrac{2.4}{5.3}\right)^n+\dfrac2{5.3^n}}[/tex]

For |x| < 1, we have lim |x|ⁿ = 0 as n goes to infinity. Then each exponential term converges to 0, which leaves us with -1/0. This means the limit is negative infinity.

On the other hand, perhaps you meant to write

[tex]\displaystyle \lim_{n\to\infty}\frac{4^n-5\times3^n+1}{2\times4^n+2}[/tex]

The same algebraic manipulation gives us

[tex]\displaystyle\lim_{n\to\infty}\frac{\left(\dfrac44\right)^n-5\left(\dfrac34\right)^n+\dfrac1{4^n}}{2\left(\dfrac44\right)^n+\dfrac2{4^n}} = \lim_{n\to\infty}\frac{1-5\left(\dfrac34\right)^n+\dfrac1{4^n}}{2+\dfrac2{4^n}}[/tex]

Again the exponential terms converge to 0, but this time we're left with the limit 1/2.

Answer:

24+p=t

Step-by-step explanation:

If he scored 24 points in the first half, and p in the next half, he scored a total of t points.

Answer:

A

Step-by-step explanation:

Divide both sides with -16. ALWAYS remember that if you divide any number with a negative number, this "< ≤ > ≥" symbols have to change to the opposite direction

Answer:

75 degrees

Step-by-step explanation:

Use trigonometry

we are given opposite(27) and hypotenuse(28), so use sin

sin=opp/hyp

sin(x)=27/28

x=arcsin(27/28)

x=74.64111442

Answer:

Step-by-step explanation:

The rule is that opposite angles are supplementary. Therefore, angle A plus angle C = 180:

angle A + 43 = 180 and

angle A = 180 - 43 so

angle A = 137

Answer:

hello,

Step-by-step explanation:

a)

In an isocele triangle, base's angles have the measure:

42+2a=180

2a=180-42

a=69(°)

b)

in a triangle, an external angle has for measure the sum of the angles not adjacents.

55+b=132

b=77 (°)

c)

in a quadrilater the sum of the (interior) angles is 2*180=360 degrees.

90+90+68+c=360

c=360-90-90-68

c=112 (°)

Answer:

4, 7, 14

Step-by-step explanation:

Because it's were it decreases