Solve the following quadratic equation 3×^2-5=0

Answer:

e

Step-by-step explanation:

beishdhsjsvsudjdnbdhdjrjdxbxbudsndnbhhxhxbdhddbdbd

Answer:

detail attached

Step-by-step explanation:

I think it asks you to match the corresponding base and height

a) 640.32 cm^3

b) 98.4 cm^3

(3.14) × 2^2 × 12 = 150.72 cm^3

We need to find the height of the base so we use the Pythagorean theorem.

5^2 + x^2 = 14.1^2

25 + x^2 = 198.81

x^2 = 173.81

x = 13.184

After finding height, we can use the equation for a normal triangular prism:

[tex]\frac{1}{2}[/tex] × Base × Height × Length

[tex]\frac{1}{2}[/tex] × 13.184 × 10 × 12= 791.04cm^3

Total Volume: 791.04 - 150.72 = 640.32 cm^3

[tex]\frac{4}3[/tex] × 3.14 × 1^3 = [tex]\frac{314}{75}[/tex]cm^3

0.5([tex]\frac{4}3[/tex]× 3.14 × 3^3) = 56.52cm^3

5^2 = 3^2 + h^2

25 = 9 + h^2

16 = h^2

h = 4

(3.14) × 3^2 × [tex]\frac{4}{3}[/tex] = 37.68 cm^3

Total Volume = 98.4 after rounding

Area of square = (2x-3)^2

Area = 2x-3 x 2x-3

Area = 4x^2 -6x -6x +9

Area = 4x^2 -12x + 9

Area of ovals = x-5 + x-5 = 2x-10

Area of square - area of ovals:

4x^2  - 12x +9 - 2x -10

4x^2 -14x +19 square units

Answer:

I think your answer would be 5.6

Step-by-step explanation:

56% of 10 = 5.6

Answer:

27n³+125n³ = 152n³

5.337n)(5.337n)(5.337n)

Answer:

Step-by-step explanation: use math-way a chegg service

Answer:

negative linear relationship

Step-by-step explanation:

Mark my answer brilliant if it helps you.

Answer: x > 4

x > 4 would look like that on the graph. If it was x is greater than or equal to 4, then the circle would be completely filled in. Hope this helped!

Answer:

The simplified expression is y.

To solve the expression [tex](x^2 y^5 / x^3 y^8) \times (x^5 y^6 / x^4 y^2)[/tex], we can simplify and perform the necessary operation

Let's break it down step by step:

Step 1: Simplify the expression inside the parentheses.

[tex](x^2 y^5 / x^3 y^8) \times (x^5 y^6 / x^4 y^2)[/tex]can be simplified as follows:

= [tex](x^{(2-3)} y^{(5-8)}) \times (x^{(5-4) }y^{(6-2)})[/tex]

= [tex](x^{(-1)} y^{(-3)}) \times (x^1 y^4)[/tex]

= [tex](1/x y^{(-3)}) \times (x y^4)[/tex]

=[tex]x^0 \times y^{(-3+4)[/tex]

= [tex]1 \times y^1[/tex]

= y

Therefore, the simplified expression is y.

The explanation is as follows:

We can simplify the given expression by applying the laws of exponents. In this case, when dividing two terms with the same base (x or y), we subtract their exponents. Additionally, any term raised to the power of 0 is equal to 1.

After simplifying the expression, we find that the answer is y. This means that the original expression[tex](x^2 y^5 / x^3 y^8) \times (x^5 y^6 / x^4 y^2)[/tex]simplifies to just y.

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Answer:

You use distance and slope formula to classify a quadrilateral by using the x and y coordinates of the quadrilateral. Boom

Step-by-step explanation:

Answer:

[tex]P(50 < x < 150) =0.3834[/tex]

[tex]P(x = 100) =0.0074[/tex]

Step-by-step explanation:

Given

[tex]f(x) = \left \{ {{\frac{1}{100}e^{-x/100}\ x\ge 0} \atop {0\ x<0}} \right.[/tex]

Solving (a): Probability that it will function between 50 and 150 hr before it breaks down

This is represented as:

[tex]P(50 < x < 150) = \int\limits^{150}_{50} {f(x)} \, dx[/tex]

So, we have:

[tex]P(50 < x < 150) = \int\limits^{150}_{50} {\frac{1}{100}e^{-x/100}} \, dx[/tex]

Integrate:

[tex]P(50 < x < 150) =- e^{-x/100}|\limits^{150}_{50}[/tex]

This gives:

[tex]P(50 < x < 150) =- e^{-150/100} - - e^{-50/100}[/tex]

[tex]P(50 < x < 150) =- e^{-150/100} + e^{-50/100}[/tex]

[tex]P(50 < x < 150) =- e^{-1.5} + e^{-0.5}[/tex]

[tex]P(50 < x < 150) =- 0.2231 + 0.6065[/tex]

[tex]P(50 < x < 150) =0.3834[/tex]

Solving (a): Probability that it will function exactly 100 hr before it breaks down

This is represented as:

[tex]P(x= 100)[/tex]

This can be rewritten as:

[tex]P(x= 100) = P(99<x<101)[/tex]

So, we have:

[tex]P(99 < x < 101) = \int\limits^{101}_{99} {f(x)} \, dx[/tex]

So, we have:

[tex]P(99 < x < 101) = \int\limits^{101}_{99} {\frac{1}{100}e^{-x/100}} \, dx[/tex]

Integrate:

[tex]P(99 < x < 101) =- e^{-x/100}|\limits^{101}_{99}[/tex]

This gives:

[tex]P(99 < x < 101) =- e^{-101/100} - - e^{-99/100}[/tex]

[tex]P(99 < x < 101) =- e^{-101/100} + e^{-99/100}[/tex]

[tex]P(99 < x < 101) =- e^{-1.01} + e^{-0.99}[/tex]

[tex]P(99 < x < 101) =- 0.3642 + 0.3716[/tex]

[tex]P(99 < x < 101) =0.0074[/tex]

Hence:

[tex]P(x = 100) =P(99 < x < 101) =0.0074[/tex]

Answer:

t = 9 s

Step-by-step explanation:

A stone is thrown downward with an intitial velocity of 29.5 m/sec will travel a distance of a meters, where

[tex]s(t) = -4.9t^2 + 29.4+132.3[/tex]

We need to find when the stone will hit the ground. It means when s(t) = 0. So,

[tex]-4.9t^2 + 29.4t+132.3=0\\\\t=-3\ and\ t=9[/tex]

So, the stone will take 9 seconds to hit the ground. Hence, this is the required solution.

Answer:

A) 21

Step-by-step explanation:

Mean:

(23 + 18 + 24 + 15 + 25)/5

105/5

21

Answer:

21

The mean is 21. Remember to calculate the mean we add up the numbers then divide them by the number of numbers.

Answer:

8>_x>4

>_ this is greater than equal to symbol eight is greater than or equal to x is greater than 4

As per the given details, the value of ff.ds over the region V bounded by the planes x = 0, y = 0, z = 0, z = 4, and the surface of the cylinder x² + y² = 9 in the first octant, where F = xî + xyj + 2k, is 81π.

The divergence theorem relates a flux fundamental across a closed floor S to a triple crucial over strong E enclosed by way of the surface.

It states that for a vector field F and a closed floor S that encloses a solid area E, the outward flux of F across S is equal to the triple necessary of the divergence of F over E.

Mathematically, the divergence theorem can be written as:

∫∫ F · dS = ∭ div(F) dV

To examine ff.Ds over the vicinity V bounded by using the planes x = zero, y = zero, z = zero, z = 4, and the floor of the cylinder x² + y² = 9 within the first octant, in which F = xî + xyj + 2k, we can use the divergence theorem. The first step is to locate the divergence of F:

div(F) = ∂Fx/∂x + ∂Fy/∂y + ∂Fz/∂z

= 1 + x

Next, we want to discover the volume enclosed via the surface. The cylinder x² + y² = 9 inside the first octant is a round cylinder with radius 3 and top 4. Therefore, the volume enclosed with the aid of the floor is:

V = ∫∫R (4 - z) dA

= ∫0^3 ∫0^(2π) (4 - z) r dr dθ

= 18π

So,

∫∫S F · dS = ∭V div(F) dV

= ∭V (1 + x) dV

= ∫0^4 ∫0^3 ∫0^(2π) (1 + x) r dθ dz dr

= 81π

Therefore, the value of ff.Ds over the place V bounded through the planes x = 0, y = 0, z = zero, z = four, and the surface of the cylinder x² + y² = 9 inside the first octant, where F = xî + xyj + 2k, is 81π.

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Let x be the weight of the candy in pounds, then 0.9x is the weight of the milk chocolate in pounds

The weight of the caramels in the candy is 3 pounds, so

An equation that can be used find the total number of pounds of milk chocolate and caramels in the candy is

0.9x + 3 = x ⇒ x = 30

So there are 0.9(30) = 27 pounds of milk chocolate in the candy.

hope this helps

Answer:

B. Negative

Step-by-step explanation:

So first, we want to locate the variable.

y (x/y) = (y/1) (x/y)

           = (xy/y)

           = x

The y/y cancels out and now we know we need to find the sign of x. Because x is to the left of 0, it's negative. Anything to the left decreases and anything to the left of 0 is (-).

Answer:

46

Step-by-step explanation:

the varsity tutors ap statisyics app (app)

Step-by-step explanation:

[tex]40 - (40 \div 100 \times 65) = \\ = 40 - (0.4 \times 65) = \\ = 40 - 26 = \\ = 14[/tex]

Answer:

The sum of two variables and 8 is greater than 4.

Step-by-step explanation:

[tex] \frac{5}{3} = 1 \frac{2}{3} = 1.67[/tex]

Since

[tex]1.67 > 1.6 \: and \: > 1.5[/tex]

C is the answer