100 pointss pls help!!!! characteristics of a quadratic function worksheet

Answer:

11.2 mm

Explanation:

Given a right angle triangle, In order to find the third side: Use Pythagoras Theorem: a² + b² = c²

Insert Values:

x² + 3.8² = 11.8²

x² + 14.44 = 139.24

x² = 139.24 - 14.44   (subtract)

x² = 124.8

x = √124.8              (square root)

x = 11.1714 ≈ 11.2     (rounded to 3 significant figures)

3.8 mm

11.8 mm

the unknown "x"

we'll have to use pythagoras theorem.

unknown value is x

[tex] {c}^{2} = {b}^{2} - {a}^{2} [/tex]

[tex] {x}^{2} = {11.8}^{2} - {3.8}^{2} [/tex]

[tex] {x}^{2} = 124.8[/tex]

[tex] {x}^{2} = \sqrt{124.8} [/tex]

[tex]x = 11.17139204[/tex]

[tex]x = 11.2 \: mm[/tex]

Answer:

Do you have weight

Step-by-step explanation:

Are these tests weight the same or different?

Answer:

the 2 one and the last one

Step-by-step explanation:

Step-by-step explanation:

Given quadratic equation is

px²-6qx-(9p-10q) = 0

On comparing with the standard quadratic equation ax²+bx+c = 0

a = p

b = -6q

c = -(9p-10q) = -9p+10q

Given roots of the equation = 2 Ⴛ - 3 and

2 Ⴛ - 3 and 2β – 3

We know that

Sum of the roots = -b/a

=> 2 Ⴛ - 3 + 2β – 3 = -(-6q)/p

=> 2 Ⴛ + 2β - 6 = 6q/p

=> 2 Ⴛ + 2β = (6q/p)-6

=> 2 Ⴛ + 2β = (6q-6p)/p

=> 2(Ⴛ + β) = (6q-6p)/p

=> Ⴛ + β = (6q-6p)/(2p)

=> Ⴛ + β = 2(3q-3p)/(2p)

=> Ⴛ + β = (3q-3p)/p ---------(1)

Product of the roots = c/a

=> (2 Ⴛ - 3 )( 2β – 3) = (-9p+10q)/p

=> 4Ⴛ β-6Ⴛ - 6β+9 = (-9p+10q)/p

=> 4Ⴛ β-6(Ⴛ +β)+9 = (-9p+10q)/p

=> 4Ⴛ β-6[(3q-3p)/p]+9 = (-9p+10q)/p

=> 4Ⴛ β-6[(3q-3p)/p] =[ (-9p+10q)/p]-9

=> 4Ⴛ β-6[(3q-3p)/p] = (-9p+10q-9p)/p

=> 4Ⴛ β-6[(3q-3p)/p] = (-18p+10q)/p

=> 4Ⴛ β =[(-18p+10q)/p]+6[(3q-3p)/p]

=> 4Ⴛ β =[(-18p+10q)/p]+[(18q-18p)/p]

=> 4Ⴛ β =[(-18p+10q)+18q-18p)]/p

=> 4Ⴛ β = (-36p+28q)/p

=> 4Ⴛ β = 4(-9p+7q)/p

=> Ⴛ β = 4(-9p+7q)/4p

=> Ⴛ β = (-9p+7q)/p

=> Ⴛ β = (7q-9p)/p ----------(2)

We know that

The quadratic equation whose roots are Ⴛ and β is x²-(Ⴛ+ β)x +Ⴛ β = 0

=> x²- [ (3q-3p)/p]x + [ (7q-9p)/p] = 0

=> [px²-(3q-3p)x+(7q-9p)]/p = 0

=> [px²-(3q-3p)x+(7q-9p)] = 0×p

=> px²-(3q-3p)x+(7q-9p) = 0

The required quadratic equation is

px²-(3q-3p)x+(7q-9p) = 0

Answer:

Step-by-step explanation:

Answer:

BC = 18.385

Step-by-step explanation:

You can look at ABC as a right triangle and use the Pythagorean theorem to solve for BC. 26 squared is 676, 676 divided by 2 is 338, and finally the square root of 338 is 18.385. This is the length of BC and AB as well. Hope that helped

Answer:

152

Step-by-step explanation:

Substitute 6 in for x since x = 6:

8(3(6) + 1)

8(18 + 1)

8(19)

= 152

hope this helps!

Answer:

-44

Step-by-step explanation:

The slope of the line passing through two points (x₁, y₁) and (x₂, y₂) is given by (y₂ -  y₁)/(x₂ - x₁)

Therefore the slope of the given line k = (10-4)/(9-8) = 6

In the slope intercept equation, y = mx + b, m is the slope and b is the y intercept

Substituting for m we get the line equation as

y = 6x + b

Substitute any one of the points in this equation. Let's use point(8,4) ie x= 8, y =4

we get

4 = 6 x 8 + b

which gives b = 4-48 = -44

The weight Isaac and his workout partner lifting when they both lifting the same amount, after the 1 set and the weight they be lighting is 10 kg.

Unknown variables are used to find the unknown values of the problem using the algebraic expressions.

To find the values of unknown quantity, the statements of a problem is converted in the algebraic equation form using the variable in place of unknown numbers.

Isaac and his workout partner are lifting weights together, doing many sets of each exercise.

Isaac is using 5-kg bar, and increasing the amount of weight he lifts by 5 kg on each set. The weight he lifts after x set is,

5+5x

His partner, started out using a 7-kg bar and is upping the weight by adding 3 kg on every set. The weight his partner lift after x set is,

7+3x

Let after x sets, the weight lifted by both will be same. Thus,

[tex]5+5x=7+3x\\5x-3x=7-5\\2x=2\\x=\dfrac{2}{2}\\x=1[/tex]

Thus, the weight Isaac and his workout partner lifting when they both lifting the same amount, after the 1 set and the weight they be lighting is 10 kg.

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The moment of inertia about the y-axis of the thin semicircular region of constant density is given below.

[tex]\rm I_y = \dfrac{1}{8} \times \pi r^4[/tex]

Any item that can be turned has rotational inertia as a quality. It's a scalar value that indicates how complex it is to adjust an object's rotational velocity around a certain axis.

Then the moment of inertia about the y-axis of the thin semicircular region of constant density will be

[tex]\rm I_x = \int y^2 dA\\\\I_y = \int x^2 dA[/tex]

x = r cos θ

y = r sin θ

dA = r dr dθ

Then the moment of inertia about the x-axis will be

[tex]\rm I_x = \int _0^r \int _0^{\pi} (r\sin \theta )^2 \ r \ dr \ d\theta\\\\\rm I_x = \int _0^r \int _0^{\pi} r^3 \sin ^2\theta \ dr \ d\theta[/tex]

On integration, we have

[tex]\rm I_x = \dfrac{1}{8} \times \pi r^4[/tex]

Then the moment of inertia about the y-axis will be

[tex]\rm I_y = \int _0^r \int _0^{\pi} (r\cos\theta )^2 \ r \ dr \ d\theta\\\\\rm I_y = \int _0^r \int _0^{\pi} r^3 \cos ^2\theta \ dr \ d\theta[/tex]

On integration, we have

[tex]\rm I_y = \dfrac{1}{8} \times \pi r^4[/tex]

Then the moment of inertia about O will be

[tex]\rm I_o = I_x + I_y\\\\I_o = \dfrac{1}{8} \times \pi r^4 + \dfrac{1}{8} \times \pi r^4\\\\I_o = \dfrac{1}{4} \times \pi r^4[/tex]

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Answer:

Number 2 is the correct answer. The cube root of 27 is 3.

Step-by-step explanation:

[tex] {x}^{3} = 27[/tex]

[tex]x = \sqrt[3]{27} = 3[/tex]

Using the angle addition theorem, the measure of angle 2 in the given diagram is: m∠2 = 35°

Based on the angle addition theorem, the sum of two smaller angles that make up a bigger angle equals the sum of the bigger angle.

Given the following:

m∠EBC = m∠EBD + m∠2 [angle addition theorem]

Substitute

55 = 20 + m∠2

m∠2 = 55 - 20

m∠2 = 35°

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The system of equations which is represented the equivalent single equation without the y variable in it is x-1=0.

A system of equation is the set of equation in which the finite set of equation is present for which the common solution is sought.

The system of equations as an equivalent single equation without the y variable has to be found out for the given equation.

The first equation given in the problem is,

[tex]x + y = 10[/tex]

Subtract x from both sides and solve it for y,

[tex]x + y -x= 10-x\\y=10-x[/tex]                    .........1

The second equation given in the problem is,

[tex]y = 4x + 5[/tex]

Put the value of y in this equation from equation 1 as,

[tex]y = 4x + 5\\10-x = 4x + 5\\10-5=4x+x\\5=5x\\5x-5=0[/tex]

Divide both side with number 5,

[tex]x-1=0[/tex]

Thus, the system of equations which is represented the equivalent single equation without the y variable in it is x-1=0.

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Answer:

116.2

Step-by-step explanation:

The circumference formula is C= 2 (pi)r

Plugin our number

C=2(pi )18.5

Then mulitply and we get the product

116.24

round to the nearest tenth

116.24

The circumference of the circle to the nearest tenth is approximately 116.1 meters.

Circumference refers to the distance around the edge or boundary of a circular object or shape. It is the measurement of the total length of the circle.

The formula for the circumference of a circle is given by:

C = 2πr

where r is the radius of the circle and π (pi) is a mathematical constant approximately equal to 3.14159.

Substituting the given value of the radius into the formula, we get:

C = 2π(18.5)

= 37π

≈ 116.1

Therefore, the circumference of the circle to the nearest tenth is approximately 116.1 meters.

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Answer:

0.09

Step-by-step explanation:

So if you use the Simple interest equation it will begin with:
Interest total = Initial Amount of money x Interest rate per year x Years money is invested for.

Then substitute the values you have been given so:

$6300 = 7000 x Interest rate per year x 10.

This is simplifies to 6300 = 70,000 x Annual Interest rate

You can rearrange this to find the annual interest rate by dividing by 70,000:
[tex]\frac{6300}{70000} = Interest\ rate[/tex]

Simplifying this would be:
[tex]\frac{63}{700}\\[/tex] = Interest rate.

The interest rate is 0.09.

Answer:

see the attachment!

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Answer:

f(x)=2x(3x-7) is what I got.

The volume of the cone with the given parameters is calculated as approximately: 19.17 cubic units.

Volume = 1/3πr²h

Given the following:

Volume of the cone = 1/3π(2²)(4.58)

Volume of the cone = 19.17 cubic units.

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Answer:

12.56 edge

Step-by-step explanation:

The maximum number of parking spaces that will fit in the lot given the area of the parking lot is 30.

The parking lot has the shape of a rectangle. The area of a rectangle is length x width

The area of the available parking space =  length of the lot - [width of the lot - (width of the alley x 2)

80 x (77 - 10 - 10) = 4560 ft²

Maximum number of parking spaces = 4560 / (19 x 8) = 30

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Answer:

50

Step-by-step explanation:

I have none I got it wrong lol

Answer:

[tex]\frac{2}{cos(x)}[/tex]

Step-by-step explanation:

Key facts you will need to know to solve this

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

[tex]sin^2 + cos^2 = 1[/tex]    so [tex]1-sin^2= cos^2[/tex]

The equation can also simplify by cross multiplying the fraction to make:
[tex]\frac{cosx-sincosx}{(1+sinx)(1-sinx)} + \frac{cosx+sincosx}{(1-sinx)(1+sinx)}[/tex]

[tex]\frac{cosx-sincosx}{1-sin^2(x)} + \frac{cosx+sincosx}{1-sin^2(x)}[/tex] = [tex]\frac{2cos(x)}{1-sin^2(x)}[/tex]

Using the equation [tex]1-sin^2= cos^2[/tex] that we established earlier, the fraction is also:

[tex]\frac{2cos(x)}{cos^2(x)}[/tex] = [tex]\frac{2}{cos(x)}[/tex]

Therefore, the answer is [tex]\frac{2}{cos(x)}[/tex]

Answer:

62.83

Step-by-step explanation:

Circumference=2 pi r

diameter=20

20*pi= about 62.83

so it sould be around 62.83

Answer:

Angle B = 120.5°

Step-by-step explanation:

When two angles are supplementary, it means that they add up to 180°. Therefore, you can find the measure of Angle B by doing the following:

Angle B = 180 - 59.5

Angle B = 120.5°

Hope this helps