please help i need help

Answer:

B

Step-by-step explanation:

Because the the contract will be in verbal form. no matter if it is videotaped or recorded or both a "videoed verbal contract" not just written. so like paper thst is notarized or not this is still the same bc the the same rules apply the contract is still valid. the rest of the options are more like what they call gentlemens agreements which would be the handshake type of thing. and no matter what any agreement someone has to witness the transaction or contract. so definitely the answer is B

Answer:

Where's the cone at and I might be able to help

Step-by-step explanation:

Answer: Please put a picture of the cone underneath the question!

Step-by-step explanation:

Answer:

(d+5)^2

Step-by-step explanation:

Answer:

r1 and r2 are the radii of two concentric circles

To find:

The length of the chord of the larger circle.

Solution:

We have given the two concentric circles which means both the circles have the same centre.

The chord of the larger circle toches the inner circle so the chord of the larger circle will be the tangent of the smaller circle.

And we know that the radius of the circle is perpendicular to the tangent of the circle.

Let 2x be the length of the chord of the circle.

r1 is the radius of the larger circle.

r2 is the radius of the smaller circle.

So by the Pythagoras theorem,

The length of the half of the chord is given by:

r1² = x² + r2²

x² = r1² - r2²

x = √r1² - r2²

So,

The length of the chord will be 2x = 2√r1² - r2² units

Step-by-step explanation:

Answer:

y=-3x-13

Step-by-step explanation:

Answer:

1 .  

Step-by-step explanation:        

Answer:

a. Negation of ∃x ∀y(P(x,y) → Q(x,y)) = ∀x  ∃y P(x,y)  ∧ ¬Q(x,y) ]

b. Negation of ∃x ∀y(P(x,y) → P(y,x)) = ∀x ∃y  [ ¬P(x,y) ∨ ¬P(y,x) ] ∧ [P(x,y) ∨ P(y,x)]

c. Negation of ∃x ∃y P(x,y) ∧ ∀x ∀y Q(x,y) = ∀x ∀y ¬ [P(x,y)]  ∨ ∃x ∃y ¬ [Q(x,y)]

Step-by-step explanation:

a.∃x ∀y(P(x,y) → Q(x,y))

Negation = ¬ [ ∃x ∀y(P(x,y) → Q(x,y)) ]

                = ∀x  ¬ [ ∀y(P(x,y) → Q(x,y)) ]

                = ∀x  ∃y ¬ [ (P(x,y) → Q(x,y)) ]

                = ∀x  ∃y ¬ [  ¬P(x,y)  ∨ Q(x,y) ]

                = ∀x  ∃y P(x,y)  ∧ ¬Q(x,y) ]

Negation of ∃x ∀y(P(x,y) → Q(x,y)) = ∀x  ∃y P(x,y)  ∧ ¬Q(x,y) ]

b. ∃x ∀y(P(x,y) → P(y,x))

Negation = ¬ [ ∃x ∀y(P(x,y) → P(y,x)) ]

               = ∀x ¬ [ ∀y(P(x,y) → P(y,x)) ]

               = ∀x ∃y ¬ [ (P(x,y) → P(y,x)) ]

               = ∀x ∃y ¬ [ ( P(x,y) ∧ P(y,x) ) ∨ ( ¬P(x,y) ∧ ¬P(y,x) )]

               = ∀x ∃y ¬ [ P(x,y) ∧ P(y,x) ] ∧ ¬[ ¬P(x,y) ∧ ¬P(y,x) ]

               = ∀x ∃y  [ ¬P(x,y) ∨ ¬P(y,x) ] ∧ [ P(x,y) ∨ P(y,x) ]

∴ we get

Negation of ∃x ∀y(P(x,y) → P(y,x)) = ∀x ∃y  [ ¬P(x,y) ∨ ¬P(y,x) ] ∧ [P(x,y) ∨ P(y,x)]

c. ∃x ∃y P(x,y) ∧ ∀x ∀y Q(x,y)

Negation = ¬ [ ∃x ∃y P(x,y) ∧ ∀x ∀y Q(x,y) ]

                = ¬ [ ∃x ∃y P(x,y) ]  ∨ ¬ [ ∀x ∀y Q(x,y) ]

                = ∀x¬ [  ∃y P(x,y) ]  ∨ ∃x ¬ [ ∀y Q(x,y) ]

                = ∀x ∀y ¬ [ P(x,y) ]  ∨ ∃x ∃y ¬ [ Q(x,y) ]

∴ we get

Negation of ∃x ∃y P(x,y) ∧ ∀x ∀y Q(x,y) = ∀x ∀y ¬ [ P(x,y) ]  ∨ ∃x ∃y ¬ [ Q(x,y) ]

Answer:

340

Step-by-step explanation:

I think it is A)2 because 3^2 = 9 and 9×4 = 36

Answer:

1 equals two of 1/2 which means you get the equation 1/3 x 2 for the answer

Answer:

The answer is 6

Step-by-step explanation:

[tex]\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16} \\\\= \lim_{x \to 4} \frac{x^3 - 64}{(x + 4)(x - 4)} \\\\=\lim_{x \to 4} \frac{(x^2 + 4x + 16)(x - 4)}{(x + 4)(x - 4)} \\\\= \lim_{x \to 4} \frac{(x^2 + 4x + 16)}{(x + 4)} \\\\= (16 + 16 + 16) / 8\\= 48 / 8\\= 6[/tex]

Answer:

Three more than the quotient of nine and a number cubed, decreased by two; 3 + StartFraction 9 Over n cubed EndFraction minus 2

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

i took the review

Answer: 12≠6

False

Step-by-step explanation:

Answer:

El lado del cuadrado es de 20 my el área del cuadrado es de 400 metros cuadrados. Paso a paso ... El perímetro de un cuadrado es 80 m. ¿Cuál es la longitud de cada lado del cuadrado ... Para celosía & lt; L, * & gt; si a, b, ce L thena + (a e b) = a se llama.

Step-by-step explanation:

Answer:

You know that 5 g of sugar form 12%. Let x g be the amount of water needed. The total amount of syrup will be x+5 g. Then

5 g - 12%,

x+5 g - 100%.

mathematically you can write a proportion;

5/x+5=12/100

=>500=12(x+5)

,500=12x+60

,12x=400,

x=440/12

=110/3

36.66 grams of water is added  to make 12% sugar syrup using 5g sugar.

A ratio is an ordered pair of numbers a and b, written a / b where b does not equal 0.

Let the amount of water is added be 'x'.

Amount of sugar = 5 g

We need to make 12% of sugar syrup.

x grams of water is added to 5 g of sugar, to 12% of sugar syrup.

5/5+x=12/100

Apply cross multiplication

5×100=12(5+x)

500=60+12x

Subtract 60 from both sides

440=12x

Divide both sides by 12

x=36.66

Hence, 36.66 grams of water is added  to make 12% sugar syrup using 5g sugar.

To learn more on Ratios click:

https://brainly.com/question/1504221

#SPJ2

Answer:

4 1/3(b-21) im pretty sure

Answer:

two numbers are 37 and 20

Step-by-step explanation:

x+y=57

let x be the smaller number

y=x+17

x + x+17=57

2x=57-17

2x=40

x=20

y=x +17

y=20+17

y=37

Answer:

a) x ≤ -8

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Step-by-step explanation:

Step 1: Define

-3 + x ≤ -11

Step 2: Solve for x

Here we see that any value x smaller than or equal to -8 would work as a solution to the inequality.

Answer:

x=0 the value of y is 1/1 so this line cuts the y axis at y=1.00000. y=0 the value of x is 1/5 our line therefore cuts the axis at x=0.20000

Answer:

[tex]f(x) = 2x - 11 \\ f (x)= 2( - 17) - 11 \\ \\ f(x) = - 34 - 11 \\ f(x) = - 45[/tex]

Answer:

Perpendicular

Answer:

1/how ever many students are going to Paris

Step-by-step explanation:

Answer:

The equation is y = (2/3)x + (4/3).

Step-by-step explanation:

First, you have to change the equation into slope-intecept formula to find its gradient (slope) :

[tex]3x + 2y = 12[/tex]

[tex]2y = - 3x + 12[/tex]

[tex]y = - \frac{3}{2}x + 6[/tex]

Given that when a line is perpendicular to the other line, their slope value multiplied, will form -1. Next, we have to find the slope for line :

[tex]m1 \times m2 = - 1[/tex]

[tex] - \frac{3}{2} \times m2 = - 1[/tex]

[tex]m2 = - 1 \div - \frac{3}{2} [/tex]

[tex]m2 = \frac{2}{3} [/tex]

Lastly, we have to subatitute the x and y values into the equation, to find its intercept value :

[tex]y = \frac{2}{3}x + b[/tex]

[tex]let \: x = 1,y = 2[/tex]

[tex]2 = \frac{2}{3} (1) + b[/tex]

[tex]b = 2 - \frac{2}{3} [/tex]

[tex]b = \frac{4}{3} [/tex]

[tex]y = \frac{2}{3} x + \frac{4}{3} [/tex]

Answer:

Option C is the correct answer.

Comment for explaination

Answer:

y = (x – 6) / -3 = [2 – x/3]

Step-by-step explanation:

y = f(x) = -3x + 6

f⁻¹(x) → x = -3y + 6 → (x – 6) / -3 → x / -3 + -6/-3 → -x / 3 + 2 → 2 – x/3

Answer:

60 divided by 8

Step-by-step explanation:

60 divided by 8

Step-by-step explanation:

[tex] \underline{ \underline{ \text{Given}}} : [/tex]

[tex] \underline{ \underline{ \text{To \: find}}} : [/tex]

[tex] \underline{ \underline{ \text{Solution}}} : [/tex]

~ Plug the value of x and then multiply !

⇾ [tex] \tt{6x = 6 \times 6 \times = \boxed{36}}[/tex]

[tex] \red{ \boxed{ \boxed{ \text{Our \: final \: answer : \boxed{ \tt{36}}}}}}[/tex]

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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Answer:

[tex](-2-2\sqrt{2},3+2\sqrt{2})\,,\,(-2+2\sqrt{2},3-2\sqrt{2})[/tex]

Step-by-step explanation:

Given:

[tex](x+2)^2+(y-3)^2=16\\x+y-1=0[/tex]

To find: value of [tex]x,y[/tex]

Solution:

[tex](x+2)^2+(y-3)^2=16\,\,\,...(i)\\x+y-1=0\,\,\,(ii)[/tex]

From (ii),

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

Put this value of [tex]x[/tex] in (i)

[tex](1-y+2)^2+(y-3)^2=16\\(3-y)^2+(y-3)^2=16\\(y-3)^2+(y-3)^2=16\\2(y-3)^2=16\\(y-3)^2=8[/tex]

[tex]y-3=[/tex] ± [tex]\sqrt{8}[/tex] = ± [tex]2\sqrt{2}[/tex]

[tex]y=3[/tex] ± [tex]2\sqrt{2}[/tex]

At [tex]y=3[/tex] + [tex]2\sqrt{2}[/tex] ,

[tex]x=1-(3+2\sqrt{2})=1-3-2\sqrt{2}=-2-2\sqrt{2}[/tex]

At [tex]y=3-2\sqrt{2}[/tex] ,

[tex]x=1-(3-2\sqrt{2})=1-3+2\sqrt{2}=-2+2\sqrt{2}[/tex]

Solutions are [tex](-2-2\sqrt{2},3+2\sqrt{2})\,,\,(-2+2\sqrt{2},3-2\sqrt{2})[/tex]