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2 years ago

Write an expression for 4 snacks at n pence each

1 answer:

5 0

<span>The expression for the total price paid for all snacks (P pence)= 4n Pence will be; P=4n. Calculation: let the total price for the four snacks be P pence. let the four snacks be s1,s2,s3, and s4; their prices are: s1=n pence; s2= n pence; s3=n pence; s4= n pence. therefore; the total price paid for all snacks (P pence)= s1 price +s2 price +s3 price +s4 price = n+n+n+n= 4n hence; P=4n i.e. the total price paid for all snacks (P pence)= 4n Pence</span>

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David bought a baseball card for $40. the value of the card increased by 25%. what is the new value of the card?

kenny6666 [7]

Answer:

65 i think

Step-by-step explanation:

5 0

2 years ago

A bridge is built in the shape of a parabolic arch. The bridge arch has a span of 166 feet and a maximum height of 40 feet. Find

scoray [572]

Answer:

38.27775 feet

Step-by-step explanation:

The bridge has been shown in the figure.

Let the highest point of the parabolic bridge (i.e. vertex of the parabola) be at the origin, $O(0,0)$ in the cartesian coordinate system.

As the bridge have the shape of an inverted parabola, so the standard equation, which describes the shape of the bridge is

$x^2=4ay\;\cdots(i)$

where $a$ is an arbitrary constant (distance between focus and vertex of the parabola).

The span of the bridge = 166 feet and

Maximum height of the bridge= 40 feet.

The coordinate where the bridge meets the base is $A(83, -40)$ and $B(-83, -40).$

There is only one constant in the equation of the parabola, so, use either of one point to find the value of $a$ .

Putting $A(83,-40)$ in the equation (i) we have

$83^2=4a(-40)$

$\Rightarrow a=-43.05625$

So, on putting the value of $a$ in the equation (i), the equation of bridge is

$x^2=-172.225y$

From the figure, the distance from the center is measured along the x-axis, x coordinate at the distance of 10 feet is, $x=\pm 10$ feet, put this value in equation (i) to get the value of y.

$(\pm10)^2=-172.225y$

$\Rightarrow y=-1.72225$ feet.

The point $P_1(10,-1.72225)$ and $P_2(-10,-1.72225)$ represent the point on the bridge at a distance of 10 feet from its center.

The distance of these points from the x-axis is $d=1.72225$ feet and the distance of the base of the bridge from the x-axis is $h=40$ feet.

Hence, height from the base of the bridge at 10 feet from its center

$= h-d$

$=40-1.72225=38.27775$ feet.

8 0

2 years ago

For x, y ∈ R we write x ∼ y if x − y is an integer. a) Show that ∼ is an equivalence relation on R. b) Show that the set [0, 1)

vodomira [7]

Answer:

A. It is an equivalence relation on R

B. In fact, the set [0,1) is a set of representatives

Step-by-step explanation:

A. The definition of an equivalence relation demands 3 things:

The relation being reflexive (∀a∈R, a∼a)
The relation being symmetric (∀a,b∈R, a∼b⇒b∼a)
The relation being transitive (∀a,b,c∈R, a∼b^b∼c⇒a∼c)

And the relation ∼ fills every condition.

∼ is Reflexive:

Let a ∈ R

it´s known that a-a=0 and because 0 is an integer

a∼a, ∀a ∈ R.

∼ is Reflexive by definition

∼ is Symmetric:

Let a,b ∈ R and suppose a∼b

a∼b ⇒ a-b=k, k ∈ Z

b-a=-k, -k ∈ Z

b∼a, ∀a,b ∈ R

∼ is Symmetric by definition

∼ is Transitive:

Let a,b,c ∈ R and suppose a∼b and b∼c

a-b=k and b-c=l, with k,l ∈ Z

(a-b)+(b-c)=k+l

a-c=k+l with k+l ∈ Z

a∼c, ∀a,b,c ∈ R

∼ is Transitive by definition

We´ve shown that ∼ is an equivalence relation on R.

B. Now we have to show that there´s a bijection from [0,1) to the set of all equivalence classes (C) in the relation ∼.

Let F: [0,1) ⇒ C a function that goes as follows: F(x)=[x] where [x] is the class of x.

Now we have to prove that this function F is injective (∀x,y∈[0,1), F(x)=F(y) ⇒ x=y) and surjective (∀b∈C, Exist x such that F(x)=b):

F is injective:

let x,y ∈ [0,1) and suppose F(x)=F(y)

[x]=[y]

x ∈ [y]

x-y=k, k ∈ Z

x=k+y

because x,y ∈ [0,1), then k must be 0. If it isn´t, then x ∉ [0,1) and then we would have a contradiction

x=y, ∀x,y ∈ [0,1)

F is injective by definition

F is surjective:

Let b ∈ R, let´s find x such as x ∈ [0,1) and F(x)=[b]

Let c=║b║, in other words the whole part of b (c ∈ Z)

Set r as b-c (let r be the decimal part of b)

r=b-c and r ∈ [0,1)

Let´s show that r∼b

r=b-c ⇒ c=b-r and because c ∈ Z

r∼b

[r]=[b]

F(r)=[b]

∼ is surjective

Then F maps [0,1) into C, i.e [0,1) is a set of representatives for the set of the equivalence classes.

4 0

2 years ago

50+x(greater than or equal to)300

nikdorinn [45]

Answer:

x ≥ 250

Step-by-step explanation:

Note that "greater than or equal to" looks like ≥

50 + x ≥ 300

Isolate the variable. What you do to one side, you do to the other. Subtract 50 from both sides.

x + 50 (-50) ≥ 300 (-50)

x ≥ 300 - 50

x ≥ 250

x ≥ 250 is your answer.

4 0

2 years ago

How do I check 102÷17=6 without multiplying

insens350 [35]

You could add 6 togetber 17 times? Sorry for the stupid answer I couldn't think of anything else I searched it up and I couldn't find anything at all

3 0

2 years ago