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2 years ago

2 pounds equals how many onces

Mathematics

2 answers:

pochemuha2 years ago

8 0

Answer: two pounds is equivalent to 32 ounces.

Step-by-step explanation: 1 pound is equal to 16 ounces, and since 2 is 1 two times, just multiply 16 by 2, 16x2=32 ounces

Semenov [28]2 years ago

4 0

There are 16 oz in a pound, so 2 lbs is 32 oz.

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If x and y vary directly and x=25 when y=100 find x when y=24.<br> PLEASE HELP FASTTTTTTT

Veseljchak [2.6K]

Answer:

x = 6

Step-by-step explanation:

Given that x and y vary directly then the equation relating them is

y = kx ← k is the constant of variation

To find k use the condition x = 25 when y = 100 , then

100 = 25k ( divide both sides by 25 )

4 = k

y = 4x ← equation of variation

When y = 24 , then

24 = 4x ( divide both sides by 4 )

6 = x

6 0

2 years ago

Pleaseee help me out !!!

Dahasolnce [82]

Answer:

y = 16

Step-by-step explanation:

y = 11 - x

of x is -5, subtitute it into the equation

y = 11 - (-5)

y = 11 + 5

y = 16

8 0

1 year ago

Read 2 more answers

1.What are the zeros of the polynomial function?

Lorico [155]

Let's to the first example:

f(x) = x^2 + 9x + 20

Ussing the formula of basckara

a = 1
b = 9
c = 20

Delta = b^2 - 4ac

Delta = 9^2 - 4.(1).(20)

Delta = 81 - 80

Delta = 1

x = [ -b +/- √(Delta) ]/2a

Replacing the data:

x = [ -9 +/- √1 ]/2

x' = (-9 -1)/2 <=> - 5

Or

x" = (-9+1)/2 <=> - 4
_______________

Already the second example:

f(x) = x^2 -4x -60

Ussing the formula of basckara again

a = 1
b = -4
c = -60

Delta = b^2 -4ac

Delta = (-4)^2 -4.(1).(-60)

Delta = 16 + 240

Delta = 256

Then, following:

x = [ -b +/- √(Delta)]/2a

Replacing the information

x = [ -(-4) +/- √256 ]/2

x = [ 4 +/- 16]/2

x' = (4-16)/2 <=> -6

Or

x" = (4+16)/2 <=> 10
______________

Now we are going to the 3 example

x^2 + 24 = 14x

Isolating 14x , but changing the sinal positive to negative

x^2 - 14x + 24 = 0

Now we can to apply the formula of basckara

a = 1
b = -14
c = 24

Delta = b^2 -4ac

Delta = (-14)^2 -4.(1).(24)

Delta = 196 - 96

Delta = 100

Then we stayed with:

x = [ -b +/- √Delta ]/2a

x = [ -(-14) +/- √100 ]/2

We wiil have two possibilities

x' = ( 14 -10)/2 <=> 2

Or

x" = (14 +10)/2 <=> 12
________________

To the last example will be the same thing.

f(x) = x^2 - x -72

a = 1
b = -1
c = -72

Delta = b^2 -4ac

Delta = (-1)^2 -4(1).(-72)

Delta = 1 + 288

Delta = 289

Then we are going to stay:

x = [ -b +/- √Delta]/2a

x = [ -(-1) +/- √289]/2

x = ( 1 +/- 17)/2

We will have two roots

That's :

x = (1 - 17)/2 <=> -8

Or

x = (1+17)/2 <=> 9

Well, this would be your answers.

7 0

2 years ago

Use two different methods to find an explain the formula for the area of a trapezoid that has parallel sides of length a and B a

evablogger [386]

Answer:

Formula of Trapezoid:

A = (a + b) × h / 2

The formula can be derived in different ways. for now, we have discussed two ways:

1. By using the formula of a triangle

2. By dividing into different sections

Step-by-step explanation:

1. By using the formula of a triangle

One of the ways to explain a formula for an area of a trapezoid using a formula for a triangle can be as follows.

Assume a trapezoid PQRS with lower base SR and upper base PQ (they are parallel) and sides PS and QR.

The image is attached below.

Connect vertices P and R with a diagonal.

Consider triangle ΔPQR as having a base PQ and an altitude from vertex R down to point M on base PQ (RM⊥PQ).

Its area is

S1= $\frac{1}{2} *PQ*RM$

Consider triangle ΔPRS as having a base SR and an altitude from vertex P up to point N on-base SR (PN⊥SR).

Its area is

S2= $\frac{1}{2} *SR*PN$

Altitudes RM and PN are equal and constitute the distance between two parallel bases PQ and SR.

They both are equal to the altitude of the trapezoid h.

Therefore, we can represent areas of our two triangles as

S1= $\frac{1}{2}*PQ*h$

S2= $\frac{1}{2}*SR*h$

Adding them together, we get the area of the whole trapezoid:

S=S1+S2= $\frac{1}{2} (PQ+SR)h,$

which is usually represented in words as "half-sum of the bases times the altitude".

2. By dividing into different sections

Trapezoid PQRS is shown below, with PQ parallel to RS.

Figure 1 - Trapezoid PQRS with PQ parallel to RS(image is attached below.)

We are going to derive the area of a trapezoid by dividing it into different sections.

If we drop another line from Q, then we will have two altitudes namely PT and QU.

Figure 2 - Trapezoid PQRS divided into two triangles and a rectangle. (image is attached below.)

From Figure 2, it is clear that Area of PQRS = Area of PST + Area of PQUT + Area of QRU. We have learned that the area of a triangle is the product of its base and altitude divided by 2, and the area of a rectangle is the product of its length and width. Hence, we can easily compute the area of PQRS. It is clear that

=> $A_{PQRS} = (\frac{ah}{2}) + b_{1}h + \frac{ch}{2}$