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mezya [45]
3 years ago
9

6. Which expression is equivalent to 22 ?​ HURRY HELP FAST

Mathematics
1 answer:
qaws [65]3 years ago
8 0

Answer: 22+(-3)

Step-by-step explanation:

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Jared made a 15 minute phone call yesterday that cost $0.90. how long will jared be on the phone today if the call cost $1.32, a
Blababa [14]
1. figure out cost per min

.90/15 =.06/min

2. figure out mins. since we know cost per min we can divide that by jared’s total cost for that call

1.32/.06= x mins ——->22 mins

7 0
2 years ago
What are my zeros (x^2 - 8x + 1)?
Pavel [41]

Answer:

x= 7.87298334621                          x=.12701665379

Step-by-step explanation:

This is in standard form so we have to factor this into

x^2-8x + 16 + 1 = 16

(x-4)^2 = 15

(x-4) = \sqrt{15\\}

x-4= ± 3.87298334621

x-4 = 3.87298334621                      x-4= -3.87298334621

x= 7.87298334621                          x=.12701665379

7 0
3 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}

Simplifying, we have

=>A= \frac{ah+2b_{1+C} }{2}

Factoring we have,

=> A_{PQRS} = (a+ 2b_{1} + c)\frac{h}{2}  \\= > {(a+ b_{1} + c) + b_{1} }\frac{h}{2}

 But, a+ b_{1} + c  is equal to b_{2}, the longer base of our trapezoid.

Hence, A_{PQRS}= (b_{1} + b_{2} )\frac{h}{2}

We have discussed two ways by which we can derive area of a trapezoid.

Read to know more about Trapezoid

brainly.com/question/4758162?referrer=searchResults

#SPJ10

5 0
2 years ago
What is this? I don’t know the answers plz help
Nataliya [291]

Answer:

a.

Step-by-step explanation:

Not defined

May be iylt help you

3 0
3 years ago
Read 2 more answers
A person is watching a boat from the top of a lighthouse. The boat is approaching the lighthouse directly. When first noticed, t
vladimir1956 [14]

Answer: 434.35 feet

Step-by-step explanation:

The angle of depression from the lighthouse is the angle of elevation from the boat. The line of sight from the top of the lighthouse, when depressed, becomes an alternate interior angle of two parallel lines, the line of sight from the top and the boat's movement.

From the boat's standpoint, which is what we want anyway, there is a right triangle, with the distance from the lighthouse the adjacent side, the opposite is 200,' and the boat the point desired.

tan 18.33 (33 minutes is 33/60 of a degree)=200/x

x=200/tan 18.33=592.58 feet

Nearer, the angle of elevation increases as the adjacent side gets smaller relative to the opposite side.

tan 51.33 (33 minutes is 33/60 of a degree)=200/x

x=200/tan 51.33=158.23 feet

That difference is 434.35 feet

Alternatively:

tan(51 33')= 200/base1

base1= 158.802

tan(18 33')=200/base2

base2=596.008

The distance travelled=596.008-158.802 =438m

5 0
4 years ago
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