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3 years ago

(pls help quick and explain how you got the answers for brainliest)

Mathematics

1 answer:

scoray [572]3 years ago

7 0

You only have to apply the theorem of Pythagoras here. Remember the square on the hypotenuse (the longest side) is equal to the sum of the squares on the other two sides :

1. AB is the hypotenuse, so, according to the theorem we can write :

AB² = AC² + CB²

c² = 5² + 4²

c²= 25 + 16

c² = 41

applying the square root of 41 we get :

c ≈ 6.40 rounded to the hundred

The next cases are exactly the same thing so there is no need for explanation :

AB is the hypotenuse here because it is the biggest side clearl :

AB² = AC² + CB²

25² = 15² + b²

Thus

b² = 25² - 15²

we just subtracted 15² on each side of the equation

b² = 625 - 225

b² = 400

applying the square root of 400 we get

b = √400 = 20

So AC = 20

3. The longest side is clearly AB = 60

AB² = AC² + CB²

60² = 40² + a²

subtracting 40² on each side of the equation we get :

a² = 60² - 40²

I let you finish this using your calculator and doing exactly like the previous cases

AB is the hypotenuse,

AB² = AC² + CB²

23² = b² + 14²

Subtracting 14² from each side of the equation we get

b² = 23² - 14²

AB is the biggest side :

AB² = AC² + CB²

29² = 23² + a²

We subtract 23² on each sides of the equation :

a² = 29² - 23²

You can finish with your calculator

AB² = AC² + BC²

78² = b² + 30²

subtraction...

b² = 78² - 30²

Good luck :)

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Anestetic [448]

Step-by-step explanation:

The equation of line passing through the two points is $\frac{y-y_{1} }{x-x_{1} }=\frac{y_{2} -y_{1}}{x_{2} -x_{1}}$

When substituted,the equation becomes $\frac{y-6}{x-2}=\frac{18-6}{6-2}$

which when simplified is $y=3\times x$

The line clearly passes through origin.

The distance between two points is $\sqrt{(y_{1}-y_{2} )^{2}+(x_{1}- x_{2} )^{2}}$

Distance between origin and $(2,6)$ is $\sqrt{40}$ .

Distance between origin and $(6,18)$ is $\sqrt{360}$

Scale factor is $\frac{distance\text{ }of\text{ }p_{2}\text{ from origin}}{distance \text{ } of \text{ } p_{1}\text{ from origin}}$

So,scale factor is $\frac{\sqrt{360} }{\sqrt{40} }$

which when simplified becomes 3.

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3 years ago

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1/9 of the original pie is each slice

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3 years ago

Tickets to a football final are selling well. On Thursday, 47 of the tickets are sold. On Friday, 14 of the tickets are sold. Wh

lbvjy [14]

The question seems incomplete ; as the total number of tickets to be sold isn't given.

Answer:

61 / X

Step-by-step explanation:

Let's take the total Number of tickets to be sold as : X

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Number sold on Friday = 14

Fraction of tickets available for sale on Saturday :

(Total number of tickets already sold) / Total number of tickets to be sold

(Thursday + Friday sales) / total number of tickets to be sold

Fraction available for sale on Saturday = (47+14) / X

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3 years ago

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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

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