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vodomira [7]
3 years ago
7

What is the surface area of this pyramid ? Thanks ! Help please. Its super urgent.

Mathematics
1 answer:
svet-max [94.6K]3 years ago
8 0

Answer:

(6 \times 6) + (2 \times 6 \times 7) \\ 36 + 84 \\ 120

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Need help with Geometry homework.
solmaris [256]

Answer:

SU = 22

Step-by-step explanation:

SR = RU

4x - 1 = x + 8

3x - 1 = 8

3x = 9

x = 3

SR =  4(3) - 1 = 11

RU =  3 + 8 = 11

SU = 11 + 11 = 22

7 0
3 years ago
In a group of 60 numbers, average of
Arada [10]

Answer:

c. 18

Step-by-step explanation:

First, you would find the total sum of the 60 numbers. The average is always the sum of the numbers divided by the number of numbers.

35*18=630

15*10=150

Now we do not know how many numbers the last group has. However, we can find it out doing 60-35-15 to see how many numbers are left. There would be 10 numbers left

10*30=300

630+150+300=1080

1080/60=18

Hence, the answer is C(18)

4 0
3 years ago
Point A is located at (-2, 2). Point B is located at (-2, 0). What is the distance between point A and point B ?
Andrew [12]

Answer:

2

Step-by-step explanation:

Both point a and b have an x value of -2 so to find the distance between the two points we simply have to find the horizontal distance. We can do this by subtract the y value of the first point by the y value of the second point.

Point A y value : 2

Point B y value : 0

Distance between Point A and Point B : 2 - 0 = 2 units

6 0
2 years ago
Can someone please help me?
Hunter-Best [27]
It’s D since volume is length x width x height
5 0
3 years ago
Read 2 more answers
How do you solve this: In ?PQR, PQ = 39 cm and PN is an altitude. Find PR if QN = 36 cm and RN = 8 cm.
ikadub [295]

Answer:

PR = 17 cm


Step-by-step explanation:

Given  :

In ΔPQR,

PQ = 39 cm

PN is an altitude.

QN = 36 cm

RN = 8 cm.

To Find : Length of PR

Solution :

Since we are given that PN is an altitude .

So, PN divides ΔPQR in two right angled triangles named as ΔPQN and ΔPRN. (Refer attached file)

So, first we find Length of PN in ΔPQN using Pythagoras theorem i.e.

(Hypotenuse)^{2}=(Perpendicular)^{2} +(Base)^{2}

(PQ)^{2}=(PN)^{2} +(QN)^{2}

(39)^{2}=(PN)^{2} +(36)^{2}

1521=(PN)^{2} +1296

1521 -1296=(PN)^{2}

225=(PN)^{2}

\sqrt{225} = PN

15 = PN

Thus, Length of PN = 15cm

Now to find length of PR we will use Pythagoras theorem in ΔPRN.

(PR)^{2}=(PN)^{2} +(NR)^{2}

(PR)^{2}=(15)^{2} +(8)^{2}

(PR)^{2}=225 +64

(PR)^{2}=289

PR= \sqrt{289}

PR= 17

Hence the length of PR = 17 cm



3 0
3 years ago
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