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Colt1911 [192]
4 years ago
8

Observe the following pattern and find the missing digits.​

Mathematics
1 answer:
madam [21]4 years ago
7 0

Answer:

<h2>1020304030201</h2>

Step-by-step explanation:

I don't know the pattern but 1010101² is 1020304030201.

I'm always happy to help :)

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Please help me with alg 2 question!
AlladinOne [14]

Answer:

J

Step-by-step explanation:

3 0
2 years ago
Isosceles right triangle ABC has a hypotenuse, AB, with a length of 18 feet.
kiruha [24]

we know that

if ABC is an isosceles right triangle

then

side AC=side BC

angle A=angle B=45 degrees

cos B=adjacent side angle B/hypotenuse

adjacent side angle B=BC

hypotenuse=AB------> 18 ft

angle B=45 degrees

cos 45°=(√2)/2

so

cos 45°=BC/AB-------> solve for BC

BC=AB*cos 45-------> BC=18*(√2)/2------> BC=9√2 ft

AC=BC--------> AC=9√2 ft

the answer part 1) is

the exact lengths of the two sides, AC and BC is

AC=9√2 ft

BC=9√2 ft

Part b) Find the Area of triangle ABC

Area=b*h/2-------> AC*BC/2-----> (9√2)*(9√2)/2--------> 81 ft²

the answer part b) is

the area of triangle ABC is equal to 81 ft²

7 0
3 years ago
*LAST QUESTION , PLEASE ANSWER TY* (: Quadrilateral ABCD is inscribed in a circle. If angle A measures (3x – 10)° and angle C me
marissa [1.9K]

\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Quadrilaterals.

Basically we know that, the sum of opposite angles of a quadrilateral inscribed in a circle is always 180°.

so applying this law here, we get as,

2X + (3X-10) = 180°

=> 5X - 10° = 180°

=> 5X = 190°

=> X = 190°/5

=> X = 38°

thus the angle X= 38°.

8 0
4 years ago
The following describes a sample. The information given includes the five number summary, the sample size, and the largest and s
lys-0071 [83]

Answer:

L= Q_1 - 1.5*IQR = 210 -1.5*40 =150

U= Q_3 + 1.5*IQR = 250 +1.5*40 =310

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

Step-by-step explanation:

For this case we have the 5 number summary

(160,210,220,250,270)

So then we have:

minimum = 160 , Q1 = 210, Q2= Median=220, Q3 = 250, Max=270

If we find the interquartile range we got:

IQR = Q_3 -Q_1 = 250-210 =40

For this case we need to find the lower and upper limit with the following formulas:

L= Q_1 - 1.5*IQR = 210 -1.5*40 =150

U= Q_3 + 1.5*IQR = 250 +1.5*40 =310

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

4 0
4 years ago
Find all the missing parts to the triangle below
Arada [10]
Use Law of Cosines                             g^2 = f^2 + h^2 -2fhCosG                     f^2 = g^2 + h^2 -2ghCosF                    h^2 = f^2 + g^2 -2fgCosH

f^2 = 28^2 + 15^2 -2*28*15Cos87        28^2 = 31^2 + 15^2 -2*31*15CosG 
f^2 = 784 + 225 - 43.96                        784 = 961+225 - 930CosG 
f^2 = 965.0378                                    784 - 1186 = -930CosG
f = 31                                                 -402 = -930CosG             Divide by -930
                                                           .432258 = CosG
                                                           Cos^-1(.432258) = G
                                                             G = 64 degrees

Angle H = 180 - 64 - 87 = 29 degrees

Side f =   31                     Angle F =  87 degrees
Side g =  28                     Angle G = 64 degrees
Side h =  15                     Angle H =29  degrees
3 0
3 years ago
Read 2 more answers
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