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larisa [96]
3 years ago
15

Which pair of triangles can be proven by congruent SAS? HELP ASAP

Mathematics
2 answers:
Y_Kistochka [10]3 years ago
7 0
The first one since the given measurements/guide is Side, Angle, Side
alina1380 [7]3 years ago
7 0
A. If you took apart both the parts and put them together they would be shapes. The triangles are 90 degrees too.
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10x + 12 = [2+3(2x-4)+1]<br> Solve for x
pogonyaev

Answer:

-21/4

Step-by-step explanation:

3 0
3 years ago
1+4=5, 2+5=12, 3+6=21, 8+11=?
MaRussiya [10]
96 is the correct answer to your question i
6 0
2 years ago
Please help!! :(((((
wel

Answer:

20+4(t)=?

Step-by-step explanation:

try using the vid :P

4 0
3 years ago
If n is a whole number, and 0.01 is between 1/n and 1/n+2, what is the value of n?
OLga [1]

The value of n is 99

Integers are the set of numbers including all natural numbers and 0. They are part of the real numbers that do not include fractions, decimals, or negative numbers. Counting numbers are also considered whole numbers.

Natural numbers together with zero (0) are referred to as whole numbers. We know that natural numbers refer to the set of counting numbers starting from 1, 2, 3, 4 and so on. Simply put, integers are a set of numbers without fractions, decimals, or even negative integers. It is a collection of positive integers and zero. Or we can say that integers are the set of non-negative integers.

Integers are the set of natural numbers togethe with the number 0. In mathematics, the set of integers is given as {0, 1, 2, 3, ...}, denoted by the symbol W.

W = {0, 1, 2, 3, 4, …}

All natural numbers are integers.

All integers are real numbers.

It is given that If n a whole number, and 0.01 is between 1/n and 1/n+2, then we need to find the value of n.

1/n+2 < 0.01 < 1/n

1/n+2 < 1/100 < 1/n

n< 100 < n +2

Put n = 99, then we get

99 < 100 < 101

Hence the value of n is 99.

<u>Learn more about whole numbers here:</u>

brainly.com/question/28052750

<u>#</u><u>S</u><u>P</u><u>J</u><u>4</u>

3 0
1 year ago
NO LINKS!! Find an nth-degree polynomial function with real coefficient satisfying the given conditions.
Sati [7]

n=3; We need a third degree polynomials with the following given zero's: 2 and 5i are zeros; f(-1)=156.

Since these are solutions

x = 2 ; x = 5i. Since imaginaries travel in pairs, the other answer is x= -5i.

We have (x-2)(x-5i)(x+5i) = 0

Now,

f(-1) = (-1-2)(-1-5i)(-1+5i) = 156.

f(-1) = (-3)(26) = -78.

But -78 x -2 = 156, so our polynomial becomes

Y= -2x (<em>x</em> - 2 ) x (<em>x </em>to the power of 2 + 25) = 0

6 0
2 years ago
Read 2 more answers
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