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2 years ago

Which postulate can be used to prove that △ABD≅△CBD?

Mathematics

1 answer:

Vinil7 [7]2 years ago

7 0

The answer to the postulate being used would be side Angle side

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Emily completed reading 24/3 books in 32/4 days. Find the number of books she would have completed in one day.

AlexFokin [52]

The number of books she would have completed in one day will be 2 books.

<h3>What is an expression?</h3>

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication and division.

Given that:-

Emily completed reading 24/3 books in 32/4 days

Emily completed 8 books in 4 days. The number of the books will be calculated by using the expression below:-

4 days = 8 books

1 day = 8 / 4 = 2 books

Therefore the number of books she would have completed in one day will be 2 books.

To know more about expression follow

brainly.com/question/723406

#SPJ1

3 0

1 year ago

A box of fixtures cost 18 dollars if you had four hundred thirty three dollars and bought as many boxes as you could how much mo

Gala2k [10]

With 433 dollars, you can buy 24 boxes of fixtures. When you buy 24 boxes, you will have 1 dollar left.

5 0

2 years ago

<img src="https://tex.z-dn.net/?f=9x%20%7B%7D%5E%7B2%7D%20y%20%7B%7D%5E%7B3%7D%20z%20%7B%7D%5E%7B5%7D%20%20%5Cdiv%206x%20%7B%7D%

dexar [7]

Answer:

3y²z³ / 2x

Step-by-step explanation:

we can take one term at a time:

9/6 can be simplified to 3/2

x²÷x³ = x^-1 which equals $\frac{1}{x}$

y³÷y = y²

z⁵÷z² = z³

put it all together:

3· $\frac{1}{x}$ ·y²·z³ / 2 or

3y²z³ / 2x

8 0

3 years ago

What is the sum of a 12-term arithmetic sequence where the last term is 13 and the common difference is -10?

Y_Kistochka [10]

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=12\\
d=-10\\
a_{12}=13
\end{cases}
\\\\\\
a_{12}=a_1+(12-1)d\implies 13=a_1+(12-1)(-10)
\\\\\\
13=a_1-110\implies \boxed{123=a_1}$

$\bf \\\\
-------------------------------\\\\
\textit{sum of a finite arithmetic sequence}\\\\
S_n=\cfrac{n}{2}(a_1+a_n)\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
a_n=\textit{value of the }n^{th}\ term\\
------------\\
n=12\\
a_1=123\\
a_{12}=13
\end{cases}
\\\\\\
S_{12}=\cfrac{12}{2}(a_1+a_{12})\implies S_{12}=\cfrac{12}{2}(123+13)$

and surely you know how much that is.

6 0

3 years ago

Aaron baked two identical cakes. He cut one into 9 equal

vichka [17]

Answer:

Step-by-step explanation:

3 0

2 years ago