Answer:
- ∠DEF ≅ ∠ABC when BH = EJ, BI = EK, and HI = JK
Step-by-step explanation:
<u>Corresponding segments are:</u>
- BH and ED, BI and EK, HI and JK
Therefore last option is correct
Answer:
5%
Step-by-step explanation:
given,
Earns= Rs 640
spends= Rs 608
saves= (Rs 640 - Rs 608)
=Rs 32
therefore, 32/640x100
answer = 5%
Answer From Gauth Math
Given:
The composite figure.
To find:
The volume of the given composite figure.
Solution:
Given composite figure contains a cuboid and a pyramid.
Length, breadth and height of the cuboid are 8, 6 and 4 respectively.
Volume of a cuboid is
![V_1=length\times breadth\times height](https://tex.z-dn.net/?f=V_1%3Dlength%5Ctimes%20breadth%5Ctimes%20height)
![V_1=(8)(6)(4)](https://tex.z-dn.net/?f=V_1%3D%288%29%286%29%284%29)
Length and breadth of the pyramid is same as the cuboid, i.e., 8 and 6 respectively.
Height of pyramid = 10 - 4 = 6
Volume of a pyramid is
![V_2=\dfrac{1}{3}\times length\times breadth\times height](https://tex.z-dn.net/?f=V_2%3D%5Cdfrac%7B1%7D%7B3%7D%5Ctimes%20length%5Ctimes%20breadth%5Ctimes%20height)
![V_2=(\dfrac{1}{3})(8)(6)(6)](https://tex.z-dn.net/?f=V_2%3D%28%5Cdfrac%7B1%7D%7B3%7D%29%288%29%286%29%286%29)
The volume of composite figure is
![V=V_1+V_2](https://tex.z-dn.net/?f=V%3DV_1%2BV_2)
![V=(8)(6)(4)+(\dfrac{1}{3})(8)(6)(6)](https://tex.z-dn.net/?f=V%3D%288%29%286%29%284%29%2B%28%5Cdfrac%7B1%7D%7B3%7D%29%288%29%286%29%286%29)
It can be written as
![V=(\dfrac{1}{3})(8)(6)(6)+(8)(6)(4)](https://tex.z-dn.net/?f=V%3D%28%5Cdfrac%7B1%7D%7B3%7D%29%288%29%286%29%286%29%2B%288%29%286%29%284%29)
Therefore, the correct option is A.
Answer:
Juan should put his papers in a place that nobody else knows excise. Like in the chimney, the secret door or the atti.
Step-by-step explanation:
you can’t just put it somewhere people see or know
Let's recall that I = Prt
so... hmm we're assuming here, it doesn't say there, but this is for 1 year period.
ok... so... let's use two quantities for such interest, say "a" is the Interest for the Principal of 6000, and let's use "b" is the Interest for the Principal of 13000.
![\bf \qquad \textit{Simple Interest Earned}\\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\to& \$6000\\ r=rate\to r\%\to& \frac{r}{100}\\ t=years\to &1 \end{cases} \\\\\\ a=6000\left( \frac{r}{100} \right)1\implies \stackrel{\textit{P=6000}}{a=60r}\\\\ -------------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7BSimple%20Interest%20Earned%7D%5C%5C%5C%5C%0AI%20%3D%20Prt%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0AI%3D%5Ctextit%7Binterest%20earned%7D%5C%5C%0AP%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cto%26%20%5C%246000%5C%5C%0Ar%3Drate%5Cto%20r%5C%25%5Cto%26%20%5Cfrac%7Br%7D%7B100%7D%5C%5C%0At%3Dyears%5Cto%20%261%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0Aa%3D6000%5Cleft%28%20%5Cfrac%7Br%7D%7B100%7D%20%5Cright%291%5Cimplies%20%5Cstackrel%7B%5Ctextit%7BP%3D6000%7D%7D%7Ba%3D60r%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C)
![\bf \qquad \textit{Simple Interest Earned}\\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\to& \$13000\\ r=rate\to (r-1)\%\to &\frac{r-1}{100}\\ t=years\to &1 \end{cases} \\\\\\ b=13000\left( \frac{r-1}{100} \right)1\implies \stackrel{\textit{P=13000}}{b=130(r-1)}](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7BSimple%20Interest%20Earned%7D%5C%5C%5C%5C%0AI%20%3D%20Prt%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0AI%3D%5Ctextit%7Binterest%20earned%7D%5C%5C%0AP%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cto%26%20%5C%2413000%5C%5C%0Ar%3Drate%5Cto%20%28r-1%29%5C%25%5Cto%20%26%5Cfrac%7Br-1%7D%7B100%7D%5C%5C%0At%3Dyears%5Cto%20%261%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0Ab%3D13000%5Cleft%28%20%5Cfrac%7Br-1%7D%7B100%7D%20%5Cright%291%5Cimplies%20%5Cstackrel%7B%5Ctextit%7BP%3D13000%7D%7D%7Bb%3D130%28r-1%29%7D)
now, we know that, whatever the interest amount of "a" is, is 220
less than whatever "b" is.
we know what "b" is,
b = 130(r-1), and 220 less than that is just b - 220, or
130(r-1) - 220.