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3 years ago

Prove that 27 to the 10th-9 to the 14th is divisible by 24

Mathematics

1 answer:

Anuta_ua [19.1K]3 years ago

3 0

Answer:

$27^{10}-9^{14}\\\\=(3^3)^{10}-(3^2)^{14}\\\\=3^{30}-3^{28}\\\\=3^{(27+3)}-3^{(27+1)}\\\\=3^{27} \cdot3^3-3^{27}\cdot3^1\\\\=3^{27}(3^3-3^1)\\\\=3^{27}(27-3)\\\\=3^{27}\cdot24$

$3^{27}\cdot24$ is divisible by 24, hence proving that $27^{10}-9^{14}$ is divisibe by 24

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Harvey bought a. Frame in which he put his family picture in what is the area of the frame not covered by the picture

9966 [12]

Answer:

Area of the frame not covered with photo is equal to 129 square inches

Step-by-step explanation:

Given: Image of a frame

To find: Area of the frame not covered by the picture

Solution:

Area of rectangle = length of the rectangle × breadth of rectangle

Area of the frame = length of frame × breadth of frame = 23 × 15 = 345 square inches

Area of the frame covered with photo = 18 × 12 = 216 square inches

Area of the frame not covered with photo = Area of the frame - Area of the frame covered with photo = 345 - 216 = 129 square inches

8 0

3 years ago

Perform the indicated operation. Be sure the answer is reduced.

avanturin [10]

<h3>Given Equation:-</h3>

$\boxed{ \rm \frac{4x^{2}y^{3}z}{9} \times \frac{45y}{8 {x}^{5} {z}^{5} }}$

<h3>Step by step expansion:</h3>

$\dashrightarrow \sf\dfrac{4x^{2}y^{3}z}{9} \times \dfrac{45y}{8 {x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{ \cancel4x^{2}y^{3}z}{9} \times \dfrac{45y}{ \cancel8 {x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{9} \times \dfrac{45y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{ \cancel9} \times \dfrac{ \cancel{45}y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{1} \times \dfrac{5y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{0}y^{3}z}{1} \times \dfrac{5y}{2{x}^{5 - 2} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}z}{1} \times \dfrac{5y}{2{x}^{3} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}z {}^{0} }{1} \times \dfrac{5y}{2{x}^{3} {z}^{3 - 1} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}}{1} \times \dfrac{5y}{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5y \times {y}^{3} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5y {}^{0} \times {y}^{3 + 1} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5 \times {y}^{4} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \bf \dfrac{5 {y}^{4} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\therefore \underline{ \textbf{ \textsf{option \red c \: is \: correct}}}$

8 0

2 years ago

8. There are approximately 2.2046 kilograms in 1 pound. There are about 0.4536 pounds in 1 kilogram. To convert 112 lxs to kilog

Zanzabum

Answer:

Step-by-step explanation:

first of all the question is wrong 1 kilogram = 2.2046 pounds. and 1 pound = 0.4536kilogram.

if we want to convert 112lb into kilogram. we multiply it by 0.4536.

ie 112lb = 112x0.4536kg = 50.8032kg

to convert lb to kg you do this; lb x 0.4536

to convert kg to lb you do this; kg x 2.2046

7 0

3 years ago

Keep Trying Write two division problems that have zeros in the quotient. One of the problems should have a remainder and the oth

atroni [7]

So basically you must first use the step that got you here

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3 years ago

Solve this equation. Enter your answer in the box<br> - e- je= -24<br> e =<br> I

Kipish [7]

E=16 hope it helps :)

5 0

3 years ago