Answer:
x=4
Step-by-step explanation:
If the framed picture is shaped like a square and has a 12 square foot surface area, then the answer is yes, it will fit flush against the edge of the crate.
Given Part A:
the volume of the cube = 64 cubic feet
therefore, ∛64 = 4 feet
hence one edge measures 4 feet.
Now for Part B:
the area of the square is 12 square feet.
hence, √12 = 3.36 feet.
we can observe that 3.46<4
which indicates that the area covered by the painting is less than that of the one side of the crate, which makes it easy for the painting to fit in the crate.
Hence the painting will fit a side of crate.
Learn more about Area and Perimeter here:
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Your question was incomplete. Please find the missing content here.
While packing for their cross-country move, the Chen family uses a that has the shape of a cube. PART A PART B If the crate has the volume V = 64 cubic feet, The Chens want to pack a large, framed painting. If an area of 12 square feet , will the painting fit flat what is the length of one edge? the framed painting has the shape of a square with against a side of the crate? Explain.
Answer:
there is no solution
Step-by-step explanation:
Expand.
15x+35+2x=7x+10x-4515x+35+2x=7x+10x−45
2 Simplify 15x+35+2x15x+35+2x to 17x+3517x+35.
17x+35=7x+10x-4517x+35=7x+10x−45
3 Simplify 7x+10x-457x+10x−45 to 17x-4517x−45.
17x+35=17x-4517x+35=17x−45
4 Cancel 17x17x on both sides.
35=-4535=−45
5 Since 35=-4535=−45 is false, there is no solution.
No Solution
Answer:
Hence the carnival game gives you better chance of winning.
Step-by-step explanation:
Let the event of win be given by 1/10 in the game of rifle then the event of loose is given by 9/10
the
Odds in favor of a game are given by = P(Event)/ 1- P(Event)
Odds in favor of winning a rifle are given by = 1/10/ 1- 1/10
=1/10/9/10
=1/9
= 0.111
The probability of winning aa rifle game is 0.111
The probability of winning the carnival game is 0.15
Comparing the two probabilities 0.111:0.15
The probability of winning carnival game is greater than winning a rifle game
0.15>0.11
Hence the carnival game gives you better chance of winning.
A partir de la definición de razón y la teoría de semejanza entre triángulos, la razón del área del triángulo AMN y el área del cuadrilátero BMNC es equivalente a 1/3.
<h3>¿Cómo determinar la medida de un lado de un triángulo desconocido?</h3>
En este problema tenemos un sistema formado por dos triángulos <em>similares</em>, la semejanza entre los dos triángulos se debe a la colinealidad entre los segmentos de línea AP' (triángulo <em>pequeño</em>) y AP'' (triángulo <em>grande</em>), así como de los lados AM y AB, así como los lados AN y AC, así como los <em>mismos</em> ángulos en la <em>misma</em> distribución. (Semejanza Lado - Ángulo - Lado)
En consecuencia, obtenemos las siguientes proporciones:
AP'/AP'' = MN/BC = 1/2 (1)
Finalmente, la proporción entre el triángulo AMN y el cuadrilátero BMNC es:
A partir de la definición de razón y la teoría de semejanza entre triángulos, la razón del área del triángulo AMN y el área del cuadrilátero BMNC es equivalente a 1/3.
Para aprender sobre triángulos semejantes: brainly.com/question/21730013
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