Scaled copies question 3 and 4

Original chess tournament evolve 64 participants the number Of participants decreased by 1/2 after each round of play. Write the

marishachu [46]

Answer:

The first three terms is 64, 32, 16

Step-by-step explanation:

Given

Number of participants = 64

Decrement = 1/2

Required

Find the first three terms

To calculate the first three terms, we make use of the following formula;

Current Term = Previous Term - 1/2 * Previous Term

The first term = Initial number of participants

First Term = 64

Second Term = First Term - 1/2 * First Term

Second Term = 64 - 1/2 * 64

Second Term = 64 - 32

Second Term = 32

Third Term = Second Term - 1/2 * Second Term

Third Term = 32 - 1/2 * 32

Third Term = 32 - 16

Third Term = 16

Hence, the first three terms is 64, 32, 16

4 0

3 years ago

A cereal box has a surface area of 280 square inches. What is the surface are of a similar box that is larger by a scale factor

ICE Princess25 [194]

Answer: 548.8 square inches

Step-by-step explanation:

From the question, we are informed that a cereal box has a surface area of 280 square inches.

The surface area of a similar box that is larger by a scale factor of 1.4 will be gotten by multiplying 280 by the scale factor's square. This will be:

= 280 × (1.4)²

= 280 × 1.96

= 548.8 square inches

5 0

3 years ago

PLEASE HELP ME WHAT AM I SUPPOSE TO DO?!

stepladder [879]

Answer:

Step-by-step explanation:

Angle 2= 50

angle 1= 180-50= 130 degrees (linear pair (sum 180))

angle 3= 130 degrees (vertically oposite angles are equal)

angle 6= 130 degrees (alternate interior angles)

angle 5= 180-130= 50 degrees (linear pair)

i hope this helps :)

8 0

3 years ago

Assume y≠60 which expression is equivalent to (7sqrtx2)/(5sqrty3)

Drupady [299]

Answer:

The equivalent will be:

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$

Therefore, option 'a' is true.

Step-by-step explanation:

Given the expression

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}$

Let us solve the expression step by step to get the equivalent

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}$

as

$\sqrt[7]{x^2}=\left(x^2\right)^{\frac{1}{7}}$ ∵ $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}$

$\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=x^{2\cdot \frac{1}{7}}$

$=x^{\frac{2}{7}}$

also

$\sqrt[5]{y^3}=\left(y^3\right)^{\frac{1}{5}}$ ∵ $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}$

$\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=y^{3\cdot \frac{1}{5}}$

$=y^{\frac{3}{5}}$

so the expression becomes

$\frac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}$

$=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$ ∵ $\:\frac{1}{y^{\frac{3}{5}}}=y^{-\frac{3}{5}}$

Thus, the equivalent will be:

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$

Therefore, option 'a' is true.

5 0

3 years ago

Show how to use substitution or elimination to solve the following system of equations. Write your solution as an ordered pair:

MAVERICK [17]

Answer:substitution

Step-by-step explanation:

baka

5 0

3 years ago