Ask question

Ask question

All categories

3 years ago

14

Someones question was deleted: Rathan thinks all factors of even numbers are even. Which explains whether Rathan is correct? Ans

wer: C. He is incorrect because the product of an even number and an odd number is even. The product of even numbers isn't always going to even, but the product of an even number and an odd number will be. Even times odd always equals even examples: 2 x 3 = 6 8 x 7 = 56 17 x 40 = 680

1 answer:

Andre45 [30]3 years ago

4 0

Answer:

as mentioned in the question

C as the answer.

Step-by-step explanation:

He is incorrect because the product of an even number and an odd number is even. The product of even numbers isn't always going to even, but the product of an even number and an odd number will be. Even times odd always equals even examples: 2 x 3 = 6 8 x 7 = 56 17 x 40 = 680

You might be interested in

jade buys a blouse and a skirt for 3/4of thier original price jade pays a total of 31.50 for the two items if the original price

aleksandrvk [35]

Answer: the original price of the skirt is $24

Step-by-step explanation:

Let x represent the original price of the skirt. if the original price of the blouse is 18, then the original price of the skirt and blouse is (x + 18)

Jade buys a blouse and a skirt for 3/4of their original price. It means that the amount at which she bought the skirt and blouse is

0.75(x + 18)

Applying the distributive property, it becomes

0.75x + 13.5

Jade pays a total of 31.50 for the two items. It means that

0.75x + 13.5 = 31.5

0.75x = 31.5 - 13.5

0.75x = 18

x = 18/0.75

x = $24

3 0

4 years ago

If using the method of completing the square to solve the quadratic equation x squared -20x+24=0, which number would have to be

Ksivusya [100]

Oksuru k diyaf ramincok ani kasilm asiyla olusan bir refle kstir.

6 0

3 years ago

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

6 0

3 years ago

write an equation that represents the following sentence? the sum of a number and two is twice the number.

Bogdan [553]

The sum of a number and 2
x + 2
is twice the number
2x
answer
<span>x + 2 = 2x</span>

5 0

4 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