All categories

2 years ago

The difference of the squares of two positive consecutive even integers is 84. Find the integers.

Mathematics

1 answer:

ziro4ka [17]2 years ago

3 0

Answer:

The two integers are 20 and 22.

Step-by-step explanation:

Let x be the first integer.

$(x+2)^{2}-x^{2} = 84\\x^{2} +4x+4-x^{2} = 84\\4x = 80\\x = 20$

You might be interested in

The line plot below shows the distances of discus throws in a contest.

Scrat [10]

I think is 40 meters

6 0

2 years ago

Where is point b on the number line?

Yuliya22 [10]

Answer:

i dont know but can you pick me brainiest

Step-by-step explanation:

4 0

2 years ago

Jackson has a gift card for $120 to use at a department store. He wants to purchase 2 shirts and 3 pairs of pants with his gift

garri49 [273]

The answers is A because it shows the domain: 0 is less than or equal to x and less that on equal to 60

6 0

3 years ago

Read 2 more answers

Are these ratios equivalent?<br> 1 teacher to 9 students<br> 5 teachers to 45 students

IrinaVladis [17]

Answer: yes the ratios 1:9 and 5:45 are equivalent.

Step-by-step explanation:

1:9*5:5=5:45

U got any more math questions that I can answer? I’m really bored and I have to stay busy!

3 0

2 years ago

(b) dy/dx = (x - y+ 1)^2

Elanso [62]

Substitute $v(x)=x-y(x)+1$ , so that

$\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}$

Then the resulting ODE in $v(x)$ is separable, with

$1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx$

On the left, we can split into partial fractions:

$\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx$

Integrating both sides gives

$\dfrac{\ln|1-v|+\ln|1+v|}2=x+C$

$\dfrac12\ln|1-v^2|=x+C$

$1-v^2=e^{2x+C}$

$v=\pm\sqrt{1-Ce^{2x}}$

Now solve for $y(x)$ :

$x-y+1=\pm\sqrt{1-Ce^{2x}}$

$\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}$

3 0

3 years ago