Ask question

Ask question

All categories

2 years ago

15

Find the difference. 2/5(d-10)-2/3(d+6)

1 answer:

Anon25 [30]2 years ago

6 0

$\dfrac 25(d-10) - \dfrac 23(d+6)\\\\\\=2 \left(\dfrac{d-10}5 - \dfrac{d+6}3 \right)\\\\\\=2\left[\dfrac{3(d-10) - 5(d+6)}{15}\right]\\\\\\=2 \left( \dfrac{3d-30-5d-30}{15} \right)\\\\\\=2\left(\dfrac{-2d-60}{15} \right)\\\\\\=2(-2) \left(\dfrac{d+30}{15} \right)\\\\\\=-4 \left(\dfrac{d+30}{15} \right)\\\\\\=-4\left(\dfrac d{15} +2 \right)\\\\\\=-\dfrac{4d}{15}-8$

You might be interested in

Please help - find the orthocenter.

Serjik [45]

I don’t even know bruh

8 0

3 years ago

Read 2 more answers

Can somebody help me plz

lina2011 [118]

Answer:

126 people

Step-by-step explanation:

9/5 is the ratio tea/coffee

let x be the one who preferred coffee and x+36 preferred tea

9/5=x+36/x

9x=5x+5(36)

9x-5x=180

4x=180

x=180/4=45

x=45 coffee

x+36=45+36=81

45+81=126

check : 81/45= 9/5

8 0

3 years ago

What goes where. JdhshshHshxjancixd

valentina_108 [34]

Answer:

$\boxed { \frac{7}{5}y } \to \: \boxed {y+ \frac{2}{5}y }$

$\boxed { \frac{3}{5}y } \to \: \boxed {y - \frac{2}{5}y }$

Step-by-step explanation:

We need to simplify

$y + \frac{2}{5}y$

We collect LCM to get;

$\frac{5y + 2y}{5} = \frac{7y}{5}$

Therefore:

$\boxed { \frac{7}{5}y } \to \: \boxed {y+ \frac{2}{5}y }$

Also we need to simplify:

$y - \frac{2}{5}y$

We collect LCM to get;

$y - \frac{2}{5}y = \frac{5y - 2y}{5} = \frac{3}{5} y$

Therefore

$\boxed { \frac{3}{5}y } \to \: \boxed {y - \frac{2}{5}y }$

5 0

4 years ago

Solve the differential equation. y' + 5xey = 0.

Gwar [14]

Answer:

The solution is $y = - ln(\frac{5}{2}x^{2} + C)$

Step-by-step explanation:

To solve the differential equation, we will find $y$

From the given equation, y' + 5xey = 0.

That is, $y' + 5xe^{y} = 0$

This can be written as

$\frac{dy}{dx} + 5xe^{y} = 0$

Then,

$\frac{dy}{dx} = - 5xe^{y}$

$\frac{dy}{e^{y}} = - 5x dx$

Then, we integrate both sides

$\int {\frac{dy}{e^{y}}} =\int {- 5x dx}$

$\int {e^{-y}dy }} =\int {- 5x dx}$

Then,

$-e^{-y} = -\frac{5}{2}x^{2} + C$

$e^{-y} = \frac{5}{2}x^{2} + C$

Then,

$ln(e^{-y}) = ln(\frac{5}{2}x^{2} + C)$

Then,

$-y = ln(\frac{5}{2}x^{2} + C)$

Hence,

$y = - ln(\frac{5}{2}x^{2} + C)$

8 0

3 years ago

What is the midpoint between (-8, 5) and (2,-2)?

Anastasy [175]

Answer:

the answer is -3, 3/2

3 0

4 years ago