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3 years ago

7

Hellllllllpppppp - Which expression is equivalent to x+x+y+y+3(2+4)?

1 answer:

Natalija [7]3 years ago

5 0

Step-by-step explanation:

x + x + y + y + 3(2 + 4)

= 2x + 2y + 3(6)

= 2x + 2y + 18

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In each box, 20% of the total candies are grape flavored.

mrs_skeptik [129]

Answer:

9

Step-by-step explanation:

20 x 5 = 100

45/5 = 9

9 to 20 = 11

45 to 100 = 55

<u>55</u> = 5<--- Linear

11

(since the total answer is linear, the answer is right)

6 0

2 years ago

Read 2 more answers

Ms. Tyson is doing an engineering challenge with her students. Each team will get a kit with bags of marshmallows and boxes of t

natka813 [3]

Answer:

Step-by-step explanation:

15

7 0

3 years ago

What are the coordinates of the point on the directed line segment from (-6, -3) to

melamori03 [73]

Given:

A point divides a directed line segment from (-6, -3) to (5,8) into a ratio of 6 to 5.

To find:

The coordinates of that point.

Solution:

Section formula: If point divides a line segment in m:n, then the coordinates of that point are

$Point=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)$

A point divides a directed line segment from (-6, -3) to (5,8) into a ratio of 6 to 5. Using section formula, we get

$Point=\left(\dfrac{6(5)+5(-6)}{6+5},\dfrac{6(8)+5(-3)}{6+5}\right)$

$Point=\left(\dfrac{30-30}{11},\dfrac{48-15}{11}\right)$

$Point=\left(\dfrac{0}{11},\dfrac{33}{11}\right)$

$Point=\left(0,3\right)$

Therefore, the coordinates of the required point are (0,3).

3 0

3 years ago

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

3 years ago

Help !! Picture ^^^^

nikitadnepr [17]

The answer is 78

If you multiply 17x6 you get 102
102 equals the whole rectangle after that you need to get rid of the part that’s not there
so you multiply 8x3 which equals 24
you then subtract 24 from 102 and get 78

6 0

3 years ago