Ask question

Ask question

All categories

3 years ago

8

Lines b and c are parallel. Which pairs of angles are supplementary?

1 answer:

Marat540 [252]3 years ago

8 0

Answer:

3 and 5

The angles that are vertically adjacent to the transversal.. when it cuts two parellel lines are always supplementary.. that is they make 180 degree.. hence, in the given pair 3 and 5 are supplementary..

as they lie on the transversal

You might be interested in

Please, help im being timed !!!

Agata [3.3K]

B (2x+5y=-10) is the answer.

3 0

3 years ago

Can somebody help me please

Mrac [35]

Answer:
A=(1/2)BxH
10x6=60m

4 0

3 years ago

Read 2 more answers

Noa hopes to pick the winning ticket during her math class’s raffle. the student who picks the ticket with a true statement will

Natali5045456 [20]

Answer: The winning ticket is $2(x-6)=-35$ is equivalent to $2x=-23$ .

Step-by-step explanation:

Let's check all the options

$4(x-5)=35$ is not equivalent to $4x=40$

$\text{since }4(x-5)=35\\\Rightarrow4x-20=35\\\Rightarrow4x=35+20\\\Rightarrow4x=55$

$8(x-7)=35$ is not equivalent to $8x=28$

$\text{since }8(x-7)=35\\\Rightarrow8x-56=35\\\Rightarrow8x=35+56\\\Rightarrow8x=91$

$2(x-6)=-35$ is equivalent to $2x=-23$

$\text{since }2(x-6)=-35\\\Rightarrow2x-12=-35\\\Rightarrow2x=-35+12\\\Rightarrow2x=-23$

$9(x-6)=-35$ is not equivalent to $9x=-29$

$\text{since }9(x-5)=-35\\\Rightarrow9x--40=-35\\\Rightarrow9x=-35+40\\\Rightarrow9x=5$

Thus, the winning ticket is $2(x-6)=-35$ is equivalent to $2x=-23$ .

4 0

3 years ago

Read 2 more answers

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

Ally sliced 32 kg of watermelon for a party. She divided the watermelon slices equally between 8 large bowls.

igor_vitrenko [27]

Answer:

4000g

Step-by-step explanation:

32kg=32000g

32000/8 bowls=

4000g

3 0

3 years ago