All categories

2 years ago

Hello..

Mathematics

1 answer:

Mariana [72]2 years ago

7 0

9 and 7 would be corresponding angles,
6 and 3 would be alternate interior angles,
5 and 4 should be same-side interior angles,
and the only exterior angle is 9.
I hope that helps.

You might be interested in

Which is the completely factored form of the following equation?<br><br> 4a4b2−21a2b+5=0

r-ruslan [8.4K]

The completely factored equation is (a²b - 5)(4a²b - 1)

<h3>How to factor equation?</h3>

4a⁴b² - 21a²b + 5 = 0

let's expand the equation

Therefore,

4a⁴b² - 21a²b + 5 = 0

4a⁴b² - 20a²b - a²b + 5 = 0

Rearrange the equation

4a⁴b² - a²b - 20a²b + 5 = 0

Let's factorise the equation

4a⁴b² - a²b - 20a²b + 5 = 0

a²b(4a²b - 1) - 5(4a²b - 1) = 0

Therefore, the completely factored equation is as follows:

a²b(4a²b - 1) - 5(4a²b - 1) = 0

(a²b - 5)(4a²b - 1)

learn more on equation here: brainly.com/question/8842252

#SPJ1

4 0

1 year ago

Plz help me solve this. and any song requests

LuckyWell [14K]

Answer:

Step-by-step explanation:

12(8)-9(8-6)-(12-9)(8)÷2=96-18-12 or 6(9)+(12-9)(8)/2=54+12

= 66 =66

3 0

2 years ago

Read 2 more answers

Find the x and y-intercepts of the following:

svet-max [94.6K]

Answer:

x-intercept (-3,0); y-intercept (0,12)

Step-by-step explanation:

To find the x-intercept --> y should equal 0

8x-2(0)=-24 substitute

8x=-24 divide

x=-3 (-3,0)

To find they-intercept --> x should equal 0

8(0)-2y=-24 substitute

-2y=-24 divide

y=12 (0,12)

5 0

2 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

2 years ago

The next student presents a monologue that is about 0.5 minutes long. What effect will this have on the graph?

Liula [17]

Answer:

what do you mean by graph

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers