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3 years ago

Can tell me the answer

Mathematics

2 answers:

9966 [12]3 years ago

8 0

The answer is 4/6 because the remaining fraction from 2/6 is 4/6 and we can check because 4/6 is greater than 2/6

bogdanovich [222]3 years ago

6 0

The fraction 4/6 could represent the green

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Help me solve (b) in this quadrilateral

Savatey [412]

Answer:

b = 87°

Step-by-step explanation:

In order to answer this question, we need to utilise an important angle fact which is angles in a quadrilateral add up to 360°

Using the information we can set up an equation to find the value of b

→ Form equation

63 + 140 + 70 + b = 360

→ Simplify

273 + b = 360

→ Minus 273 from both sides isolate b

b = 87°

3 0

3 years ago

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Identify the solutions to the quadratic equation!

Elenna [48]

The solution to the quadratic equation are option C) x= -5 and option E) x=3.

Step-by-step explanation:

The given quadratic equation is x²+2x-15 = 0.

Using the factorization method,

product of the roots should be -15.
Sum of the roots should be 2.

⇒ -15 = 5 $\times$ -3

⇒ 2 = 5+(-3)

(x+5)(x-3) = 0

Therefore, x = -5 and x = 3

8 0

3 years ago

Answer this question pls!!

Dvinal [7]

Answer:

3.5

Step-by-step explanation:

5 0

3 years ago

One gym charges a $50 sign up fee and $65 per month. Another gun charges a $90 sign up fee and $45 per month. Foe what number of

alina1380 [7]

2 months.
the equations are
65m+50 and 45m+90
you set them equal together
65m+50=45m+90
and you get 2.

8 0

3 years ago

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Find the solution of the initial value problem 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