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3 years ago

Which function has a vertex on the y-axis?

Mathematics

1 answer:

andre [41]3 years ago

7 0

F(x) = (x - 2)(x + 2)

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PLZ HELP WILL GIVE BRAINIEST 20 POINTS!!!!!

luda_lava [24]

the answer is y=22.000

3 0

3 years ago

Find three consecutive odd integers such that

Anuta_ua [19.1K]

Answer:

17, 19, 21

Step-by-step explanation:

x = first number

x+2 = second number

x+4 = third number

4x + 3(x+2) + 2(x+4) = 167

simplify:

4x + 3x + 6 + 2x + 8 = 167

combine like terms:

9x + 14 = 167

subtract 14 from each side of the equation:

9x = 153

divide both sides by 9:

x = 17

7 0

3 years ago

Please answer this question right now

Westkost [7]

Answer:

6) c

7) d

Step-by-step explanation:

6) 25 + 15= 40

40+35=75

Php. 75

7) 29•3=87

150-87=63

Php. 63

Hope this helps!

6 0

3 years ago

A car uses a litre of petrol to travel 4km. How many litres will it use to travel 32 kilometres.

lesya [120]

Given that,

A car uses a litre of petrol to travel 4 km.

To find,

How many litres will it use to travel 32 kilometres.

Solution,

We have,

1 litre = 4 km

1 km = (1/4) litre

Let it will use x litres to travel 32 km.

32 km = (32/4) litres

= 8 litres

Hence, 8 litres it will use to travel 32 km.

3 0

3 years ago

Evaluate the integral by changing to polar coordinatesye^x dA, where R is in the first quadrant enclosed by the circle x^2+y^2=2

34kurt

$\displaystyle\iint_Rye^x\,\mathrm dA=\int_{\theta=0}^{\theta=\pi/2}\int_{r=0}^{r=5}r^2\sin\theta e^{r\cos\theta}\,\mathrm dr\,\mathrm d\theta$

which follows from the usual change of coordinates via

$\begin{cases}\mathbf x(r,\theta)=r\cos\theta\\\mathbf y(r,\theta)=r\sin\theta\end{cases}$

and Jacobian determinant

$|\det J|=\left|\begin{vmatrix}\mathbf x_r&\mathbf x_\theta\\\mathbf y_r&\mathbf y_\theta\end{vmatrix}\right|=|r|$

Swap the order, so that the integral is

$\displaystyle\int_{r=0}^{r=5}\int_{\theta=0}^{\theta=\pi/2}r^2\sin\theta e^{r\cos\theta}\,\mathrm d\theta\,\mathrm dr$

and now let $\sigma=r\cos\theta$ , so that $\mathrm d\sigma=-r\sin\theta$ . Now, you have

$\displaystyle\int_{r=0}^{r=5}\int_{\sigma=0}^{\sigma=r}re^\sigma\,\mathrm d\sigma\,\mathrm dr=\int_{r=0}^{r=5}r(e^r-1)\,\mathrm dr=4e^5-\dfrac{23}2$

7 0

4 years ago