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2 years ago

Can yall help..me out a lil

Mathematics

1 answer:

marusya05 [52]2 years ago

7 0

Answer:

53 in^2

Step-by-step explanation:

Divide the arrow into two shapes, a triangle and a rectangle. Now find the area for both of them.

Rectangle: l*w

4*8=32 in^2

Triangle: 1/2bh

1/2(6)(11-4)

3*7

21 in^2

Combine them: 21+32=53

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A car requires 22 litres of petrol to travel a distance of 259.6 km.Find

Anon25 [30]

Answer:

Step-by-step explanation:

1. A car requires 22 litres of petrol to travel a distance of 259.6 km

what is the distance that the car can travel on 63 ltr of petrol

22ltr = 259.6km

63ltr=

cross multiply

{63 x 259.6}/22 = 16354.8/22 = 743.4 km

A car requires 22 litres of petrol to travel a distance of 259.6 km, it would require 63 ltr of petrol to travel 743.4km

2. To travel a distance of 2013.2 km

we would need to calculate the amount of fuel

A car requires 22 litres of petrol to travel a distance of 259.6 km

what amount of fuel would it require to travel 2013.2km

22ltr = 259.6km

xltr = 2013.2km

x is the value of petrol to cover 2013.2km

cross multiply

(2013.2 x 22)/259.6

44290.4/259.6 = 170.610169492≈170.6 ltr

A car requires 22 litres of petrol to travel a distance of 259.6 km, it would require 170.6 ltr of petrol to travel 2013.2km

if 1ltr is $1.99

170.6 ltr is (170.6 x 1.99)/1 = $339.494≈$339.5

The price of fuel consumed for 2013.2 km at 1 liter of petrol at $1.99 is $339.5

5 0

2 years ago

What is the answer? help me

puteri [66]

The answer that I would choose would be C.

5 0

2 years ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is

$\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6$

4 0

2 years ago

Solve each given equation and show your work.

Eduardwww [97]

Answer:

(a) x = 3

(b) y = 6

(d) x = 5

Step-by-step explanation:

(A) 2 + x = 5 You would subtract the 2 from both sides which gives you x = 3

(B) 8y = 48 You would divide 8 from both sides which is 6 so y = 6

(D) -7x - 3x + 2 = -8x - 8 First simplify the equation; -10x + 2 = -8x - 8. Next, you want to subtract 2 from both sides which gives us -10x = -8x - 10. Now you want to add 8x to both sides which is -2x = -10. And now you just divide -2 on both sides which gives us the answer 5. So, x = 5. I hope this helps. Let me know if one of the answers are incorrect.

3 0

3 years ago

50 points for this answer

rjkz [21]

Hi Pg1310LisaAwesome,

Solution:

= 2/3 + 1/5

= 10/15 + 3/15

= 13/15

Answer:

13/15

7 0

2 years ago