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3 years ago

How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 80% antifreeze?

Mathematics

1 answer:

Georgia [21]3 years ago

8 0

Answer:

The amount of water to be added is ¹/₄ gallon

Step-by-step explanation:

Given;

amount of pure antifreeze at 100% = 1 gallon

let the amount of water to be added with 0% antifreeze = (x) gallon

then, the amount of mixture at 80% antifreeze = (x + 1) gallon

For conservation of mass, we will have the following equation;

(1 x 1) + (0)(x) = 0.8(x +1)

1 = 0.8x + 0.8

0.8x = 1 - 0.8

0.8x = 0.2

x = 0.2 / 0.8

x = ²/₈

x = ¹/₄ gallon

Therefore, the amount of water to be added is ¹/₄ gallon

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A man and his young child have a race. The child starts out 120 feet ahead of the man. For every 5-foot stride of the man, the c

yKpoI14uk [10]

Answer:

25.92 foot strides

Step-by-step explanation:

The man will take 120/5 = 24 five foot stride to cover the initial 120 foot advantage of the child.

Within this time, the child would have covered an additional 24/5 x 2 = 9.6 extra foot stride.

Within this 9.6 food stride of the child, the man will have to take an additional 9.6/5= 1.96 strides.

Total foot stride by the man will be 24 + 1.96 = 25.92 foot strides.

5 0

3 years ago

8 over 3y equals 4 over 3y

saw5 [17]

<span>8 over 3y equals 4 over 3y then

8/3y = 4/3y</span>

6 0

4 years ago

Read 2 more answers

If f(x, y) = esin(x + y) and D = [−π, π] × [−π, π], show the following. 1 e ≤ 1 4π2 D f(x, y) dA ≤ e The area of the region D is

Digiron [165]

I'm guessing the purpose of this exercise is to find the average value of $f(x,y)=e^{\sin(x+y)}$ over the region $D$ ,

$D=\left\{(x,y)\mid (x,y)\in[-\pi,\pi]^2\right\}$

which is

$\displaystyle\frac{\displaystyle\iint_Df(x,y)\,\mathrm dx\,\mathrm dy}{\displaystyle\iint_D\mathrm dx\,\mathrm dy}$

The denominator is the measure (area) of $D$ , which is easy to compute:

$\displaystyle\iint_D\mathrm dx\,\mathrm dy=\int_{-\pi}^\pi\int_{-\pi}^\pi\mathrm dx\,\mathrm dy=(2\pi)^2=4\pi^2$

Not to be confused with the integral in the numerator:

$\displaystyle\iint_Df(x,y)\,\mathrm dx\,\mathrm dy=\int_{-\pi}^\pi\int_{-\pi}^\pi e^{\sin(x+y)}\,\mathrm dx\,\mathrm dy$

but this integral is difficult to compute, hence the inequalities. We have

$-1\le\sin(x+y)\le1$

$\implies\dfrac1e\le e^{\sin(x+y)}\le e$

$\implies\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}e\le\iint_De^{\sin(x+y)}\,\mathrm dx\,\mathrm dy\le\iint_De\,\mathrm dx\,\mathrm dy$

$\implies\displaystyle\frac{4\pi^2}e\le\iint_De^{\sin(x+y)}\,\mathrm dx\,\mathrm dy\le4\pi^2e$

3 0

3 years ago

7th grade math help me please :))

mash [69]

Answer:

8/15 = 16/30 = 24/45 = 32/60

= 40/75 = 48/90 = 56/105 = 64/120

= 72/135 = 80/150 = 88/165 = 96/180

= 104/195 = 112/210 = 120/225 = 128/240

= 136/255 = 144/270 = 152/285 = 160/300

= 168/315 = 176/330 = 184/345 = 192/360

= 200/375 = 208/390 = 216/405 = 224/420

= 232/435 = 240/450 = 248/465 = 256/480

= 264/495 = 272/510 = 280/525 = 288/540

= 296/555 = 304/570 = 312/585 = 320/600

= 328/615 = 336/630 = 344/645 = 352/660

= 360/675 = 368/690 = 376/705 = 384/720

= 392/735 = 400/750 = 408/765 = 416/780

= 424/795 = 432/810 = 440/825 = 448/840

= 456/855 = 464/870 = 472/885 = 480/900

= 488/915 = 496/930 = 504/945 = 512/960

= 520/975 = 528/990 = 536/1005 = 544/1020

= 552/1035 = 560/1050 = 568/1065 = 576/1080

= 584/1095 = 592/1110 = 600/1125 = 608/1140

= 616/1155 = 624/1170 = 632/1185 = 640/1200

= 648/1215 = 656/1230 = 664/1245 = 672/1260

= 680/1275 = 688/1290 = 696/1305 = 704/1320

= 712/1335 = 720/1350 = 728/1365 = 736/1380

= 744/1395 = 752/1410 = 760/1425 = 768/1440

= 776/1455 = 784/1470 = 792/1485 = 800/1500

Step-by-step explanation:

7 0

3 years ago

Find a basis for the orthogonal complement of the subspace of R4 spanned by the vectors. v1 = (1, 4, -5, 3), v2 = (4, 15, 0, 5),

Arlecino [84]

Answer:

W1 = ( -75, 20, 1 , 0 )

W2 = ( 25, -7 , 0, 1 )

Step-by-step explanation:

attached below is the remaining part of the solution

for a homogenous system of equation ; Ax = 0

x1 + 4x2 -5x3 + 3x4 = 0

-x2 + 20x3 -7x4 = 0 note: x3 and x4 are free variables

we can take x3 = 0 and x4 = 1 , hence ; x2 = -7

∴ x1 - 28 + 3 = 0 = x1 = 25

W2 = ( x1 ,x2, x3, x4 ) = ( 25, -7 , 0, 1 )

now lets take x3 = 1 and x4 = 0 hence x2 = 20 , x1 = -75

∴ W1 = ( x1 , x2 , x3, x4 ) = ( -75, 20, 1 , 0 )

7 0

3 years ago