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3 years ago

Evaluate the expression. 1.2 + 0.8 ÷ 4 – 3(0.7 – 0.81)

Mathematics

2 answers:

Olegator [25]3 years ago

8 0

Answer: 1.73

Step-by-step explanation:

1.2 + 0.8 ÷ 4 – 3(0.7 – 0.81)

1.2+0.8÷4−3×−0.11

1.2+0.2−3×−0.11

1.2+0.2−(−0.33)

1.2+0.2+0.33

1.4+0.33=1.73

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great Valentines Day! :-)

- Cutiepatutie ☺❀❤

Alex_Xolod [135]3 years ago

5 0

Answer:

173

——— = 1.73000

100

Step-by-step explanation:

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Sliva [168]

$a_1=11+(1-1)(4)=11+0=11\\\\a_2=11+(2-1)(4)=11+4=15$

so:

$a_2=a_1+\,?\\\\15=11+\,?\\\\\boxed{?=4}$

Answer B.

8 0

3 years ago

There are 6 red marbles, 9 blue marbles, and 10 green marbles in a bag.

Nat2105 [25]

Answer:

I think 64%

Step-by-step explanation:

6+9+10 = 25

25x4 = 100

6x4 = 24

10x4 = 40

24+40 = 64

64/100 = 64%

Sry if this is wrong but I think this should be correct

7 0

3 years ago

Read 2 more answers

38. Evaluate f (3x +4y)dx + (2x --3y)dy where C, a circle of radius two with center at the origin of the xy

lina2011 [118]

It looks like the integral is

$\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy$

where C is the circle of radius 2 centered at the origin.

You can compute the line integral directly by parameterizing C. Let x = 2 cos(t ) and y = 2 sin(t ), with 0 ≤ t ≤ 2π. Then

$\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \int_0^{2\pi} \left((3x(t)+4y(t))\dfrac{\mathrm dx}{\mathrm dt} + (2x(t)-3y(t))\frac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt \\\\ = \int_0^{2\pi} \big((6\cos(t)+8\sin(t))(-2\sin(t)) + (4\cos(t)-6\sin(t))(2\cos(t))\big)\,\mathrm dt \\\\ = \int_0^{2\pi} (12\cos^2(t)-12\sin^2(t)-24\cos(t)\sin(t)-4)\,\mathrm dt \\\\ = 4 \int_0^{2\pi} (3\cos(2t)-3\sin(2t)-1)\,\mathrm dt = \boxed{-8\pi}$

Another way to do this is by applying Green's theorem. The integrand doesn't have any singularities on C nor in the region bounded by C, so

$\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \iint_D\frac{\partial(2x-3y)}{\partial x}-\frac{\partial(3x+4y)}{\partial y}\,\mathrm dx\,\mathrm dy = -2\iint_D\mathrm dx\,\mathrm dy$

where D is the interior of C, i.e. the disk with radius 2 centered at the origin. But this integral is simply -2 times the area of the disk, so we get the same result: $-2\times \pi\times2^2 = -8\pi$ .