Ask question

Ask question

All categories

4 years ago

13

The cost of three pounds of grapes is $6.57 what is the constant proportionality that relates the cost in dollars to the number

of pounds of grapes

2 answers:

Kruka [31]4 years ago

8 0

2.19 is the answer :)) hope it helps

NemiM [27]4 years ago

7 0

$6.57/3 = $2.19 That is the answer.

You might be interested in

Select ALL the expressions that are equivalent to -4(8x-3)

Vinvika [58]

Answer:

dddddStep-by-step explanation:

8 0

3 years ago

18.4b=44.16 need help fast

MaRussiya [10]

Answer:

<u>If it's 18.4b=44.16:</u> $b=2.4$

<u>If it's 4b=44.16:</u> $b=11.04$

Step-by-step explanation:

<u>If it's 18.4b=44.16:</u>

$18.4b=44.16\\\\\frac{b}{18.4}=\frac{44.16}{18.4}\\\\b=2.4$

<u />

<u>If it's 4b=44.16:</u>

<u /> $4b=44.16\\\\\frac{b}{4}=\frac{44.16}{4}\\\\b=11.04$ <u />

4 0

3 years ago

Find the endpoint:

scZoUnD [109]

Answer:

The coordinates of Point D are: (-5,-8)

Step-by-step explanation:

The formula for mid-point is given by:

$M = (\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2})$

For the point (x1,y1) and (x2,y2)

Given that

M = (-3, -6)

C = (-1, -4) => (x1,y1)

Putting the values of given points in the formula for mid-point

$(-3,-6) = (\frac{-1+x_2}{2} , \frac{-4+y_2}{2})$

Putting respective coordinates equal

$-3 = \frac{-1+x_2}{2}\\-6 =-1+x_2\\x_2=-6+1\\x_2 = -5\\AND\\-6 = \frac{-4+y_2}{2}\\-12 = -4+y_2\\y_2 = -12+4\\y_2 = -8$

Hence,

The coordinates of Point D are: (-5,-8)

3 0

3 years ago

Please help! I’ll give BRAINLIEST and show ur work please

Darina [25.2K]

Answer:

M<E=T M<T=E

Step-by-step explanation:

As you can see in the trapezoid the show you only one two sides that are equals so that's mean that the angles are congruent each other.

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 <em>C</em> is the circle of radius 2 centered at the origin.

You can compute the line integral directly by parameterizing <em>C</em>. Let <em>x</em> = 2 cos(<em>t</em> ) and <em>y</em> = 2 sin(<em>t</em> ), with 0 ≤ <em>t</em> ≤ 2<em>π</em>. 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 <em>C</em> nor in the region bounded by <em>C</em>, 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 <em>D</em> is the interior of <em>C</em>, 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: $-2\times \pi\times2^2 = -8\pi$ .

3 0

3 years ago