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3 years ago

Which expressions are equivalent to 16 x 4 minus 64 check all that apply

Mathematics

2 answers:

I am Lyosha [343]3 years ago

5 0

I’m pretty sure it is choice 1, 3, and 4

Reika [66]3 years ago

3 0

Oof I was going to answer this but it got me confusion.

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What is the axis of symmetry of h(x)= 5x^2+40x+ 64?

andrezito [222]

Answer:

Axis of symmetry is x = -4

Step-by-step explanation:

h(x)= 5x^2+40x+ 64

Use the formula for axis of symmetry: x = -b/(2a)

Here, a = 5, b = 40, c = 64

x = -40/(2*5) = -40/10 = -4

Axis of symmetry is x = -4

3 0

3 years ago

Read 2 more answers

Logan made a profit of $210.00 as a mobile groomer. He charged $75.00 per appointment and received $35.00 in tips, but also had

Sonja [21]

Answer: Per the I agree statement you signed when you registered for your course. Selling course material to another person, student, entity, and/or uploading to a third-party vendor without the express written permission of FLVS is prohibited. Course materials include, but are not limited to, class notes, instructor's power points, course syllabi, tests, quizzes, labs, instruction sheets, homework, study guides, and handouts.

4 0

2 years ago

I need help please help. 2 Questions.

Colt1911 [192]

Answer:

Step-by-step explanation:

8 0

3 years ago

In APQR, p = 690 cm, a = 290 cm and r=450 cm. Find the measure of ZQ to the

katen-ka-za [31]

9514 1404 393

Answer:

16.8°

Step-by-step explanation:

The Law of Cosines can be rearranged to give the angle measure.

angle Q = arccos((p² +r² -q²)/(2pr))

angle Q = arccos((690² +450² -290²)/(2×690×450))

angle Q = arccos(594500/621000) ≈ 16.7985°

The measure of angle Q is about 16.8°.

3 0

3 years ago

37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the

Nastasia [14]

I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy$

and I assume C has a positive orientation in both cases

(a) It looks like the region has the curves y = x and y = x ² as its boundary***, so that the interior of C is the set D given by

$D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}$

• Compute the line integral directly by splitting up C into two component curves,

C₁ : x = t and y = t ² with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}$

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}$

(b) C is the boundary of the region

$D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}$

• Compute the line integral directly, splitting up C into 3 components,

C₁ : x = t and y = 0 with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = t with 0 ≤ t ≤ 1

C₃ : x = 0 and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}$

• Using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}$

4 0

3 years ago