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Akimi4 [234]
3 years ago
7

Jack buys a tablet that costs $99 and a memory card that costs $15. He has a coupon for a 15% discount. What is the amount of th

e discount on the two items?
Mathematics
1 answer:
Gala2k [10]3 years ago
7 0
It would be $84.15 for the tablet and $12.75 for the memory card, so $14.85 off on the tablet and $2.25 off on the card. I hope this helps!
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Step-by-step explanation:

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Solve this equation 53y - 55 = 42y
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53y-42y=55
11y=55
Y=55/11 =5
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Rectangle ABCD was dilated to create rectangle A'B'C'D. Triangle A B C D is dilated to form triangle A prime B prime C prime D.
Julli [10]

Answer:

AB = 6

Step-by-step explanation:

Given

BC = 3.8

A'B' = 15

B'C' = 9.5

Required

Determine the length of AB

Because A'B'C'D is a dilation of ABCD, then the following relationship must exist:

A'B' : AB = B'C' : BC

Substitute values for A'B', B'C' and BC

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3 years ago
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Consider F and C below. F(x, y, z) = 2xz + y2 i + 2xy j + x2 + 9z2 k C: x = t2, y = t + 3, z = 3t − 1, 0 ≤ t ≤ 1 (a) Find a func
Ray Of Light [21]

Answer:

a)∇f = 2y + 2x + 18z

b) \int\limits^._C {F} \, dr =108

Step-by-step explanation:

Given:

f (x,y,z ) = (2xz+ y^{2})i + (2xy) j +(x^{2} + 9z^{2})k

The curve C :

x=t^{2} ,\\y= t+3\\z= 3t-1

where 0 ≤ t ≤ 1

Required:

(a) F = ∇f =? (F is a vector here)

(b) \int\limits^._C {F} \, dr =?

Solution

First we will find the directional derivative F = ∇f

for that , we will use the formula :

∇f = F_{x}i+ F_{y} j+F_{z}k

Fx= δf/δx = δ/δx (2xz+ y^{2})i = 2z i

Fy= δf/δy=δ/δy (2xy)j = 2x j

Fz= δf/δz=δ/δz(x^{2} + 9z^{2})k = 18z k

∇f = (2z) i .i + (2x) j.j + (18z) k.k

∇f = 2z + 2x + 18z

<em>For part b):</em>

<em>we will use line integral formula:</em>

\int\limits^._C {F} \, dr

to calculate dr, we will need the curve C:

r = x(t)+y(t)+z(t)

r=(t^{2})i + (t+3) j +(3t-1) k

\frac{dr}{dt}=\frac{dx}{dt} +\frac{dy}{dt} + \frac{dz}{dt}

\frac{dx}{dt} = 2t

\frac{dy}{dt} = 1

\frac{dz}{dt} = 3

\int\limits^._C {F} \, dr = \int\limits^1_0 {F_{x} } \, dx+ F_{y} dy +F_{z} dz

= \int\limits^1_0 {(2z(2t) + 2x(1) + 18z (3)} \, )

put values of y, x and z

= \int\limits^1_0 {2(3t-1) + 2(t^{2}) +18 (3) (3t-1)} \,

={2t^{2} + 6t+162t -54-2}\, |^1_0

={ 2t^{2}+ 168t - 56} \,|^1_0               (Note : f(1)-f(0))

=2(1)+162(1)+2(0)+162(0)-56

= 2+162 -56

\int\limits^._C {F} \, dr =108

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Which is the general form of the equation of the circle shown ?
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Answer:

there is no circle

Step-by-step explanation:

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