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Carrots sell for $2.10 per pound, and crackers sell for $2.90 per pound. Glen bought some carrots and some crackers. The total w

musickatia [10]

The answer is
carrots - 0.8 pounds
crackers - 1.5 pounds

x - the weight of carrots
y - the weight of crackers
The total weight was 2.3 pounds: x + y = 2.3

Carrots sell for $2.10 per pound: 2.10x
Crackers sell for $2.90 per pound: 2.90y
The total cost was $6.03: 2.10x + 2.90y = 6.03

x + y = 2.3
2.10x + 2.90y = 6.03
_________
x = 2.3 - y
2.10x + 2.90y = 6.03
_________
2.10(2.3 - y) + 2.90y = 6.03
4.83 - 2.10y + 2.90y = 6.03
4.83 + 0.80y = 6.03
0.80y = 6.03 - 4.83
0.80y = 1.20
y = 1.20y : 0.80
y = 1.5 pounds

x = 2.3 - y = 2.3 - 1.5
x = 0.8 pounds

3 0

3 years ago

Read 2 more answers

Identify any extrema of the function by recognizing its given form or its form after completing the square. Verify your results

Andrews [41]

Answer:

Step-by-step explanation:

Given that,

g(x, y) = 5 - (x - 3)² - (y + 2)²

Let find the grad of the function,

The grad of a function is defined as

∇g= ∂g/∂x •i + ∂g/∂y •j + ∂g/∂z •k

∇g = gx•i + gy•j + gz•k

gx = -2(x-3) = -2x+6

gy = -2(y+2) = -2y -4

∇g = (-2x+6) •i + (-2y-4)•j

We have a maximum or a minimum If g conservative, then, ∇g = 0i +0j

Then comparing this to the grad of the function

(-2x+6) •i + (-2y-4)•j = 0i +0j

Then, -2x+6 = 0

2x=6

x = 3

Also, -2y-4=0

-2y=4

y = -2

Then, g(x, y) = 5 - (x - 3)² - (y + 2)²

g(x, y) = 5 - (3 - 3)² - (-2 + 2)²

g(x, y) = 5

So the critical point is (3, -2, 5)

gx =-2x+6

Second derivative of gx with respect to x

gxx=-2

gy=-2y-4

Second derivative of gy with respect to y

gyy=-2

Second derivative of gx with respect to y

gxy =0

d =gxxgyy - (gxy)²

Then, gyy=gxx

d = -2×-2 -0²

d = 4-0=4

Since d>0

Since d is greater than 0, then, it is not a chair points.

Then, since gxx=-2<0, gyy=-2<0

Cause the second derivative in x (or in y) is less than zero, then the point is relatively maximum

So the maximum point is (3, -2, 5).

6 0

3 years ago

Need answer ASAP please

bulgar [2K]

Answer:

0.44444444444, round it to the tenth or hundredth or wherever it tells you to round the answers to.

5 0

3 years ago

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Emay ears $15 for each car she washes. Create an equation to represent the relationship between the number of cars

emmasim [6.3K]

Answer:

$25

Step-by-step explanation:

8 0

3 years ago

Any 10th grader solve it for 50 points

kkurt [141]

Answer:

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$ is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., $a_{1}=A$ and common difference=D

The nth term can be written as

$a_{n}=A+(n-1)D$

pth term of given arithmetic progression is a

$a_{p}=A+(p-1)D=a$

qth term of given arithmetic progression is b

$a_{q}=A+(q-1)D=b$ and

rth term of given arithmetic progression is c

$a_{r}=A+(r-1)D=c$

We have to prove that

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0$

Now to prove LHS=RHS

Now take LHS

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)$

$=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)$

$=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)$

$=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}$

$=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}$

$=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}$