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3 years ago

Ribbon is sold at 3 yards for 6.33. Jackie bought 24 yards of ribbon for a project. How much did he pay?

Mathematics

2 answers:

spin [16.1K]3 years ago

7 0

Yup it 50.67 because every 3 yard that tooling alway would equal 6.33

Paladinen [302]3 years ago

3 0

$\frac{3}{6.33}=\frac{24}{n}\\\frac{3}{\frac{19}{3}}*\frac{8}{8}=\frac{24}{n}\\\frac{24}{\frac{152}{3}}=\frac{24}{n}\\\frac{\frac{152}{3}}{24}=\frac{n}{24}\\\frac{152}{3}=n$

$50.67

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Plz help asap thank youuu

wariber [46]

Answer:

1. y = 0, b/a

2. y = 0, a/b

3. y =0, c/b

4. y =0, c/a

Step-by-step explanation:

The slope formula for y = mx + b which is known as standard form is Ax + By ≥ C

slope = A/B

x-intercept = C/A

y-intercept = C/B

and now we are dueling with y, so we will use y-intercept

1. cx + ay = b

where c is x, a = y and b=c

Y-int = C/B

y = 0, b/a

2. cx + by = a

y = 0, a/b

3. ax + by = c

y =0, c/b

4. bx+ ay = c

y =0, c/a

Hope this helps

7 0

3 years ago

Round to the nearest tenths 2.52_______?

abruzzese [7]

1. 2.5

2. 96

3. 16.7

4. 6.30

5. 26.8

6. 42.2

7. 993.20

8. 10.24

9. 597.11

10. 776

11. 8.86

12. 3

5 0

3 years ago

Part A

KIM [24]

Answer:

9.275 inches or 9 11/40

Step-by-step explanation:

current depth = beginning depth - inches swam up + inches swam down

8 3/4 - 2 3/5 + 3 1/8

$9\frac{30 - 24 + 5}{40}$ = 9 11/40

8 0

2 years ago

Find the absolute maximum and minimum values of f(x, y) = x+y+ p 1 − x 2 − y 2 on the quarter disc {(x, y) | x ≥ 0, y ≥ 0, x2 +

Andreas93 [3]

Answer:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

Step-by-step explanation:

In order to find the absolute max and min, we need to analyse the region inside the quarter disc and the region at the limit of the disc:

Region inside the quarter disc:

There could be Minimums and Maximums, if:

∇f(x,y)=(0,0) (gradient)

we develop:

(1-2x, 1-2y)=(0,0)

x=1/2

y=1/2

Critic point P(1/2,1/2) is inside the quarter disc.

f(P)=1/2+1/2+p1-1/4-1/4=1/2+p1

f(0,0)=p1

We see that:

f(P)>f(0,0), then P(1/2,1/2) is a maximum relative

Region at the limit of the disc:

We use the Method of Lagrange Multipliers, when we need to find a max o min from a f(x,y) subject to a constraint g(x,y); g(x,y)=K (constant). In our case the constraint are the curves of the quarter disc:

g1(x, y)=x^2+y^2=1

g2(x, y)=x=0

g3(x, y)=y=0

We can obtain the critical points (maximums and minimums) subject to the constraint by solving the system of equations:

∇f(x,y)=λ∇g(x,y) ; (gradient)

g(x,y)=K

Analyse in g2:

x=0;

1-2y=0;

y=1/2

Q(0,1/2) critical point

f(Q)=1/4+p1

We do the same reflexion as for P. Q is a maximum relative

Analyse in g3:

y=0;

1-2x=0;

x=1/2

R(1/2,0) critical point

f(R)=1/4+p1

We do the same reflexion as for P. R is a maximum relative

Analyse in g1:

(1-2x, 1-2y)=λ(2x,2y)

x^2+y^2=1

Developing:

x=1/(2λ+2)

y=1/(2λ+2)

x^2+y^2=1

So:

(1/(2λ+2))^2+(1/(2λ+2))^2=1

$\lambda_{1}=\sqrt{1/2}*-1 =-0.29$

$\lambda_{2}=-\sqrt{1/2}*-1 =-1.71$

$\lambda_{2}$ give us (x,y) values negatives, outside the region, so we do not take it in account

For $\lambda_{1}$ : S(x,y)=(0.70, 070)

and

f(S)=0.70+0.70+p1-0.70^2-0.70^2=0.42+p1

We do the same reflexion as for P. S is a maximum relative

Points limits between g1, g2 y g3

we need also to analyse the points limits between g1, g2 y g3, that means U(0,0), V(1,0), W(0,1)

f(U)=p1

f(V)=p1

f(W)=p1

We can see that this 3 points are minimums relatives.

Conclusion:

We compare all the critical points P,Q,R,S,T,U,V,W an their respective values f(x,y). We find that:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

4 0

3 years ago

Q.no 21 solve it. Plz. Plz

tekilochka [14]

Answer:

Hamm i don't know sorry a want yo help You Buy a can't because a don't know sorry

8 0

2 years ago