All categories

3 years ago

What is the slope of the line represented by the equation y = 4/5x-3?

Mathematics

2 answers:

vekshin13 years ago

8 0

Answer:

4/5

Step-by-step explanation:

The equation for the line given is y = 4/5x - 3. This equation is in slope-intercept form, which always looks like y = mx + b. The "m" in that is the slope, so we can use this information to find the slope in the equation.

It looks like 4/5 is before the x in y = 4/5x - 3, and so 4/5 has to be our "m," and thus our slope. So, that's the answer. Hopefully that's helpful! If have any more questions, let me know. :)

dsp733 years ago

3 0

Answer:

Step-by-step explanation:

option C.

When in the form $y= mx +q$ the slope of a line is given by the number multiplying x

You might be interested in

there are 45 new houses built. Last month 2/3 of them sold. This month, 1/5 of the remaining houses were sold. How many houses l

Nady [450]

Answer: 9

Step-by-step explanation: 45 times 2/3 is 30. So, 30 houses were sold last month. 30 times 1/5 is 6. So, 6 houses were sold this month. Add them together. 30+6=36. Subtract that from 45 and you will get 9. There are 9 houses left to be sold.

7 0

3 years ago

Solve for y Please i need help

Nataly [62]

Answer:

y = 1/9

Step-by-step explanation:

Use distributive property and simplify!

5 0

3 years ago

Select all the correct graphs.

Anika [276]

Answer:

garoph 1 and 2 hope this helps

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

1. The perimeter of a piece of paper is 18 cm.

OlgaM077 [116]

Answer:

a) length = 3 and width = 6

Area of the rectangle = 3 × 6 = 18 cm²

b) length = 7 and width = 2

Area of the rectangle = 7 × 2 = 14 cm²

c) length = 5 and width = 4

Area of the rectangle = 5× 4 = 20 cm²

Step-by-step explanation:

Step(i):-

Given perimeter of the paper = 18 cm

we know that the paper has a rectangle shape

The perimeter of the rectangle = 18 cm

2( l + w ) = 18

we choose length = 3 and width = 6

The perimeter of the rectangle = 2( 3+6) = 18 cm

Area of the rectangle = 3 × 6 = 18 cm²

we choose length = 7 and width = 2

The perimeter of the rectangle = 2( 7+2) = 18 cm

Area of the rectangle = 7 × 2 = 14 cm²

we choose length = 5 and width = 4

The perimeter of the rectangle = 2( 5+4) = 18 cm

Area of the rectangle = 5× 4 = 20 cm²

3 0

3 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