All categories

2 years ago

I so need help plaese problem A and B

Mathematics

1 answer:

Allisa [31]2 years ago

4 0

( A )

In an inverse variation relationship as one quantity increases the other quantity decreases.

This is indicated in both the table and the graph in that as length increases in value , the breaking weight decreases in value

(B )

given k = WL

For a length of 4 in, from the table the corresponding value of W is 41

k = 41 × 4 = 164

For a length of 6 in, from the table the corresponding value of W is 26

k = 26 × 6 = 156

You might be interested in

What is the y-intercept of the graph of the function f(x)=x^2+3x+5?

marta [7]

Answer:

y int = (0,5)

Step-by-step explanation:

hope this helps

8 0

2 years ago

Read 2 more answers

Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=x+4y subject to the constraint x2+y2=9, if such values

Vesnalui [34]

The Lagrangian is

$L(x,y,\lambda)=x+4y+\lambda(x^2+y^2-9)$

with critical points where the partial derivatives vanish.

$L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}$

$L_y=4+2\lambda y=0\implies y=-\dfrac2\lambda$

$L_\lambda=x^2+y^2-9=0$

Substitute $x,y$ into the last equation and solve for $\lambda$ :

$\left(-\dfrac1{2\lambda}\right)^2+\left(-\dfrac2\lambda\right)^2=9\implies\lambda=\pm\dfrac{\sqrt{17}}6$

Then we get two critical points,

$(x,y)=\left(-\dfrac3{\sqrt{17}},-\dfrac{12}{\sqrt{17}}\right)\text{ and }(x,y)=\left(\dfrac3{\sqrt{17}},\dfrac{12}{\sqrt{17}}\right)$

We get an absolute maximum of $3\sqrt{17}\approx12.369$ at the second point, and an absolute minimum of $-3\sqrt{17}\approx-12.369$ at the first point.

4 0

2 years ago

caroline bought 6 pieces of gold ribbon. each piece was 3/5 of a yard long. How many yards of ribbon did she buy in all?

Zarrin [17]

Answer:

= 3 3/5 yards

Step-by-step explanation:

If 1 piece = 3/5 yard

What about 6 pieces = ?

= (6 x 3/5) ÷ 1

= 18/5 ÷ 1

= 18/5

= 3 3/5 yards

5 0

2 years ago

Solve for W.<br><br> 7=LW<br><br> W=

sladkih [1.3K]

Answer:

this is the chan of formula from 7=LW to

w =7/L

because we divided both sides by L.

4 0

3 years ago

The radius of a circle is 13miles. what is the area of a sector bounded by a 20 degree arc

Stolb23 [73]

Answer:

Area of the sector = 29.48 square miles.

Step-by-step explanation:

Given:

Radius of a circle = 13 miles

Angle bounded by the arc = 20 deg

To find

Area of the sector bounded by 20 degrees.

Note: Area of a sector = $\frac{(\theta)}{360}\times \pi\times r^2$

Plugging the values in the equation.

⇒ $\frac{(\theta)}{360}\times \pi\times r^2$

⇒ $(\frac{20}{360})\times 3.14\times 13^2$

⇒ $29.48$ square-miles

So the area of the sector bounded by 20 degree arc is 29.48 square miles.

5 0

3 years ago