All categories

3 years ago

Nel

Mathematics

1 answer:

dusya [7]3 years ago

8 0

Answer:

can you rewrite that pls I can't understand

You might be interested in

Please help me out on this one

vivado [14]

it is definitely the answer B

8 0

2 years ago

As a butcher, you sell beef in 3.5-pound packages. You anticipate needing 25 packages for the next two days. The

bekas [8.4K]

Answer:

O weight in kilograms = 25 x 3.5 ÷ 2.2

Step-by-step explanation:

As a butcher, you sell beef in 3.5-pound packages. You anticipate needing 25 packages for the next two days.

Therefore, the total number of pounds for the 25 packages =

3.5 pound packages × 25

= 87.5 pounds

We are told that : The wholesaler from whom you purchase large pieces of meat measures the weight of the beef in kilograms.

Note that:

2.2 pounds = 1 kg

Hence:

87.5 pounds =

= 87.5÷ 2.2

= 39.772727273 kg

He needs to buy 39.772727273kg from the wholesaler

The equation can you use to determine the weight of the beef, in kilograms, that you need to purchase from the wholesaler is:

O weight in kilograms = (25 x 3.5) ÷ 2.2

7 0

3 years ago

Factor 1/6 out of 1/6x+1/6

lesya692 [45]

Answer:

1/6(1x+1)

Step-by-step explanation:

8 0

2 years ago

Scarlett made 36 muffins on Tuesday, and 20 of the muffins were blueberry. What is the ratio of blueberry muffins to all muffins

Paha777 [63]

Answer:I belive it is 56 I may be wrong but it iz what it iz

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to chec

klasskru [66]

Answer:

There is a horizontal tangent at (0,-4)

The tangent is vertical at (-2,-3) and (2,-3).

Step-by-step explanation:

The given function is defined parametrically by the equations:

$x=t^3-3t$

and

$y=t^2-4$

The tangent function is given by:

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\implies \frac{dy}{dx}=\frac{2t}{3t^2-3}$

The tangent is vertical at when $\frac{dx}{dt}=0$

$\implies \frac{3t^2-3}{2t}=0$

$\implies 3t^2-3=0$

$\implies 3t^2=3$

$\implies t^2=1$

$\implies t=\pm1$

When t=1,

$x=1^3-3(1)=-2$ and $y=1^2-4=-3$

When t=-1,

$x=(-1)^3-3(-1)=2$ and $y=(-1)^2-4=-3$

The tangent is vertical at (-2,-3) and (2,-3).

The tangent is horizontal, when $\frac{dy}{dx}=0$ or $\frac{dy}{dt}=0$

$\implies 2t=0$

$\implies t=0$

When t=0,

$x=0^3-3(0)=0$ and $y=0^2-4=-4$

There is a horizontal tangent at (0,-4)

5 0

3 years ago