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3 years ago

What is half of a teaspoon of vanilla

Mathematics

1 answer:

Ksivusya [100]3 years ago

6 0

Answer:

I believe the answer would be: Half of a teaspoon of vanilla.

Step-by-step explanation:

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Anita's, a fast-food chain specializing in hot dogs and garlic fries, keeps track of the proportion of its customers who decide

Svetradugi [14.3K]

Answer: Pt = Ng / Nt

Step-by-step explanation:

Where Ng = number of customers that order food to go as reported at Anita's

Nt = total number of customers reported at Anita's

Pt = probability that a customer will order food to go at Anita's

Pt= Ng/Nt

Goodluck...

8 0

3 years ago

Breadth

Stolb23 [73]

Answer:

Volume of a cuboid = (length × breadth × height) cubic units.

= (l × b × h) cubic units.

(Since area = ℓ × b)

Volume of a cuboid = area of one surface × height cubic units

Let us look at the given cuboid.

The length of the cuboid = 5 cm

The breadth of the cuboid = 3 cm

The height of cuboid (thickness) = 2 cm

The number of 1 cm cubes in the given cuboid = 30 cubes = 5 × 3 × 2

We find that volume of the given cuboid with length 5 cm, breadth 3 cm and height 2 cm is 30 cu cm.

6 0

2 years ago

Please help asap 25 pts

yuradex [85]

Since y has to be larger, the area is above the line.

Since the boundary is included (because of ≥ in stead of >) it should be drawn as a solid line.

Brings us to answer A, right?

See plot below.

6 0

3 years ago

Use the substitution u = tan(x) to evaluate the following. int_0^(pi/6) (text(tan) ^2 x text( sec) ^4 x) text( ) dx

Rudiy27

If we use the substitution $u = \tan x$ , then $du = \sec^2 {x}\ dx$ . If you try substituting just $u$ and $du$ into the integrand, though, you'll notice that there's a $\sec^2x$ left over that we have to deal with.

To get rid of this problem, use the identity $\tan^2 x + 1 = \sec^2 x$ and substitute in the left side of the identity for the extra $\sec^2x$ , as shown:

$\int\limits^{\pi/6}_0 {tan^2 x \ sec^4 x} \, dx$
$\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx$

From there, we can substitute in $u$ and $du$ , and then evaluate:

$\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx$
$\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^2(u^2 + 1)} \, du$
$\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^4 + u^2} \, du$
$= \left.\frac{u^5}{5} + \frac{u^3}{3}\right|_0^\frac{1}{\sqrt{3}}$
$= (\frac{(\frac{1}{\sqrt{3}})^5}{5} + \frac{(\frac{1}{\sqrt{3}})^3}{3}) - (\frac{(0)^5}{5} + \frac{(0)^3}{3})$
$= \frac{1}{45\sqrt{3}} + \frac{1}{9\sqrt{3}} = \frac{6}{45\sqrt{3}} = \bf \frac{2}{15\sqrt{3}}$

8 0

3 years ago

6(0.25n−2)=n−0.5(5−2n)

Anni [7]

Answer: n= -19

Hope that helped

8 0

3 years ago