All categories

4 years ago

A 4-pack of frappuccino’s costs $10.88 how much does each individual can cost

Mathematics

1 answer:

Aleks04 [339]4 years ago

7 0

Answer:

$ 2.72

Step-by-step explanation:

divide 10.88 by 4

You might be interested in

Which side lengths form a right triangle?

Masja [62]

Answer: choice B

Explanation: Go through each answer choice and try the Pythagorean theorem to see if it works. This is because a right triangle’s side lengths are described by the Pythagorean theorem. one side^2 + second side^2 = longest side^2. In a right triangle, this equation will be true. Let me show you what I mean

For answer choice A:

(root 2)^2 + (root 3)^2 = (root 4)^2

2 + 3 = 4

5 = 4

5 does not equal 4. Therefore it cannot be A.

For answer choice B:

(root 8)^2 + (3)^2 = (root 17)^2

8 + 9 = 17

17 = 17

this is true! therefore this is a right triangle and it’s B

3 0

3 years ago

HELP PLEASE! BRAINLIEST ANSWER! A bag contains a total of 25 marbles. Five of the marbles are blue, 3 are red, 8 are green, and

Andrew [12]

Maybe 50 divided by 25 is 2 so the answer is 2

hope i helped :D : D

4 0

3 years ago

A carton measuring 72 cm by 48 cm by 75 cm was packed with cyndrical tins of radius 6 cm and height 15 cm standing upright.How m

Ierofanga [76]

Answer:

milk

Step-by-step explanation:

7 0

3 years ago

Identify the coefficient(s) of the variable in the expression below.

tankabanditka [31]

Answer:

Step-by-step explanation:

25 and 6

Coefficient is the number in front of the unknown variable

3 0

3 years ago

Read 2 more answers

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago