All categories

3 years ago

DUE AT 8PM!!! HELP ASAP!!!

Mathematics

2 answers:

Murrr4er [49]3 years ago

7 0

Answer:

Option B cost less per ounce

Step-by-step explanation:

A - 3.30/16 = .21

B - 8.64/48 = .18

mr Goodwill [35]3 years ago

7 0

Option b. It costs less per ounce

You might be interested in

What is an example of 2 numbers that both have 6 digits but the greater number is determined by the hundreds place

romanna [79]

273,123
273,723

The second number is the same as the first one with only the hundreds place different

4 0

3 years ago

Read 2 more answers

PLEASE HELP ME IT IS DUE NOW

Goryan [66]

Answer: X is 3 here. Haven't done this math (don't know how to graph) but I found x so it should be hard from here :)

Good Luck

4 0

3 years ago

si en una caja hay 5 pelotas una verde una amarilla.una azul.una negra y una roja ya a salido la rija y la verde cual es la posi

Leokris [45]

Answer:

uno a tres

Step-by-step explanation:

Creo que es correcto

Avísame si estoy equivocado

Y que tengas un feliz día de acción de gracias

3 0

3 years ago

The segment GH and GI are tangent

vodka [1.7K]

Answer:

OG = 8.7

Step-by-step explanation:

The angle a tangent makes with a radius at the point of tangency is a right angle. That means ΔOGH is a right triangle.

The tangents to a circle from an external point are congruent, so GH = GI = 6.3. Now we know the sides of the right triangle are of lengths 6 and 6.3. The hypotenuse, OG is found using the Pythagorean theorem.

OG² = OH² +GH²

OG² = 6² +6.3² = 75.69

OG = √75.69 = 8.7

The length of OG is 8.7 units.

______

<em>Additional comment</em>

You might recognize this triangle as a 0.3 multiple of the Pythagorean triple (20, 21, 29).

8 0

3 years ago

A quantity with an initial value of 6200 decays continuously at a rate of 5.5% per month. What is the value of the quantity afte

ELEN [110]

Answer:

410.32

Step-by-step explanation:

Given that the initial quantity, Q= 6200

Decay rate, r = 5.5% per month

So, the value of quantity after 1 month, $q_1 = Q- r \times Q$

$q_1 = Q(1-r)\cdots(i)$

The value of quantity after 2 months, $q_2 = q_1- r \times q_1$

$q_2 = q_1(1-r)$

From equation (i)

$q_2=Q(1-r)(1-r) \\\\q_2=Q(1-r)^2\cdots(ii)$

The value of quantity after 3 months, $q_3 = q_2- r \times q_2$

$q_3 = q_2(1-r)$

From equation (ii)

$q_3=Q(1-r)^2(1-r)$

$q_3=Q(1-r)^3$

Similarly, the value of quantity after n months,

$q_n= Q(1- r)^n$

As 4 years = 48 months, so puttion n=48 to get the value of quantity after 4 years, we have,

$q_{48}=Q(1-r)^{48}$

Putting Q=6200 and r=5.5%=0.055, we have

$q_{48}=6200(1-0.055)^{48} \\\\q_{48}=410.32$

Hence, the value of quantity after 4 years is 410.32.

4 0

3 years ago

Read 2 more answers