All categories

3 years ago

Can someone help me with my math problem pls my teacher is on my back about this

Mathematics

2 answers:

Dafna11 [192]3 years ago

8 0

Answer:

1. 12 sugar cubes

2. 24 sugar cubes

3. 72 by 54 by 144, because it gives you the dimensions for one box so if you need something that will fit 36 boxes then you need to multiply the dimensions by 36.

Step-by-step explanation:

1. multiply all the dimensions given to find volume

2. you could fit twice as many sugar cubes if they are half the original size so just multiply the 12 by 2

Karo-lina-s [1.5K]3 years ago

5 0

Answer:

A: 8

B: 96

I'm not sure on C, sorry.

You might be interested in

Please Do These Three Equations And Show Work

german

2) Add 0.1 to both sides. v/2.2=7.5

Then multiply by 2.2 on both sides.

v=16.5

4) Subtract 1.9 from both sides. -1.3g=-13

Divide both sides by -1.3

g=10

6) Add 1.4 to both sides. -12.9=-3d

Divide -3 on both sides.

d=4.3

3 0

2 years ago

Explain the steps you would take to find the area of the following composite shape.

julsineya [31]

What composite shape? Where is the shape?

7 0

3 years ago

Factor the trinomial by using the factor X method 3n-8n+4

Alex73 [517]

3n² - 8n + 4
3n² - 6n - 2n + 4
(3n - 2)(n - 2)

7 0

3 years ago

Which is the better buy? 50 fl oz of detergent for $2.49 or <br> 32 fl oz of detergent for $1.99

slamgirl [31]

50 fl oz of detergent

3 0

3 years ago

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

3 0

3 years ago