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3 years ago

15

Which number is larger -3.4 or -3 1/2 (BTW this is a fraction)? How will graphing these numbers on a number line help you determ

ine which number is larger?
PLEASE HELP ME OUT!!!!!!

1 answer:

valkas [14]3 years ago

3 0

Answer:

-3.4

Step-by-step explanation:

negative numbers are reverse counting on the line graph.

-3 1/2 is -3.5

-3.4 VS. -3.5

since it is reverse, -3.4 is greater

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Expand the following.... please help

Evgesh-ka [11]

= (3/8) (16x + - 24)
=(3/8) (16x) +(3/8)(-24)
=6x - 9

4 0

3 years ago

A group of friends were working on a student film. They spent all their budget on props and costumes. They spent $243 on props,

SVETLANKA909090 [29]

Answer:

The total budget for their student film was $540

Step-by-step explanation:

The total percentage of any quantity is 100%

∵ A group of friends was working on a student film

∵ They spent all their budget on props and costumes

∴ The total budget of the props and costumes = 100%

∵ They spent $243 on props

∵ They spent 55% of the total budget on costumes

→ Find the percentage of the props

∴ The percentage of the props = 100% - 55%

∴ The percentage of the props of the total budget = 45%

→ That means 45% of the total budget = 243

∵ 45% × total budget = 243

→ Change the percent to a number by divide it by 100

∵ 45% = 45 ÷ 100 = 0.45

∴ 0.45 × total budget = 243

→ Divide both sides by 0.45

∴ The total budget = 540

∴ The total budget for their student film was $540

4 0

3 years ago

Read 2 more answers

A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t

Licemer1 [7]

If $A(t)$ is the amount of salt in the tank at time $t$ , then the rate at which this amount changes over time is given by the ODE

$A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$

$A'(t)+\dfrac1{100}A(t)=15$

We're told that the tank initially starts with no salt in the water, so $A(0)=0$ .

Multiply both sides by an integrating factor, $e^{t/100}$ :

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}$
$\left(e^{t/100}A(t)\right)'=15e^{t/100}$
$e^{t/100}A(t)=1500e^{t/100}+C$
$A(t)=1500+Ce^{-t/100}$

Since $A(0)=0$ , we have

$0=1500+C\implies C=-1500$

so that the amount of salt in the tank over time is given by

$A(t)=1500(1-e^{-t/100})$

After 10 minutes, the amount of salt in the tank is

$A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}$

8 0

3 years ago

Prove that if one solution for a quadratic equation of the form x 2 + bx + c = 0 is rational (where b and c are rational), then

erica [24]

$x^2+bx+c = (x-r)(x-s)\\
x^2+bx+c = x^2-(r+s)x+rs\\\\
b = -(r+s)\\ c = rs$

Say r is rational. Suppose for a second, that s is not. Then, r+s is irrational. But this contradicts the fact that b is rational.

So, if one root is rational, then the other root is also rational

7 0

3 years ago

The other side is 10 yds

Vesnalui [34]

The correct answer is 900yds.

6 0

3 years ago