What does 7/3 + (-3/8) equal?

Mathematics

1 answer:

AlekseyPX3 years ago

3 0

The correct answer is 1 23/24. Hope this helps.

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Can someone please help me witch is the answer 20 points to the one who can answer first

ladessa [460]

First, let's find the slope.

Slope = (7 - (-3)) / (2 - 0) = (7 + 3) / 2 = 10 / 2 = 5

So far our equation looks like: y = 5x + b

Leta plug one of the points into our equation to find b.

-3 = 5(0) + b

Simplify.

-3 = b

The equation is: y = 5x - 3

Answer choice B is correct.

8 0

3 years ago

A comparison of the measures of central angles and inscribed angles of circles to their intercepted arcs are listed in the table

geniusboy [140]

A central angle means an angle that is at the center of a circle and whose legs are intersecting the circle.

<h3>What is a central angle?</h3>

Your information is incomplete as the data is not given. Therefore, an overview will be given.

Central angles are shaped by an arc between the two points. When a central angle and the inscribed angle intercept the same arc, then the central angle would be the double of the inscribed angles while the inscribed angle will be half of the central angle.

Learn more about central angle on:

brainly.com/question/19758686

3 0

2 years ago

A small military base housing 1,000 troops, each of whom is susceptible to a certain virus infection. Assuming that during the c

slava [35]

Answer:

$I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$

Step-by-step explanation:

The rate of infection is jointly proportional to the number of infected troopers and the number of non-infected ones. It can be expressed as follows:

$\frac{dI}{dt}=a*I*(1000-I)$

Rearranging and integrating

$\frac{dI}{dt}=a*I*(1000-I)\\\\\frac{dI}{I*(1000-I)}=a*dt\\\\\int\frac{dI}{I*(1000-I)}=\int a*dt\\\\-\frac{ln(1000/I-1)}{1000}+C=a*t$

At the initial breakout (t=0) there was one trooper infected (I=1)

$-\frac{ln(1000/1-1)}{1000}+C=0\\\\-0,006906755+C=0\\\\C=0,006906755$

In two days (t=2) there were 5 troopers infected

$-\frac{ln(1000/5-1)}{1000}+0,006906755=a*2\\\\-0,005293305+0,006906755=2*a\\a = 0,00161345 / 2 = 0,000806725$

Rearranging, we can model the number of infected troops (I) as

$-\frac{ln(1000/I-1)}{1000}+0,006906755=0,000806725*t\\\\-\frac{ln(1000/I-1)}{1000}=0,000806725*t-0,006906755\\-ln(1000/I-1)=0,806725*t-0.6906755\\\\\frac{1000}{I}-1=exp^{0,806725*t-0.6906755} \\\\\frac{1000}{I}=exp^{0,806725*t-0.6906755}+1\\\\I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$