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3 years ago

(07.05) Graph the first six terms of a sequence where a = -10 and d = 3.

Mathematics

1 answer:

butalik [34]3 years ago

7 0

Answer:

Step-by-step explanation:

Using the formular, Sn = n/2(2a + (n - 1)d)

where, n = 6: Sn = 6/2( -20 + 5(3)) = 3(15 - 20)

∴ Sn = 15

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APQR is inscribed in a circle with P=60, Q = 80, and R = 40. What are the measures of PQ, QR, and PR?

Sergeeva-Olga [200]

Answer:

PQ:80

QR:160

PR:120

Step-by-step explanation:

$PQ=2x$ ∠ $R$

$=2$ × $40$

= $80$

$QR=2$ × $m$ ∠ $p$

$=20$ × $80$

$=160$

$PR=2$ × $m$ ∠ $Q$

= $2$ × $60\\$

= $120$

<em>hope it helps </em>

<em>have a great day!!</em>

5 0

2 years ago

A bacteria culture starts with 400 bacteria and grows at a rate proportional to its size. After 4 hours, there are 9000 bacteria

Kaylis [27]

Answer:

A) The expression for the number of bacteria is $P(t) = 400e^{0.7783t}$ .

B) After 5 hours there will be 19593 bacteria.

C) After 5.55 hours the population of bacteria will reach 30000.

Step-by-step explanation:

A) Here we have a problem with differential equations. Recall that we can interpret the rate of change of a magnitude as its derivative. So, as the rate change proportionally to the size of the population, we have

$P' = kP$

where $P$ stands for the population of bacteria.

Writing $P'$ as $\frac{dP}{dt}$ , we get

$\frac{dP}{dt} = kP$ .

Notice that this is a separable equation, so

$\frac{dP}{P} = kdt$ .

Then, integrating in both sides of the equality:

$\int\frac{dP}{P} = \int kdt$ .

We have,

$\ln P = kt+C$ .

Now, taking exponential

$P(t) = Ce^{kt}$ .

The next step is to find the value for the constant $C$ . We do this using the initial condition $P(0)=400$ . Recall that this is the initial population of bacteria. So,