Ask question

Ask question

All categories

olga nikolaevna [1]

2 years ago

11

Find the area of the figure. 5ft 85° 9ft

1 answer:

Natali5045456 [20]2 years ago

8 0

Answer:

where is your figure

Step-by-step explanation:

You might be interested in

Help please see picture

kondaur [170]

Answer:

B: 6/4 or 1 2/4

Step-by-step explanation:

5 0

3 years ago

Which is the least expensive per oz? 12 oz for $5.28 or 18 oz for $7.96. round to the nearest cent?

blondinia [14]

5.28 / 12 = 0.44 per oz
7.96 / 18 = 0.4422 rounds to 0.44 per oz

technically, by rounding to the nearest cent, they are the same cost.

6 0

2 years ago

Read 2 more answers

Please help logarithms!

nlexa [21]

Given:

$\log_34\approx 1.262$

$\log_37\approx 1.771$

To find:

The value of $\log_3\left(\dfrac{4}{49}\right)$ .

Solution:

We have,

$\log_34\approx 1.262$

$\log_37\approx 1.771$

Using properties of log, we get

$\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_349$ $\left[\because \log_a\dfrac{m}{n}=\log_am-\log_an\right]$

$\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_37^2$

$\log_3\left(\dfrac{4}{49}\right)=\log_34-2\log_37$ $[\log x^n=n\log x]$

Substitute $\log_34\approx 1.262$ and $\log_37\approx 1.771$ .

$\log_3\left(\dfrac{4}{49}\right)=1.262-2(1.771)$

$\log_3\left(\dfrac{4}{49}\right)=1.262-3.542$

$\log_3\left(\dfrac{4}{49}\right)=-2.28$

Therefore, the value of $\log_3\left(\dfrac{4}{49}\right)$ is $-2.28$ .

5 0

2 years ago

Ralph is leasing a $38,000 car for 36 months. The terms of his lease include an 8.5% interest rate (money factor of 0.00354) and

Alex_Xolod [135]

Answer:

hey, the answer is D 443.73

Step-by-step explanation:

3 0

2 years ago

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

8 0

2 years ago