All categories

4 years ago

Round 43.586 to the nearest tenth. Pleaseee help ASAP

Mathematics

2 answers:

Dahasolnce [82]4 years ago

3 0

Answer:

43.6

Step-by-step explanation:

I hope this helps you :)

den301095 [7]4 years ago

3 0

Answer:

43.59

Step-by-step explanation:

43.59 because it is

You might be interested in

According to the given information, segment uv is parallel to segment wz, while angles squ and vqt are vertical angles. angle vq

svp [43]

Angle vqt is congruent to Angle squ by the vertical angles theorem. because angles squ and wrs are corresponding angles, they are congruent according to the Corresponding angles theorem.

Option b is true.

Here,

Segment uv is parallel to segment wz,

Angles squ and vqt are vertical angles.

Angle vqt is congruent to Angle squ by the Vertical angles theorem.

What is corresponding angles?

Any pair of angles each of which is on the same side of one of two lines cut by a transversal and on the same side of the transversal.

Now,

Angle vqt is congruent to Angle squ by the vertical angles theorem.

When, two angles are corresponding angle then, Angle vqt is congruent to Angle squ by the Vertical angles theorem.

And, they are congruent according to the Corresponding angles theorem.

Hence, Segment uv is parallel to Segment wz, while Angles squ and vqt are Vertical angles. Angle vqt is congruent to Angle squ by the vertical angles theorem. Because angles squ and wrs are corresponding angles, they are congruent according to the corresponding angle theorem. finally, Angle vqt is congruent to Angle wrs by the transitive property of equality.

So, Option b is true.

Learn more about the corresponding angles visit:

brainly.com/question/22190474

#SPJ4

4 0

2 years ago

Precalculus. Please look at the picture

erik [133]

Answer:

<u>Exponential model</u>

$y=Ae^{rt}$

where:

y = value at "t" time
A = initial value
r = rate of growth/decay
t = time (in years)

Given:

y = 100 g
t = 0 years

Substituting given values into the formula and solving for A:

$\begin{aligned}y & =Ae^{rt}\\\implies 100 & = Ae^{r \times 0}\\100 & = Ae^0\\100 & = A(1)\\A & = 100\end{aligned}$

Given:

A = 100 g
y = 50 g when t = 30.17

Substituting the given values into the equation and solving for r:

$\begin{aligned}y& =Ae^{rt}\\\\\implies 50 & =100e^{30.17r}\\\\\dfrac{1}{2} & = e^{30.17r}\\\\ln \dfrac{1}{2} & = \ln e^{30.17r}\\\\\ln 1-\ln2 & =30.17r \ln e\\\\0-\ln 2 & =30.17r(1)\\\\-\ln 2 & =30.17r\\\\r & = \dfrac{-\ln 2}{30.17}\end{aligned}$

Therefore, the final equation is:

$y=100e^{\left(-\dfrac{\ln 2}{30.17}\right)t}$

<h3><u>Question 1</u></h3>

Q: From 100g how much remains in 80 years?

$\begin{aligned}t=80 \implies y & =100e^{\left(-\dfrac{\ln 2}{30.17}\right)80}\\& = 15.91389949 \: \sf g\end{aligned}$

Q: How long will it take to have 10% remaining?

10% of 100 g = 10 g

$\begin{aligned}y=10 \implies 10 & =100e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\dfrac{1}{10} & =e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln \dfrac{1}{10} & =\ln e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln 1 - \ln 10 & =\left(-\dfrac{\ln 2}{30.17}\right)t\ln e\\\\0 - \ln 10 & =\left(-\dfrac{\ln 2}{30.17}\right)t(1)\\\\-\ln 10 & =\left(-\dfrac{\ln 2}{30.17}\right)t\\\\t & = \dfrac{- \ln 10}{\left(-\dfrac{\ln 2}{30.17}\right)}\\\\t & = 100.2225706\: \sf years\end{aligned}$

<h3><u>Question 2</u></h3>

Q: How much remains after 50 years (time)?

<u></u>

$\begin{aligned}t=50 \implies y & =100e^{\left(-\dfrac{\ln 2}{30.17}\right)50}\\& = 31.70373153 \: \sf g\end{aligned}$

Q: How long to reach 20 g (amount remaining)?

<u></u> $\begin{aligned}y=20 \implies 20 & =100e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\dfrac{1}{5} & =e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln \dfrac{1}{5} & =\ln e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln 1 - \ln 5 & =\left(-\dfrac{\ln 2}{30.17}\right)t\ln e\\\\0 - \ln 5 & =\left(-\dfrac{\ln 2}{30.17}\right)t(1)\\\\-\ln 5 & =\left(-\dfrac{\ln 2}{30.17}\right)t\\\\t & = \dfrac{- \ln 5}{\left(-\dfrac{\ln 2}{30.17}\right)}\\\\t & = 70.05257062\: \sf years\end{aligned}$