All categories

2 years ago

Joseph used base 10 blocks to find 2.1÷3 he stated that 2.1÷3 = 0.6 find his mistake and corrected. Please help!

Mathematics

1 answer:

garri49 [273]2 years ago

7 0

Answer:

Its 0.7

Step-by-step explanation:

You might be interested in

Let h(x)=20e^kx where k ɛ R (Picture attached. Thank you so much!)

zloy xaker [14]

Answer:

$k=0$

$\displaystyle \begin{aligned} 2k + 1& = 2\ln 20 + 1 \\ &\approx 2.3863\end{aligned}$

$\displaystyle \begin{aligned} k - 3&= \ln \frac{1}{2} - 3 \\ &\approx-3.6931 \end{aligned}$

Step-by-step explanation:

We are given the function:

$\displaystyle h(x) = 20e^{kx} \text{ where } k \in \mathbb{R}$

Given that h(1) = 20, we want to find k.

h(1) = 20 means that h(x) = 20 when x = 1. Substitute:

$\displaystyle (20) = 20e^{k(1)}$

Simplify:

$1= e^k$

Anything raised to zero (except for zero) is one. Therefore:

$k=0$

Given that h(1) = 40, we want to find 2k + 1.

Likewise, this means that h(x) = 40 when x = 1. Substitute:

$\displaystyle (40) = 20e^{k(1)}$

Simplify:

$\displaystyle 2 = e^{k}$

We can take the natural log of both sides:

$\displaystyle \ln 2 = \underbrace{k\ln e}_{\ln a^b = b\ln a}$

By definition, ln(e) = 1. Hence:

$\displaystyle k = \ln 2$

Therefore:

$2k+1 = 2\ln 2+ 1 \approx 2.3863$

Given that h(1) = 10, we want to find k - 3.

Again, this meas that h(x) = 10 when x = 1. Substitute:

$\displaystyle (10) = 20e^{k(1)}$

Simplfy:

$\displaystyle \frac{1}{2} = e^k$

Take the natural log of both sides:

$\displaystyle \ln \frac{1}{2} = k\ln e$

Therefore:

$\displaystyle k = \ln \frac{1}{2}$

Therefore:

$\displaystyle k - 3 = \ln\frac{1}{2} - 3\approx-3.6931$

3 0

2 years ago

What fraction is equivalent to 0.997

Katena32 [7]

It’s is 997/1000. Just put the number over the highest decimal place

8 0

2 years ago

Is negative 1 4/5 greater than negative 1 7/10

Yakvenalex [24]

Answer:

Step-by-step explanation:

1 4/5 = 1 8/5 if you make a common denominator

3 0

2 years ago

If a dice is thrown, then choose the correct statement from below.

olasank [31]

Where is the question

7 0

3 years ago

Amanda and Stephen wrote the following proofs to prove that vertical angles are congruent. Who is correct? Line segment NT inter

jolli1 [7]

In fact, both Amanda's and Stephen's profs are correct; they are just using different supplementary angles. Amanda's is using the supplementary angles ∠1 and ∠4, and ∠3 and ∠4, whereas Stephen is using ∠1 and ∠2, and ∠2 and ∠3. Please check the picture to visualize this more effectively.

6 0

3 years ago

Read 2 more answers