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2 years ago

Dan made $5000 in taxable income last year.

1 answer:

8 0

well, we know is 10% for the first 7500, but Dan only made 5000 of taxable income, so he's in that range of 0 - 7500, so he only gets to pay 10% of 5000.

$\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{10\% of 5000}}{\left( \cfrac{10}{100} \right)5000}\implies 500$

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(01.02 LC)Simplify 9 − (−4).

levacccp [35]

Answer:

Step-by-step explanation:

When we subtract a negative, it is like adding

9 - -4

9 +4

3 0

3 years ago

Read 2 more answers

Here's the question.

OleMash [197]

Answer:

The value of T₂₀ - T₁₅ is -20.

Step-by-step explanation:

Given :

>> If for an A.P, d = -4

To Find :

>> T₂₀ - T₁₅

Using Formula :

General term of an A.P.

$\star{\small{\underline{\boxed{\sf{\red{ T_n = a + (n - 1)d}}}}}}$

>> Tₙ = nᵗʰ term
>> a = first term
>> n = no. of terms
>> d = common difference

Solution :

Firstly finding the A.P of T₂₀ by substituting the values in the formula :

${\dashrightarrow{\pmb{\sf{ T_n = a + (n - 1)d}}}}$

${\dashrightarrow{\sf{ T_{20} = a + (20 - 1) d}}}$

${\dashrightarrow{\sf{ T_{20} = a + (19)d}}}$

${\dashrightarrow{\sf{ T_{20} = a + 19 \times d}}}$

${\dashrightarrow{\sf{ T_{20} = a + 19d}}}$

${\star \: {\underline{\boxed{\sf{\pink{ T_{20} = a + 19d}}}}}}$

Hence, the value of T₂₀ is a + 19d.

$\rule{190}1$

Secondly, finding the A.P of T₁₅ by substituting the values in the formula :

${\dashrightarrow{\pmb{\sf{ T_n = a + (n - 1)d}}}}$

${\dashrightarrow{\sf{ T_{15}= a + (15 - 1) d}}}$

${\dashrightarrow{\sf{ T_{15}= a + (14) d}}}$

${\dashrightarrow{\sf{ T_{15}= a + 14 \times d}}}$

${\dashrightarrow{\sf{ T_{15}= a + 14d}}}$

${\star{\underline{\boxed{\sf \pink{ T_{15}= a + 14d}}}}}$

Hence, the value of T₁₅ is a + 14d

$\rule{190}1$

Now, finding the difference between T₂₀ - T₁₅ :

${\dashrightarrow{\pmb{\sf{T_{20} - T_{15}}}}}$

${\dashrightarrow{\sf{(a + 19d) - (a + 14d)}}}$

${\dashrightarrow{\sf{a + 19d - a - 14d}}}$

${\dashrightarrow{\sf{a - a + 19d - 14d}}}$

${\dashrightarrow{\sf{0+ 19d - 14d}}}$

${\dashrightarrow{\sf{19d - 14d}}}$

${\dashrightarrow{\sf{5 \times - 4}}}$

${\dashrightarrow{\sf{ - 20}}}$

${\star \: \underline{\boxed{\sf{\pink{T_{20} - T_{15} = - 20}}}}}$

Hence, the value of T₂₀ - T₁₅ is -20.

$\underline{\rule{220pt}{3.5pt}}$

3 0

2 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$