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2 years ago

Can you help me with5 1/4 × 3 7/8

Mathematics

2 answers:

Sholpan [36]2 years ago

4 0

In decimal form the answer is 20.34375

SashulF [63]2 years ago

3 0

19.57
because 1/4 × 7/8=4.57. 5×3=15+4.57=19.57

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What is the mode of the following numbers?<br> 67, 76, 67, 76, 67, 23, 32, 23, 32<br> 0

AleksAgata [21]

Answer:

Step-by-step explanation:

67, 76, 67, 76, 67, 23, 32, 23, 32 ,0

Put the numbers in order from smallest to largest

0,23, 23, 32,32, 67, 67, 67, 76, 76,

The mode is the number that appears most often

67 appears 3 times so it is the mode

5 0

2 years ago

Mr Chua borrows a sum of money from the bank which charges a compound interest of 4.2% per annum, compounded 6.Given that Mr Chu

Inessa [10]

Answer:

Original sum of money = $2246.51

Step-by-step explanation:

Interest = $96.60

Interest is compounded 6 times in a year ; n = 6

time = 1 year ; Rate of interest (r) = 4.2%

Interest = Future Value - Principal Value ...........(1)

$\text{Future Value = }Principal\cdot (1+\frac{r}{100\times n})^{n\cdot t}\\\\\text{Substituting this value in equation (1) , We get }\\\\Interest=Principal\cdot (1+\frac{r}{100\times n})^{n\cdot t}-Principal\\\\\implies 96.60=Principal[\cdot (1+\frac{4.2}{100\times 6})^{6\cdot 1}-1]\\\\\implies Principal=\$2246.51$

Hence, the original sum of money borrowed = $2246.51

4 0

2 years ago

The area of the conference table in Mr. Nathan’s office must be no more than 175 ft2. If the length of the table is 18 ft more t

Usimov [2.4K]

The answer to the question is 300 square feet

4 0

2 years ago

Question:twice as much as the sum of eighteen and eleven<br><br> in numbers and don't solve

Dmitrij [34]

2(18+11)
i think this is the expression you are looking for

3 0

2 years ago

Read 2 more answers

Let P2 be the vector space of all polynomials of degree 2 or less, and let H be the subspace spanned by 10x2+4xâ1, 3xâ4x2+3, and

lord [1]

I suppose

$H=\mathrm{span}\{10x^2+4x-1,3x-4x^2+3,5x^2+x-1\}$

The vectors that span $H$ form a basis for $P_2$ if they are (1) linearly independent and (2) any vector in $P_2$ can be expressed as a linear combination of those vectors (i.e. they span $P_2$ ).

Independence:

Compute the Wronskian determinant:

$\begin{vmatrix}10x^2+4x-1&3x-4x^2+3&5x^2+x-1\\20x+4&3-8x&10x+1\\20&-8&10\end{vmatrix}=-6\neq0$

The determinant is non-zero, so the vectors are linearly independent. For this reason, we also know the dimension of $H$ is 3.

Span:

Write an arbitrary vector in $P_2$ as $ax^2+bx+c$ . Then the given vectors span $P_2$ if there is always a choice of scalars $k_1,k_2,k_3$ such that

$k_1(10x^2+4x-1)+k_2(3x-4x^2+3)+k_3(5x^2+x-1)=ax^2+bx+c$

which is equivalent to the system

$\begin{bmatrix}10&-4&5\\4&3&1\\-1&3&-1\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$

The coefficient matrix is non-singular, so it has an inverse. Multiplying both sides by that inverse gives

$\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}-\dfrac{6a-11b+19c}3\\\dfrac{3a-5b+2c}3\\\dfrac{15a-26b+46c}3\end{bmatrix}$

so the vectors do span $P_2$ .

The vectors comprising $H$ form a basis for it because they are linearly independent.

4 0

2 years ago