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3 years ago

Pls help with question.

Mathematics

1 answer:

BigorU [14]3 years ago

4 0

Answer:

4a) 4 1/6

4b) 2

5a) 36.424

5b) 5.4

Step-by-step explanation:

When multiplying or dividing by mixed numbers, convert them into improper and just do it from left to right for each numerator and denominator

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What is the equation of the line, in general form, that passes through the point (-1, -1) and is parallel to the line whose equa

Ksenya-84 [330]

Answer:

Step-by-step explanation:

y = -x + 3

y + 1 = -(x + 1)

y + 1 = -x - 1

y = -x - 2

x + y + 2 = 0 is the solution

5 0

3 years ago

(b) Ten men can assemble 400 cycles in 8 days. How many cycles 5 men will

Deffense [45]

Answer:

400

Step-by-step explanation

First, find how many cycles 5 man can make in 8 days.

400/2=200 (/ means divide)

Next, mutliply the answer to find out how many cycles 5 man can make in 16 days.

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8 0

4 years ago

An animal shelter spends $3.00 per day to care for each cat and $6.50 per day to care for each dog. Amelia noticed that the shel

Ede4ka [16]

Answer:

There were 15 cats at the shelter on Friday.

Step-by-step explanation:

Cost of caring each cat = $3.00

Cost for caring each dog = $6.50

Total money spent by shelter = $84.00

Total Cats and Dogs in the shelter = 21

Let the number of cats in the shelter = k

So, the number of dogs in shelter = 21 -k

Also according to the question:

Total expenditure on cats + total expenditure on dogs = $84.00

or, k ($3.00) + (21-k)($6.50) = $84.00

or, 3k + 136.5 - 6.5 k = 84

or, -3.5 k = 84 - 136.5 = -52.5

⇒3.5 k = 52.5 or k = 52.5/3.5 = 15

⇒ k = 15

Hence, the number of cats in the shelter was 15 on Friday.

7 0

3 years ago

HELP FAST!!! WILL UPVOTE! What is the slope of this line? Enter your answer in the box.

alexandr402 [8]

The answer is zero there is no slope

3 0

3 years ago

Please help if you can pic attatched

MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

$CI=\left[\begin{array}{ccc}1&6&0\\0&1&2\\1&-1&3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

Subtract row 3 from row 1:

$\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&7&-3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\1&0&-1\end{array}\right]$

Subtract row 3 from 7 times row 2:

$\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&17\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-1&7&1\end{array}\right]$

Divide row 3 by 17:

$\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]$

Subtract 2 of row 3 from row 2:

$\left[\begin{array}{ccc}1&6&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]$

Subtract 6 of row 2 from row 1:

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}\frac{5}{17}&\frac{-18}{17}&\frac{12}{17}\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]$

$C^{-1}=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]$

$C^{-1}b=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]\left[\begin{array}{c}10&1&3\end{array}\right]=\left[\begin{array}{c}4&1&0\end{array}\right]$

3 0

3 years ago