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2 years ago

Please Help I have an exam due TODAY And 15 brainly points!

Mathematics

1 answer:

Dmitry_Shevchenko [17]2 years ago

4 0

Answer:

whats difference

Step-by-step explanation:

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Can anyone do 4,5 and 6 for me plz

cricket20 [7]

For question 4, $sinA =\frac{a}{c}, cosA= \frac{b}{c}, tan B = \frac{b}{a}, sin J = \frac{j}{l}, cosK = \frac{j}{l}, tanK = \frac{k}{j}.$

Question 5. Option a and question 6. Option j

Step-by-step explanation:

Step 1:

The three basic formula needed to solve these questions are:

$sin\theta = \frac{oppositeside}{hypotenuse} , cos\theta = \frac{adjacentside}{hypotenuse}, tan\theta= \frac{opposite side}{adjacent side}.$

Step 2:

Using the above formula, we solve the following values

$sinA = \frac{oppositeside}{hypotenuse}$ $=\frac{a}{c}.$

$cosA = \frac{adjacentside}{hypotenuse}$ $= \frac{b}{c}.$

$tanB= \frac{opposite side}{adjacent side}$ $= \frac{b}{a}.$

$sinJ = \frac{oppositeside}{hypotenuse}$ $= \frac{j}{l}.$

$cosK = \frac{adjacentside}{hypotenuse}$ $= \frac{j}{l}.$

$tanK= \frac{opposite side}{adjacent side}$ $= \frac{k}{j}.$

Step 3:

For question 5, The triangle's angle = 23°, opposite side = BC inches and hypotenuse = 4 inches.

$sin\theta= \frac{opposite side}{hypotenuse}. sin 23^{\circ}= \frac{BC}{4}, sin23^{\circ} = 0.3907,BC = (0.3907)(4) = 1.5628.$

SO BC is 1.5628 inches, rounding this off to the nearest tenth, we get BC = 1.6 inches which is option a.

Step 4:

For question 6, The triangle's angle = 50°, opposite side = QR m the adjacent side = 8.1 m.

$tan\theta= \frac{opposite side}{adjacentside}. tan 50^{\circ}=\frac{QR}{8.1}, tan50^{\circ} = 1.1917,QR = (1.1917)(8.1) = 9.65277$

SO QR is 9.65277 meters, rounding this off to the nearest tenth, we get QR = 9.7 inches which is option j.

7 0

3 years ago

George Kyparisis makes bowling ball in his Miami plant. With recent increases in his cost, he has a newfound interest in efficie

Harlamova29_29 [7]

We compute the productivity percentage change of each category using the formula:

$\text{\% Change }=\frac{\text{ Value now - Value last year}}{\text{Value last year}}\cdot100\%.$

Productivity percentage change for each category

1) Units produced:

$\frac{1,000-1,000}{1,000}\cdot100\%=0\cdot100\%=0\%.$

2) Last (hours)

$\frac{275-300}{300}\cdot100\%=-\frac{25}{300}\cdot100\%=-8.33\%.$

3) Resin (pounded)

$\frac{45-50}{50}\cdot100\%=-\frac{5}{50}\cdot100\%=-10\%.$

4) Capital invested ($)

$\frac{11,000-10,000}{10,000}\cdot100\%=\frac{1,000}{10,000}\cdot100\%=10\%.$

5) Energy (BTU)

$\frac{2,850-3,000}{3,000}\cdot100\%=\frac{-150}{3,000}\cdot100\%=-5\%.$

2) Improvement for labour hours

From the previous computations, we see that the percentage change of the category "Last (hours)" has a negative value of -8.33%, which means that this value decreased by 8.33% with respect to the last year.

Answer

1) Productivity percentage change for each category

0. Units produced: ,0%

1. Last (hours): , -8.33%

2. Resin (pounded):, -10%

3. Capital invested ($): ,10%

4. Energy (BTU): ,-5%

2) Improvement for labour hours

The percentage change of the category "Last (hours)" has a negative value of -8.33%, which means that this value decreased by 8.33% with respect to the last year.

4 0

2 years ago

Pls help with explanation

skad [1K]

Answer:

what i would do is just divide each of the fractions until you get 0.6.

for example if you divide 2/3 you get 0.6

which is your answer, or B

Step-by-step explanation:

4 0

3 years ago

PLEASE HELP

Murrr4er [49]

75 divided by 3 because for every 75 fish 3 are tagged 1,217 (rounded up from 1,216.6 repeating)

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2 years ago

∆ADB ≅ ∆CDB by the _____

aleksley [76]

The marked angles and side BD make up two adjacent angles and a side not between them. The applicable theorem is ...
AAS Theorem

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4 years ago

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