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3 years ago

Find the lateral surface area of the can of mandarin oranges. Round your answer to the nearest tenth.

Mathematics

1 answer:

Mkey [24]3 years ago

4 0

Answer:

AL=2πrh

Step - by - step explanation:

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3 years ago

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.

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2 years ago