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3 years ago

Find the value of each variable using sine and cosine. Round your answers to the nearest tenth.

Mathematics

1 answer:

Alisiya [41]3 years ago

5 0

Answer:

x ≈ 9.5 , y ≈ 15.3

Step-by-step explanation:

sin32° = $\frac{opposite}{hypotenuse}$ = $\frac{x}{18}$ ( multiply both sides by 18 )

18 × sin32° = x , then

x ≈ 9.5 ( to the nearest tenth )

cos32° = $\frac{adjacent}{hypotenuse}$ = $\frac{y}{18}$ ( multiply both sides by 18 )

18 × cos32° = y , then

y ≈ 15.3 ( to the nearest tenth )

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34kurt

Answer:

Step-by-step explanation:

1 yard = 3 feet

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6yd 1ft = (6)(3)ft + 1ft = 19ft

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Substitute w = 19ft and l = 17ft:

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3 years ago

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2 years ago

What is the probability that there is at most one purple marble? 0.100 0.301 0.770 0.971

andriy [413]

The probability that there is at most one purple marble is 0.9701.

<h3>What is the probability about?</h3>

Probability is a term that connote the possibility of an event or something occurring or not.

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Based on the fact that there is a bag that has eight yellow marbles, nine green marbles, three purple marbles, and five red marbles, the probability that there is at most one purple marble is 0.9701.

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2 years ago

Please help me !! find the x intercept and the y intercept

klemol [59]

X intercept: when y = 0
y intercept: when x = 0
Plug info inside the equation
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2 years ago

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A square pyramid has a base with an area of 20 square meters, and its lateral faces have a slant height of x meters. Sydney is c

svetoff [14.1K]

1. Consider the pyramid ABCDE in the figure 1 attached:

Each side of the square base equals $\sqrt{20} = \sqrt{4*5} = \sqrt{4}* \sqrt{5}=2 \sqrt{5}$ (units)

Area of 1 lateral side of pyramid ABCDE is $\frac{1}{2}* 2 \sqrt{5}*x=\sqrt{5}x$

so the total lateral area is $4\sqrt{5}x$ (units squared)

and the total surface area is $20+4\sqrt{5}x$ (units squared)

2. Now consider the second figure KLMNP, Sydney's pyramid:

The lateral area is $4* \frac{1}{2}* 4 \sqrt{5} *2x=16 \sqrt{5}x$

so the total area of the second pyramid is $20+16 \sqrt{5}x$

So 2 things can be said about the areas:

i) the lateral area of Sydney's pyramid is $\frac{16 \sqrt{5}x}{4\sqrt{5}x} =4$ times larger than the lateral area of the original pyramid.

ii) The total area of Sydney's pyramid is $(20+16 \sqrt{5}x)-(20+4\sqrt{5}x)=12\sqrt{5}x$ units squared larger than the total area of the original pyramid.

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3 years ago

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