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3 years ago

Please help ASAP I’m very unsure on how to do these 2 questions please give answers for both :)

Mathematics

2 answers:

Temka [501]3 years ago

8 0

Answer:

1) Option 2

2) Option 1

Step-by-step explanation:

1) 3a-2b

Where a = 3, b = 2

=> 3(3)-2(2)

=> 9-4

=> 5

2) 5c+3d

Where c = 4, d = -5

=> 5(4) + 3(-5)

=> 20 - 15

=> 5

Orlov [11]3 years ago

3 0

Answers:

5, 5

Step-by-step explanation:

Put a as 3 and b as 2 and evaluate:

3(3) - 2(2)

9 - 4

= 5

Put c as 4 and d as -5 and evaluate:

5(4) + 3(-5)

20 - 15

= 5

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Use De Moivre's theorem to express cos 5θ and sin 5θ in terms of sin θ and cos θ.

Sonja [21]

From the de Moivre's we have,

(cosθ+isinθ)^n=cos(nθ)+isin(nθ)

Therefore,

R((cosθ+isinθ)^5)=cos(5θ)I((cosθ+isinθ)^5)=sin(5θ)

Simplifying,

cos^5(θ)−10(sin^2(θ))(cos^3(θ))+5(sin^4(θ))(cosθ)=cos(5θ)

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3 years ago

And right triangle ABC shown, what is the length of AC?

Vera_Pavlovna [14]

Simple pythagorus theorem with the equation a^2=b^2+c^2
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4 0

2 years ago

The length of a parasite in experiment A is 2.5 × 10–3 inch. The length of a parasite in experiment B is 1.25 × 10–4 inch. How m

torisob [31]

We have been given that :-

The length of a parasite in experiment A is $2.5 \times 10^{-3}$

The length of a parasite in experiment B is $1.25 \times 10^{-4}$

Let us write the the length of the parasite in experiment A in the exponent of -3.

$1.25 \times 10^{-4}= 0.125 \times 10^{-3}$

Clearly, the length of parasite in experiment A is greater than the length of parasite in experiment B.

The difference in the length is given by

$2.5 \times 10^{-3} -0.125 \times 10^{-4}$

$=2.375 \times 10^{-3}$

Therefore, the length of the parasite in experiment A is $=2.375 \times 10^{-3}$ inches greater than the length of the parasite in experiment B.

3 0

3 years ago

Read 2 more answers

IF A SUBSCRIPTION IS $499 PLUS 8% TAX FOR 30 DAYS, BUT IS BEING PRORATED FOR 7 DAYS, PLUS THERE IS A $10 OFF COUPON, WHAT'S THE

ELEN [110]

Answer:

Your final answer is $528.92

Step-by-step explanation:

$499 + 8\% = 538.92$

$538.92 - 10 = 528.92$

5 0

3 years ago

Read 2 more answers

(I've been trying to figure this out for 3 days and I really need help)

liq [111]

Check the picture below.

since the diameter of the cone is 6", then its radius is half that or 3", so getting the volume of only the cone, not the top.

$\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=4 \end{cases}\implies V=\cfrac{\pi (3)^2(4)}{3}\implies V=12\pi \implies V\approx 37.7$

now, the top of it, as you notice in the picture, is a semicircle, whose radius is the same as the cone's, 3.

$\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=3 \end{cases}\implies V=\cfrac{4\pi (3)^3}{3}\implies V=36\pi \\\\\\ \stackrel{\textit{half of that for a semisphere}}{V=18\pi }\implies V\approx 56.55$

well, you'll be serving the cone and top combined, 12π + 18π = 30π or about 94.25 in³.

pretty much the same thing, we get the volume of the cone and its top, add them up.

$\bf \stackrel{\textit{cone's volume}}{\cfrac{\pi (3)^2(8)}{3}}~~~~+~~~~\stackrel{\stackrel{\textit{half a sphere}}{\textit{top's volume}}}{\cfrac{4\pi 3^3}{3}\div 2}\implies 24\pi +18\pi \implies 42\pi ~~\approx~~131.95~in^$

8 0

3 years ago