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3 years ago

A cow ate 38 pounds of grass, the cow now weighs 394 pounds. How much would the cow weigh if it did not eat the grass?

Mathematics

2 answers:

MrRa [10]3 years ago

7 0

The cow before eating will weigh 352 pounds

frutty [35]3 years ago

6 0

Since it ate 38 pounds so just minus 394 by 38 so you get the weight before it eats. the answer should be 356 pounds

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3 years ago

Write an equation for the direct variation (y varies directly with x) if y=12 when x=4.

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3 years ago

Find the polynomial equation of least degree with roots -1, 3, and (+/-)3i

jasenka [17]

Each of these roots can be expressed as a binomial:

(x+1)=0, which solves to -1
(x-3)=0, which solves to 3
(x-3i)=0 which solves to 3i
(x+3i)=0, which solves to -3i
There are four roots, so our final equation will have x^4 as the least degree

Multiply them together. I'll multiply the i binomials first:
(x-3i)(x+3i) = x²+3ix-3ix-9i²
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x²+9 [since i²=-1]

Now I'll multiply the first two binomials together:
(x+1)(x-3) = x²-3x+x-3
x²-2x-3
Lastly, we'll multiply the two derived terms together:

(x²+9)(x²-2x-3) [from the binomial, I'll distribute the first term, then the second term, and I'll stack them so we can simply add like terms together]

x^4 -2x³-3x²
<u> +9x²-18x-27</u>
x^4-2x³+6x²-18x-27

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3 years ago

How to solve and graph y>-1 and y>3

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Open circle on (0,-1) and arrow going strait up

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4 0

4 years ago

State the number of possible triangles that can be formed using the given measurements.

romanna [79]

Answer: 39) 1 40) 2

41) 1 42) 0

<u>Step-by-step explanation:</u>

39) ∠A = ? ∠B = ? ∠C = 129°

a = ? b = 15 c = 45

Use Law of Sines to find ∠B:

$\dfrac{\sin B}{b}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin B}{15}=\dfrac{\sin 129}{45}\rightarrow \quad \angle B=15^o\quad or \quad \angle B=165^o$

If ∠B = 15°, then ∠A = 180° - (15° + 129°) = 36°

If ∠B = 165°, then ∠A = 180° - (165° + 129°) = -114°

Since ∠A cannot be negative then ∠B ≠ 165°

∠A = 36° ∠B = 15° ∠C = 129° is the only valid solution.

40) ∠A = 16° ∠B = ? ∠C = ?

a = 15 b = ? c = 19

Use Law of Sines to find ∠C:

$\dfrac{\sin A}{a}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin 16}{15}=\dfrac{\sin C}{19}\rightarrow \quad \angle C=20^o\quad or \quad \angle C=160^o$

If ∠C = 20°, then ∠B = 180° - (16° + 20°) = 144°

If ∠C = 160°, then ∠B = 180° - (16° + 160°) = 4°

Both result with ∠B as a positive number so both are valid solutions.

Solution 1: ∠A = 16° ∠B = 144° ∠C = 20°

Solution 2: ∠A = 16° ∠B = 4° ∠C = 160°

41) ∠A = ? ∠B = 75° ∠C = ?

a = 7 b = 30 c = ?

Use Law of Sines to find ∠A:

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{7}=\dfrac{\sin 75}{30}\rightarrow \quad \angle A=13^o\quad or \quad \angle A=167^o$

If ∠A = 13°, then ∠C = 180° - (13° + 75°) = 92°

If ∠A = 167°, then ∠C = 180° - (167° + 75°) = -62°

Since ∠C cannot be negative then ∠A ≠ 167°

∠A = 13° ∠B = 75° ∠C = 92° is the only valid solution.

42) ∠A = ? ∠B = 119° ∠C = ?

a = 34 b = 34 c = ?

Use Law of Sines to find ∠A:

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{34}=\dfrac{\sin 119}{34}\rightarrow \quad \angle A=61^o\quad or \quad \angle A=119^o$

If ∠A = 61°, then ∠C = 180° - (61° + 119°) = 0°

If ∠A = 119°, then ∠C = 180° - (119° + 119°) = -58°

Since ∠C cannot be zero or negative then ∠A ≠ 61° and ∠A ≠ 119°

There are no valid solutions.

6 0

3 years ago