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3 years ago

How do you solve this?

Mathematics

1 answer:

posledela3 years ago

4 0

Notice that the given info consists of boundary lines for an area in the xy plane. We are interested ONLY in values of x and y that are 0 or greater (positive). Graph 5x + 3y $\leq$ 37 and 3x + 5y less than or equal to 35.

Find the points of intersection of all four straight lines (including the x- and y-axes). There will be 4 such points (incl. the origin).

Next, evaluate the objective function 2x + 14y at each of these 4 points. Which of the four results is the largest? the smallest? Label them as 'maximum' and 'minimum.'

Questions? Just comment on this discussion.

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Multiply: (2x2 − 5x)(4x2 − 12x + 10)

11Alexandr11 [23.1K]

(2x^2-5x)(4x^2-12x+10) = 8x^4 - 44x^3 + 80x^2 - 50x

So, the answer is option C.

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3 years ago

What are the first and second steps performed in solving the inequality 2(4w + 8) – 2w + 42 > 78? Name the property used.

Phoenix [80]

Hello :
2(4w + 8) – 2w + 42 > 78
8w+16-2w+42 > 78......(distributivity × for + )
6w > 20
w > 20/6

6 0

3 years ago

Determine which side is the smallest in the diagram

AnnyKZ [126]

the shortest side is always opposite the smallest angle.

In triangle ABC the shortest side is AC

In triangle ADC the shortest side is DC

DC<AC since they are in the same triangle.

5 0

3 years ago

Simplify 7y-4y-5y Thanks :)!

Jobisdone [24]

7y-4y-5y
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7 0

2 years ago

Read 2 more answers

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

Step-by-step explanation:

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

2 years ago