We read your photo as:

A trader mixes 350kg of beans type A costing sh 84 per Kg with 140kg of beans type B costing sh 105 per Kg. Calculate the selling price per Kg of a mixture if the trader makes a profit of 25%. 16. Solve for X in the equation 10 Sin2x - 7 Cos X = 2 for 270 <= X <= 360.

Mathematics

ScanSolve AI Answer

Answer

sh 112.50\text{sh } 112.50

Step-by-step solution

9 steps

Hapa kuna suluhisho la maswali yako:

15.

  1. Kokotoa gharama ya maharage ya aina A na B.

    Gharama ya maharage aina A = 350 kg×sh 84/kg=sh 29400350 \text{ kg} \times \text{sh } 84/\text{kg} = \text{sh } 29400.
    Gharama ya maharage aina B = 140 kg×sh 105/kg=sh 14700140 \text{ kg} \times \text{sh } 105/\text{kg} = \text{sh } 14700.

  2. Kokotoa jumla ya gharama na jumla ya uzito wa mchanganyiko.

    Jumla ya gharama = sh 29400+sh 14700=sh 44100\text{sh } 29400 + \text{sh } 14700 = \text{sh } 44100.
    Jumla ya uzito = 350 kg+140 kg=490 kg350 \text{ kg} + 140 \text{ kg} = 490 \text{ kg}.

  3. Kokotoa bei ya gharama kwa Kg ya mchanganyiko.

    Bei ya gharama kwa Kg = sh 44100490 kg=sh 90/kg\frac{\text{sh } 44100}{490 \text{ kg}} = \text{sh } 90/\text{kg}.

  4. Kokotoa bei ya kuuza kwa Kg ili kupata faida ya 25%.

    Bei ya kuuza kwa Kg = Bei ya gharama kwa Kg ×(1+asilimia ya faida)\times (1 + \text{asilimia ya faida})
    Bei ya kuuza kwa Kg = sh 90×(1+0.25)\text{sh } 90 \times (1 + 0.25)
    Bei ya kuuza kwa Kg = sh 90×1.25=sh 112.50\text{sh } 90 \times 1.25 = \text{sh } 112.50.
    Bei ya kuuza kwa Kg ya mchanganyiko ni sh 112.50\boxed{\text{sh } 112.50}.

    16.

  5. Tumia utambulisho wa pembe mbili kwa sin⁡(2x)\sin(2x).

    Tunajua kuwa sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2 \sin(x) \cos(x). Badilisha hii kwenye mlinganyo:

    10(2sin⁡(x)cos⁡(x))−7cos⁡(x)=210 (2 \sin(x) \cos(x)) - 7 \cos(x) = 2 20sin⁡(x)cos⁡(x)−7cos⁡(x)=220 \sin(x) \cos(x) - 7 \cos(x) = 2
  6. Panga upya mlinganyo na toa cos⁡(x)\cos(x) kama kigezo cha kawaida.

    cos⁡(x)(20sin⁡(x)−7)=2\cos(x) (20 \sin(x) - 7) = 2
  7. Chunguza ishara za cos⁡(x)\cos(x) na sin⁡(x)\sin(x) katika safu iliyotolewa.

    Safu iliyotolewa ni 270∘≤x≤360∘270^\circ \le x \le 360^\circ, ambayo ni roboduara ya nne.
    Katika roboduara ya nne:

    • cos⁡(x)\cos(x) ni chanya (>0> 0).
    • sin⁡(x)\sin(x) ni hasi (<0< 0).
  8. Tathmini ishara ya upande wa kushoto wa mlinganyo.

    Kwa kuwa sin⁡(x)<0\sin(x) < 0, basi 20sin⁡(x)20 \sin(x) ni hasi.
    Hii inamaanisha 20sin⁡(x)−720 \sin(x) - 7 pia ni hasi.
    Kwa hivyo, upande wa kushoto wa mlinganyo ni bidhaa ya nambari chanya (cos⁡(x)\cos(x)) na nambari hasi (20sin⁡(x)−720 \sin(x) - 7).
    Bidhaa ya nambari chanya na nambari hasi daima ni nambari hasi.
    Hivyo, cos⁡(x)(20sin⁡(x)−7)<0\cos(x) (20 \sin(x) - 7) < 0.

  9. Linganisha na upande wa kulia wa mlinganyo.

    Upande wa kulia wa mlinganyo ni 2, ambayo ni nambari chanya.
    Kwa hivyo, tunayo:

    (nambari hasi)=(nambari chanya)(\text{nambari hasi}) = (\text{nambari chanya})

    Huu ni utata, kwani nambari hasi haiwezi kuwa sawa na nambari chanya.
    Kwa hivyo, hakuna suluhisho la xx katika safu 270∘≤x≤360∘270^\circ \le x \le 360^\circ linalokidhi mlinganyo huu.
    Hakuna suluhisho halisi kwa X katika safu iliyotolewa.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp