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AMNP is an isosceles triangle with |MN| = 26 cm and |NP| = 20 cm. Find the area of the triangle.

Mathematics

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Answer

D

Step-by-step solution

16 steps

Here are the solutions to the questions:

Part 1· 3 steps

An isosceles triangle MNPMNP has sides MN=MP=26MN = MP = 26 cm and NP=20NP = 20 cm. To find the area, we first need to find the height.

  1. Draw an altitude from MM to the base NPNP.

    Let the point where the altitude meets NPNP be QQ.
    In an isosceles triangle, the altitude to the base bisects the base. So, NQ=QP=NP2=202=10NQ = QP = \frac{NP}{2} = \frac{20}{2} = 10 cm.

  2. Use the Pythagorean theorem in the right-angled triangle △MQN\triangle MQN to find the height MQMQ.

    MN2=MQ2+NQ2MN^2 = MQ^2 + NQ^2 262=MQ2+10226^2 = MQ^2 + 10^2 676=MQ2+100676 = MQ^2 + 100 MQ2=676−100MQ^2 = 676 - 100 MQ2=576MQ^2 = 576 MQ=576MQ = \sqrt{576} MQ=24 cmMQ = 24 \text{ cm}

    The height of the triangle is 24 cm.

  3. Calculate the area of the triangle.

    Area=12×base×height\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} Area=12×NP×MQ\text{Area} = \frac{1}{2} \times NP \times MQ Area=12×20 cm×24 cm\text{Area} = \frac{1}{2} \times 20 \text{ cm} \times 24 \text{ cm} Area=10 cm×24 cm\text{Area} = 10 \text{ cm} \times 24 \text{ cm} Area=240 cm2\text{Area} = 240 \text{ cm}^2

    The correct option is D.

    D. 240 cm2\boxed{\text{D. } 240 \text{ cm}^2}

Part 2· 3 steps

Given that (x−2)(x-2) is a factor of 8−2qx−x28 - 2qx - x^2, we need to find the value of qq.

  1. Apply the Factor Theorem.

    If (x−2)(x-2) is a factor of a polynomial P(x)P(x), then P(2)=0P(2) = 0.
    Let P(x)=8−2qx−x2P(x) = 8 - 2qx - x^2.

  2. Substitute x=2x=2 into the polynomial and set the expression to zero.

    P(2)=8−2q(2)−(2)2=0P(2) = 8 - 2q(2) - (2)^2 = 0 8−4q−4=08 - 4q - 4 = 0

  3. Solve for qq.

    4−4q=04 - 4q = 0 4=4q4 = 4q q=44q = \frac{4}{4} q=1q = 1

    The correct option is D.

    D. 1\boxed{\text{D. } 1}

Part 3· 3 steps

Write 6.02517 to three significant figures.

  1. Identify the first three significant figures.

    The first significant figure is 6.
    The second significant figure is 0.
    The third significant figure is 2.

  2. Look at the fourth significant figure to determine rounding.

    The fourth significant figure is 5. Since it is 5 or greater, we round up the third significant figure.
    The third significant figure (2) becomes 3.

  3. Write the number rounded to three significant figures.

    6.036.03

    The correct option is D.

    D. 6.03\boxed{\text{D. } 6.03}

Part 4· 3 steps

Factorize completely (h2−k2)−p(h+k)(h^2 - k^2) - p(h+k).

  1. Recognize the difference of squares pattern in the first term.

    h2−k2=(h−k)(h+k)h^2 - k^2 = (h-k)(h+k)

  2. Substitute this into the given expression.

    (h−k)(h+k)−p(h+k)(h-k)(h+k) - p(h+k)

  3. Factor out the common term (h+k)(h+k).

    (h+k)[(h−k)−p](h+k)[(h-k) - p] (h+k)(h−k−p)(h+k)(h-k-p)

    The correct option is D.

    D. (h+k)(h−k−p)\boxed{\text{D. } (h+k)(h-k-p)}

Part 5· 4 steps

Solve the compound inequality 2x−3≤4x+5<3x+72x - 3 \le 4x + 5 < 3x + 7.

  1. Split the compound inequality into two separate inequalities.

    Inequality 1: 2x−3≤4x+52x - 3 \le 4x + 5
    Inequality 2: 4x+5<3x+74x + 5 < 3x + 7

  2. Solve Inequality 1.

    2x−3≤4x+52x - 3 \le 4x + 5

    Subtract 2x2x from both sides:

    −3≤2x+5-3 \le 2x + 5

    Subtract 5 from both sides:

    −3−5≤2x-3 - 5 \le 2x −8≤2x-8 \le 2x

    Divide by 2:

    −4≤xorx≥−4-4 \le x \quad \text{or} \quad x \ge -4

  3. Solve Inequality 2.

    4x+5<3x+74x + 5 < 3x + 7

    Subtract 3x3x from both sides:

    x+5<7x + 5 < 7

    Subtract 5 from both sides:

    x<7−5x < 7 - 5 x<2x < 2

  4. Combine the solutions from both inequalities.

    We have x≥−4x \ge -4 and x<2x < 2.
    Combining these gives:

    −4≤x<2-4 \le x < 2

    The correct option is D.

    D. −4≤x<2\boxed{\text{D. } -4 \le x < 2}
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