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Alina [70]
2 years ago
6

Martha bought a rectangular piece of hand-made paper of area 120 square inches to wrap a gift. Later she found she needed a foil

of exactly double the dimensions to wrap the gift. What is the area of the foil she needs?
A. 260
B. 480
C. 550
D. 640
Mathematics
1 answer:
White raven [17]2 years ago
8 0

Answer:

B. 480

Step-by-step explanation:

Prime Example[s]:

2[1]2[120] → [2][240] >> 480

2[5]2[24] → [10][48] >> 480

I am joyous to assist you anytime.

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Suppose an opaque jar contains 3 red marbles and 10 green marbles. The following exercise refers to the experiment of picking tw
kow [346]

Answer:

\frac{5}{13}

Step-by-step explanation:

Given,

Red marbles = 3,

Green marbles = 10,

So, the total marbles = 3 + 10 = 13,

\because \text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}

Since, here replacement is not allowed,

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= first red and second green + first green second red

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Note : The probability of getting a green marble first and a red marble second

= \frac{10}{13}\times \frac{3}{12}

=\frac{30}{156}

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4 0
3 years ago
|. Identify the following Pōints of each values.Write your ans
Dmitry_Shevchenko [17]
<h2>✒️VALUE</h2>

\\ \quad  \begin{array}{c} \qquad \bold{Distance \: \green{ Formula:}}\qquad\\ \\ \boldsymbol{ \tt d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}} \end{array}\\  \begin{array}{l} \\ 1.)\: \bold{Given:}\: \begin{cases}\tt D(- 5,6), E(2.-1),\textsf{ and }F(x,0) \\ \tt DF = EF \end{cases} \\ \\  \qquad\bold{Required:}\:\textsf{ value of }x \\ \\ \qquad \textsf{Solving for }x, \\ \\  \tt  \qquad DF = EF \\ \\  \implies\small \tt{\sqrt{(x -(- 5))^2 + (0 - 6)^2} = \sqrt{(x - 2)^2 + (0 - (-1))^2}} \\ \\   \implies\tt\sqrt{(x + 5)^2 + 36 } = \sqrt{(x - 2)^2 + 1 } \\ \\ \textsf{Squaring both sides yields} \\ \\  \implies\tt (x + 5)^2 + 36 = (x - 2)^2 + 1 \\ \\  \implies\tt x^2 + 10x + 25 + 36 = x^2 - 4x + 4 + 1 \\ \\ \implies \tt x^2 + 10x + 61 = x^2 - 4x + 5 \\ \\   \implies\tt10x + 4x = 5 - 61 \\ \\   \implies\tt14x = -56 \\ \\  \implies \red{\boxed{\tt x = -4}}\end{array}  \\  \\  \\  \\\begin{array}{l} \\ 2.)\: \bold{Given:}\: \begin{cases}\tt P(6,-1), Q(-4,-3),\textsf{ and }R(0,y) \\ \tt PR = QR \end{cases} \\ \\ \bold{Required:}\:\textsf{ value of }y \\ \\  \qquad\textsf{Solving for }y, \\ \\  \qquad\tt PR = QR \\ \\  \implies \tt\small{\sqrt{(0 - 6)^2 + (y - (-1))^2} = \sqrt{(0 - (-4))^2 + (y - (-3))^2}} \\ \\   \implies\tt\sqrt{36 + (y + 1)^2} = \sqrt{16 + (y + 3)^2 } \\ \\ \textsf{Squaring both sides yields} \\ \\  \implies \tt \: 36 + (y + 1)^2 = 16 + (y + 3)^2 \\ \\  \implies\tt 36 + y^2 + 2y + 1 = 16 + y^2 + 6y + 9 \\ \\  \implies \tt \: y^2 + 2y + 37 = y^2 + 6y + 25 \\ \\  \implies \tt \: 2y - 6y = 25 - 37 \\ \\ \implies \tt -4y = -12 \\ \\   \implies\red{\boxed{ \tt y = 3}} \end{array}  \\  \\  \\ \begin{array}{l} \\ 3.)\: \bold{Given:}\: \begin{cases}\: A(4,5), B(-3,2),\textsf{ and }C(x,0) \\ \: AC = BC \end{cases} \\ \\ \bold{Required:}\:\textsf{ value of }x \\ \\  \qquad\textsf{Solving for }x, \\ \\   \qquad\tt AC = BC \\ \\ \implies\tt\small{\sqrt{(x - 4)^2 + (0 - 5)^2} = \sqrt{(x - (-3))^2 + (0 - 2)^2}} \\ \\ \implies\tt\sqrt{(x - 4)^2 + 25} = \sqrt{(x + 3)^2 + 4} \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies\tt\:(x - 4)^2 + 25 = (x + 3)^2 + 4 \\ \\ \implies\tt\:x^2 - 8x + 16 + 25 = x^2 + 6x + 9 + 4 \\ \\ \implies\tt\:x^2 - 8x + 41 = x^2 + 6x + 13 \\ \\ \implies\tt-8x - 6x = 13 - 41 \\ \\\implies\tt -14x = -28 \\ \\ \implies\red{\boxed{\tt\:x = 2}} \end{array}

#CarryOnLearning

#BrainlyMathKnower

#5-MinutesAnswer

7 0
2 years ago
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