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3 years ago

Area of a rectangle 8cm long and 4cm in height. please tell me how to do this also, I'm so confused.

Mathematics

1 answer:

Tema [17]3 years ago

6 0

The area of a rectangle is easy to compute.
Area = length times height
Area = 8 cm * 4 cm = 32 square centimeters

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Agnes has 23 collectible stones, all of

Svetllana [295]

Answer:

The answer is she sold 20 labradorite crytals and 3 galena crystals.

Step-by-step explanation:

First you times 20 x 20 and get 400 and then times 13 x 3 and get 39 and 400 plus 39 equals $439.

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3 years ago

Write a system of equations to describe the situation below, solve using substitution, and fill in the blanks.Josiah has punch c

love history [14]

Let x be the number of weeks; therefore, the two linear equations are

$\begin{gathered} T=2+7x \\ \text{and} \\ C=11+4x \end{gathered}$

Where T stands for the punches on the tea punch card and C for the punches on the coffee punch card.

Solving by substitution. Set T=C, then

$\begin{gathered} T=C \\ \Rightarrow2+7x=11+4x \\ \Rightarrow3x=9 \\ \Rightarrow x=\frac{9}{3}=3 \\ \Rightarrow x=3 \end{gathered}$

Thus, substitute the value of x=3 into the first equation,

$\begin{gathered} x=3 \\ \Rightarrow T=2+7\cdot3=23 \\ \Rightarrow T=23 \end{gathered}$

Thus, after 3 weeks, Josiah will have the same number of punches on each card, and he will have 23 punches on each card.

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1 year ago

Please help?

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

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3 years ago

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Answer:

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Step-by-step explanation:

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