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3 years ago

Which shapes are quadrilaterals with at least 1 right angle?

Mathematics

2 answers:

Nataly_w [17]3 years ago

8 0

Answer:

1 and 4

Step-by-step explanation:

quadrilaterals have 4 sides. Quad=4 of something.

Gwar [14]3 years ago

5 0

D. Only 1 and 4

2, 3 aren't quadrilateral

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Answer:

Start at the Origin, and then go up 5, Right 1.

Step-by-step explanation:

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3 years ago

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Suppose three moons A, B, and C orbit 100,000 kilometers above the surface of a planet. Suppose m ZABC = 90°, and the planet is

Vesna [10]

Answer:

Step-by-step explanation:

Please see diagram attached.

Moon A and C are 100,000km + 10,000 = 110,000km from center of planet.

So they are 110,000 + 110,000 = 220,000km apart.

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2 years ago

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Write an equivalent<br> expression for<br> 7(3 – X) + 15x

Ne4ueva [31]

Answer:

21+8x

Step-by-step explanation:

7(3 – X) + 15x

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3 years ago

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Verdich [7]

It should be 1/-4

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3 years ago

Find the volume of the garden shed

Leno4ka [110]

Answer:

47.3 m³

Step-by-step explanation:

The garden shed is made up of a rectangular prism and a pyramid.

<h3><u>Volume of a rectangular prism</u></h3>

$\begin{aligned}\textsf{Volume of a rectangular prism}&=\sf width \times length \times height\\&=4 \times 4 \times 2\\&=16 \times 2\\&=32\;\; \sf m^3\end{aligned}$

<h3><u>Volume of a pyramid</u></h3>

<u />

From inspection of the given diagram, the slant height of the pyramid is 3.5 m.

Calculate the perpendicular height of the pyramid using Pythagoras Theorem:

$\begin{aligned}\implies a^2+b^2&=c^2\\2^2+h^2&=3.5^2\\h^2&=8.25\\h&=\sqrt{8.25}\;\; \sf m\end{aligned}$

Therefore:

$\begin{aligned}\textsf{Volume of a pyramid}&=\dfrac{\sf length \times width \times height}{3}\\\\&=\dfrac{\sf 4 \times 4 \times \sqrt{8.25}}{3}\\\\& = 15.31883372\;\; \sf m^3\end{aligned}$

<h3><u>Volume of the garden shed</u></h3>

$\begin{aligned}\implies \textsf{Volume of shed}&=\textsf{Volume of rectangular prism}+\textsf{Volume of pyramid}\\&=32+15.31883372\\& = 47.3\;\; \sf m^3\;(nearest\;tenth)\end{aligned}$

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1 year ago

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