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2 years ago

A pyramid with a rectangular base has how many more edges than vertices? List how many edges and vertices there are.

Mathematics

1 answer:

Aleksandr-060686 [28]2 years ago

4 0

Edges:8
Vertices:5
8-5=3. There are 3 more edges than vertices.

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I ordered some books online for myself and friends. Each book costs $13 and the store charges a flat rate for shipping of $20. W

Anna35 [415]

Answer:

13n+20

Step-by-step explanation:

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3 years ago

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The figure shows a circle inscribed in a triangle. To construct the inscribed circle, angle bisectors were first constructed at

klemol [59]

Answer:

Segments perpendicular to the sides of the triangle through the intersection of the angle bisectors were constructed.

Step-by-step explanation:

The above choice represents a bit of excess work. Actually, only one such perpendicular line segment needs to be constructed in order to determine the radius of the inscribed circle.

Once you know the center and radius, you can construct the inscribed circle.

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3 years ago

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A cook has 6 and 1/2 cup of sliced carrots how many 3/4 cup servings of carrots is this

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Answer:

8 and 2/3

Step-by-step explanation:

Divide 6.5 by 0.75 and you get $26/3$ , or 8 and 2/3

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2 years ago

Can some explain steps

Mashutka [201]

Answer:

r = 3.

Step-by-step explanation:

16 = 10 + √(3r + 27)

√(3r + 27) = 6

Square both sides:

3r + 27 = 36

3r = 36 - 27 = 9

r = 3.

Check the result:

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= 10 + √36 = 16

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2 years ago

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Which is the approximate solution to the system y = 0.5x + 3.5 and y = − 2/3 x + 1/3 shown on the graph? (–2.7, 2.1) (–2.1, 2.7)

AlekseyPX

Answer:

The approximate solution to the system is $(-2.7, 2.1)$ .

Step-by-step explanation:

To solve the system of equations $\begin{bmatrix}y=0.5x+3.5\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}$ you must:

$\mathrm{Rationalize\:equations}\\\\\begin{bmatrix}y=\left(\frac{1}{2}\right)x+\left(\frac{7}{2}\right)\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}$

$\mathrm{Subsititute\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\begin{bmatrix}-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\end{bmatrix}$

$\mathrm{Isolate}\:x\:\mathrm{for}\:-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\\\\-\frac{2}{3}x=\frac{19}{6}+\frac{1}{2}x\\\\-\frac{7}{6}x=\frac{19}{6}\\\\6\left(-\frac{7}{6}x\right)=\frac{19\cdot \:6}{6}\\\\-7x=19\\\\x=-\frac{19}{7}\approx-2.7$

$\mathrm{For\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\mathrm{Subsititute\:}x=-\frac{19}{7}\\\\y=-\frac{2}{3}\left(-\frac{19}{7}\right)+\frac{1}{3}\\\\y=\frac{15}{7}\approx 2.1$

The approximate solutions to the system of equations are:

$x=-2.7 ,\:y=2.1$

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3 years ago

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