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Triss [41]
3 years ago
12

A prism is a 3d shape with the same _______ all the way through

Mathematics
1 answer:
Aloiza [94]3 years ago
8 0

Answer:

cross-section

Step-by-step explanation:

prism is a type of three-dimensional (3D) shape with flat sides.

It has the same cross-section all along the shape from end to end; that means if you cut through it you would see the same 2D shape as on either end.

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The_______ coordinate located at the highest or lowest point on a parobla​
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Vertex

Step-by-step explanation:

6 0
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Below, enter x If the graph of the given equation is symmetric with respect to the x-axis, y if it is symmetric with respect to
ololo11 [35]

Answer:

Step-by-step explanation:

You can tell by looking at the form of each equation.

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This is an up-opening parabola with vertex at (-½,-¼). Not symmetrical to either axis nor to the origin.

Can't tell what "y = (1 + x2)3" means. Which terms are exponents?

5 0
3 years ago
Evaluate the sum of the following finite geometric series.
rjkz [21]

Answer:

\large\boxed{\dfrac{156}{125}\approx1.2}

Step-by-step explanation:

<h3>Method 1:</h3>

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}

<h3>Method 2:</h3>

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}

a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1

\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}

5 0
3 years ago
How do I solve for x
skad [1K]

Answer:

<h3>x = 40/7</h3>

Step-by-step explanation:

Look at the picture.

If ABC is any triangle and AD bisects (cuts in half) the angle BAC, then

\dfrac{AB}{BD}=\dfrac{AC}{DC}

In our triangle we have the proportion:

\dfrac{x}{8}=\dfrac{10-x}{6}            <em>cross multiply</em>

6x=8(10-x)               <em>use distributive property</em>

6x=(8)(10)+(8)(-x)

6x=80-8x                   <em>add 8x to both sides</em>

14x=80            <em>divide both sides by 14</em>

x=\dfrac{80}{14}

\boxed{x=\dfrac{40}{7}}

4 0
3 years ago
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