All categories

2 years ago

Name the geometric term(s) modeled by each object

Mathematics

1 answer:

anyanavicka [17]2 years ago

3 0

A line is a one-dimensional shape that is straight, has no thickness, and extends in both directions indefinitely. The geometric term(s) modeled by each object are Hexagon, point, line, line, and plane, respectively.

A line is a one-dimensional shape that is straight, has no thickness, and extends in both directions indefinitely. A line segment is also a line but the line with finite length and having fixed endpoints.

The geometric term(s) modeled by each object

9. Stop sign ⇒ Hexagon

10. Tip of pin ⇒ Point

11. Strings of a violin ⇒ Line

12. A car antenna ⇒ Line

13. A library card ⇒ Plane

Hence, The geometric term(s) modeled by each object are Hexagon, point, line, line, and plane, respectively.

Learn more about Line here:

brainly.com/question/21511618

#SPJ1

You might be interested in

What is the area of the trapezoid? 68 54 76 108

iren [92.7K]

108 is the answer for this question

6 0

3 years ago

Read 2 more answers

The difference of two numbers is 9 and the sum of the numbers is 55. what are the numbers?

yKpoI14uk [10]

The two numbers I think is 3&6 because this is a trick question

3 0

3 years ago

What is the common ratio of the following sequence? 80, 20, 5, ... A) -4 B) 1/4 C) 4

Stolb23 [73]

$\bf 80~~,~~20~~,~~5~\hfill \cfrac{20}{80}\implies \boxed{\cfrac{1}{4}}~~,~~ \cfrac{5}{20}\implies \boxed{\cfrac{1}{4}}~\hfill \stackrel{\textit{common ratio}}{\boxed{\cfrac{1}{4}}}$

6 0

3 years ago

Violet's Botanical Garden produced flowers throughout the year. A graph demonstrating how many flowers she produced over a 12-mo

Lady bird [3.3K]

Answer:

The domain represents a 12-month period of flower production

Step-by-step explanation:

The domain is the horizontal extent of the graph. Both the problem statement and the graph itself tell you that is a 12 month time period.

7 0

3 years ago

Read 2 more answers

If (x^2 - 1) is a factor of ax^4 + bx^3 + cx^2 + dx + e, show that a + c + e = b + d = 0

alisha [4.7K]

$x^2-1=x^2-1^2=(x-1)(x+1)$

If x²-1 is a factor of the polynomial, both x-1 and x+1 are factors of it.

According to the remainder theorem, if a binomial x-a is a factor of a polynomial p(x), then p(a)=0.

If x-1 and x+1 are factors of the polynomial p(x)=ax⁴+bx³+cx²+dx+e, then p(1)=0 and p(-1)=0.

$p(1)=a \times 1^4+b \times 1^3 + c \times 1^2+ d \times 1+e \\
p(-1)=a \times (-1)^4 + b \times (-1)^3 + c \times (-1)^2 + d \times (-1)+e \\ \\
p(1)=a+b+c+d+e \\
p(-1)=a-b+c-d+e \\ \\
p(1)=0 \\
p(-1)=0 \\ \\ \hbox{add both equations:} \\
a+b+c+d+e=0 \\
\underline{a-b+c-d+e=0} \\
2a+2c+2e=0 \\
2(a+c+e)=0 \\
a+c+e=0 \\ \\
\hbox{substitute 0 for a+c+e in the first equation:} \\
a+b+c+d+e=0 \\
(a+c+e)+b+d=0 \\
0+b+d=0 \\
b+d=0 \\ \\
\boxed{a+c+e=b+d=0} \\
\hbox{proved } \checkmark$

8 0

3 years ago