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Fantom [35]
3 years ago
12

A ____________ has no dimension and is represented by a small dot.

Mathematics
1 answer:
monitta3 years ago
7 0

Answer:

Point

Step-by-step explanation:

If you are referring to geometry

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How would I go about solving 12,15,18 thanks
grigory [225]
For 12, you would do 90=4x+42, and 48=4x, and x=12
for 15, you would do 90=8x+66, and 24=8x, and x=3
for 18, you would do 90=3x+57, and 33=3x, and x=11
3 0
3 years ago
What is 18k + 30fk in like terms. Will give brainliest to correct answer.
oee [108]

Answer:

6k(5f+3)

Step-by-step explanation:

18k+30fk

Factor out 6.

6(3k+5fk)

Consider 3k+5fk. Factor out k.

k(3+5f)

Rewrite the complete factored expression.

6k(5f+3)

4 0
3 years ago
Read 2 more answers
Simplify the expression 1.5y2+6.3+6y-3.9-4y+y2​
Lena [83]

Answer: 2.5y^{2} +2y+2.4

Step-by-step explanation:

8 0
3 years ago
Expand the binomial (2x+5)^5
Whitepunk [10]
That would simplify to 32x^5+400x^4+200x^3+5000x^2+6250x+3125
You would do that because you have to write it repeatedly as (2x+5)(2x+5) and so on.
3 0
3 years ago
Which of the equations below could be the equation of this parabola?
nirvana33 [79]

Answer:

 y=-4x^2  is the equation of this parabola.

Step-by-step explanation:

Let us consider the equation

y=-4x^2

\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

\mathrm{Range\:of\:}-4x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:0]\end{bmatrix}

\mathrm{Axis\:interception\:points\:of}\:-4x^2:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)

As

\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=a\left(x-m\right)\left(x-n\right)

\mathrm{is\:the\:average\:of\:the\:zeros}\:x_v=\frac{m+n}{2}

y=-4x^2

\mathrm{The\:parabola\:params\:are:}

a=-4,\:m=0,\:n=0

x_v=\frac{m+n}{2}

x_v=\frac{0+0}{2}

x_v=0

\mathrm{Plug\:in}\:\:x_v=0\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}

y_v=-4\cdot \:0^2

y_v=0

Therefore, the parabola vertex is

\left(0,\:0\right)

\mathrm{If}\:a

\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}

a=-4

\mathrm{Maximum}\space\left(0,\:0\right)

so,

\mathrm{Vertex\:of}\:-4x^2:\quad \mathrm{Maximum}\space\left(0,\:0\right)

Therefore,  y=-4x^2  is the equation of this parabola. The graph is also attached.

7 0
3 years ago
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