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tigry1 [53]
3 years ago
13

3x(x²-5) what's the product​

Mathematics
2 answers:
Aleksandr-060686 [28]3 years ago
6 0
3x(x^2 - 5)
(3x)(x^2 + 5)
(3x)(x^2) + (3x)(-5)
Thus, your answer would be: 3x^3 -15x
fenix001 [56]3 years ago
3 0

3x(x²-5)

Multiply the bracket by 3x

(3x)(x^2)=3x^3

(3x)(-5)=15x

Answer: 3x^3+15x

You might be interested in
tom has a cube a rectangular prism and a triangular prism. which of the shapes have more than 6 vertices?
Ierofanga [76]

Answer:

<h3>A cube</h3>

Step-by-step explanation:

Vertices of a solid shape is a point where 2 lines or edges of the shape intersects. If tom has a cube a rectangular prism and a triangular prism, then the cube has 8 vertices, a rectangular prism has 5 vertices (a rectangle has 4 vertices plus the point where all the lines coming from the vertices of the rectangle intersects).

Also, a triangular prism has 4 vertices(a triangle has 3 vertices plus the point where all the lines coming from the vertices of the triangle intersects)

Hence the only shape that has more than 6 vertices is a cube.

4 0
3 years ago
Simply (4x-5)+(3x+6)
STatiana [176]

Answer:

7x+1

Step-by-step explanation:

Let's simplify step-by-step.

4x−5+3x+6

=4x+−5+3x+6

Combine Like Terms:

=4x+−5+3x+6

=(4x+3x)+(−5+6)

=7x+1

Hope this helps!

5 0
3 years ago
Read 2 more answers
What are the x-intercepts of x^2– 3x = 4?
Sindrei [870]

Answer:

x^2-3x=4

x^2-3x-4=0

x^2-4x+x-4=0

x(x-4)+1(x-4)=0

(x+1)(x-4)=0

x=-1,4

x- intercepts in the point (-1,0) and (4,0)

Step-by-step explanation:

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=12%20%5Cfrac%7B3%7D%7B6%7D%20%2B%2014%5Cfrac%7B4%7D%7B6%7D%20" id="TexFormula1" title="12 \fra
Vikki [24]
27 1/6 or 163/6 or 27.16
6 0
3 years ago
Read 2 more answers
Mr wells drew a plan for a rectangle dog run three of the vertices are (2 1/3, 7 1/2), (12, 7 1/2), and (12, 1) what are the coo
melisa1 [442]

Answer:

Step-by-step explanation:

let 4th fourth vertex be D (x,y)

A(2 1/3,7 1/2),B(12, 7 1/2),C(12,1)

or A(7/3,15/2),B(12,15/2),C(12,1)

Mid-point of diagonal AC is  P((7/3+12)/2,(15/2+1))

Mid -point of diagonal BD is P((12+x)/2,(15/2+y)/2)

(Because mid -point of diagonal is same.)

(12+x)/2=(7/3+12)/2

12+x=7/3+12

x=7/3

(15/2+1)/2=(15/2+y)/2

15/2+1=15/2+y

y=1

so fourth vertex is (7/3,1)

8 0
3 years ago
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