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Anarel [89]
3 years ago
15

Name of the geometric solid suggested by a number cube

Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
6 0
I think its d.... but dont listen to me.<span>
</span>
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5. Combien de paires de nombres naturels peut-on former dont : a) la somme est 23? b) la différence est 100?​
Lilit [14]

Answer: a) 24 b) infini

Step-by-step explanation: a) parce que tu peut faire que 24 somme pour faire 23. B) infini car tout plus haut 100 fonctione

3 0
1 year ago
Solve the system by substitution.
tino4ka555 [31]
As we know y=3/4x-3
Let's put it in the second equation
3/4x-3=1/4x+1
3/4 x -1/4 x = 1+3
2/4 x = 4
X= 4*4/2
x=8
Put x= 8 in first equation
y= 3/4 *8 -3
y=6-3
y=3
Check the answer
3=3/4 *8 -3
3=6-3
3=3
Correct
So(y, x) = (3,8)
Because x=8 and y=3
3 0
3 years ago
Determine as a linear relation in x, y, z the plane given by the vector function F(u, v) = a + u b + v c when a = 2 i − 2 j + k,
Ostrovityanka [42]

Answer:

2x - y - 3z = 0

Step-by-step explanation:

Since the set

{i, j}  = {(1,0), (0,1)}

is a base in \mathbb{R}^2

and F is linear, then

<em>{F(1,0), F(0,1)}  </em>

would be a base of the plane generated by F.

F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k

F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k

Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k

We need a normal vector which is the cross product of 3i+2k and 4i-j+3k

(3i+2k)X(4i-j+3k) = 2i-j-3k

The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by

2(x-3) -1(y-0) -3(z-2) = 0

or what is the same

2x - y - 3z = 0

3 0
3 years ago
Match the phenomenal expression on the left with the simplified version on the right
nalin [4]

Given the polynomial expression:

(y + 5)²

(y - 5)(y + 5)

Let's simplify each of the given expression:

a.) (y + 5)²

The given equation is a factor of a perfect square trinomial. For this type of expression, the following is the formula for expanding it.

\text{ (a+b)}^2\text{ = }a^2\text{ + 2ab + }b^2

We get,

\mleft(y+5\mright)^2=(y)^2+2(y)(5)+(5)^2(y+5)^2=y^2+10y+25

b.) (y - 5)(y + 5)

To be able to simplify the following expression. We will be using the formula for the difference of two squares.

(a+b)(a-b)=a^2-b^2

We get,

\mleft(y-5\mright)\mleft(y+5\mright)=(y)^2-(5)^2(y-5)(y+5)=y^2-25^{}

8 0
1 year ago
Can someone answer plzzzz
telo118 [61]

Answer:

6,882,000 miles

Step-by-step explanation:

8 0
3 years ago
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