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Blababa [14]
3 years ago
13

this is a confusing question which polynomial is written in standard form x^3y^2+4x^4y+10x^9 or X^4y^2+4x^4y^5 + 10x^2 or xy^2+4

x^3y^8+10x^2 or x^4y^5+4x^3y^2+10x^2
Mathematics
1 answer:
kherson [118]3 years ago
7 0
Satandard form is when the powers are written in decreasing order


first one
we see at x^9 after x^3, that is unacceptable

2nd one
y^5 is near the end after y^2, unacceptable

3rd one
x^3 is after x^1

4th one
correct

answer is last one
x^4y^5+4x^3y^2+10x^2
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BROO HELP ME PLEASE I DONT UNDERSTAND AND IM STRESSING OUT IT WOULD MEAN ALOT THANK YOUUUU
Tema [17]

Answer: 1a. $190

b. The total cost of making 20 bracelets.

c. $350

d. The total cost of making 100 bracelets.

e. $1150

f. The total cost of making 500 bracelets.

g. 150 bracelets

h. 425 bracelets

Step-by-step explanation:

1a. C(20)=150+2(20)

    C(20)=150+40

    C(20)=$190

b. The total cost of making 20 bracelets.

c. C(100)=150+2(100)

   C(100)=150+200

   C(100)=$350

d. The total cost of making 100 bracelets.

e. C(500)=150+2(500)

   C(500)=150+1000

   C(500)=$1150

f. The total cost of making 500 bracelets.

g. 450=150+2b

   2b=300

     b=150 bracelets

h. 1000=150+2b

   2b=850

     b=425 bracelets

5 0
2 years ago
An integer smaller than -5 but greater than -8​
labwork [276]

Answer:

so what is the question?

7 0
3 years ago
9-6a-24a^2 factor completely
love history [14]
There answer would be 24a^2-6a+9 because you would put the greatest exponer first then start to decrease from your terms.
6 0
3 years ago
Read 2 more answers
Question
krok68 [10]

Answer:

Acute angle between the two planes: approximately 43^\circ.

Step-by-step explanation:

Find the normal vector of each plane:

  • The normal vector of the plane x - 2\, y + 5\, z = 3 is \displaystyle \begin{bmatrix}1 \\ -2 \\ 5\end{bmatrix}.
  • The normal vector of the plane 2\, x + y - 3\, z = 15 is \displaystyle \begin{bmatrix}2 \\ 1\\ -3\end{bmatrix}.

As the name suggests, there is a 90^\circ angle between a plane and its normal vector. The following four angles will correspond to the vertices of a quadrilateral:

  • The 90^\circ angle between the first plane and its normal vector.
  • The angle between the normal vector of each plane.
  • The 90^\circ angle between the second plane and its normal vector.
  • The smallest angle between these two planes.

The sum of these four angles should be 360^\circ. Two of these four angles were known to be 90^\circ. Once the third angle (the angle between the two normal vectors) is found, subtractions would give the measure of the other angle (the smallest angle between these two planes.)

Make use of the dot product to find the angle between these two normal vectors. Let \theta denote the angle between these two vectors.

\displaystyle \cos \theta = \frac{\begin{bmatrix}1 \\ -2 \\ 5\end{bmatrix} \cdot \begin{bmatrix}2 \\ 1 \\ -3\end{bmatrix}}{\sqrt{1^2 + (-2)^2 + 5^2} \cdot \sqrt{2^2 + 1^2 + (-3)^2}} \approx -0.73193.

Before continuing, notice that the smallest angle between the two planes would be 360^\circ - 90^\circ - 90^\circ - \theta = 180^\circ - \theta.

Consider the identity: \cos\left(180^\circ - \theta\right) = -\cos \theta.

In other words, \cos\left(180^\circ - \theta\right), the cosine of the smallest angle between the two planes (which the question is asking for) will be the opposite of \cos \theta, the cosine of the angle between the two normal vectors.

Therefore, the cosine of the smallest angle between the two planes will be -(-0.73193) = 0.73193.

Apply the inverse cosine function to find the size of that angle:

\arccos(0.73193) \approx 43^\circ.

8 0
3 years ago
Is 0.98 larger rhan 1.97?
Aleks04 [339]

Answer: No

Step-by-step explanation:

1.97 is larger than 0.98

4 0
3 years ago
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