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jeyben [28]
3 years ago
12

2. Which answer is equivalent to "y less than 2 times 7"?

Mathematics
1 answer:
mart [117]3 years ago
5 0
The answer is to the equation is b
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Suppose, instead, that Bruce used 6 full boxes of tiles plus an additional 2 tiles, and Felicia used 5 full boxes of tiles plus
Kitty [74]

Answer: there were 15 tiles in each box

Step-by-step explanation:

Let the number of tiles in each full box be x

Suppose, instead, that Bruce used 6 full boxes of tiles plus an additional 2 tiles. It means that he used a total of

(6x + 2) tiles

Suppose, that Felicia used 5 full boxes of tiles plus an additional 17 tiles. It means that he used a total of

(5x + 2) tiles.

If they each used the same number of tiles, it means that the total number of tiles used by Bruce = total number of tiles used by Felicia. Therefore,

6x + 2 = 5x + 17

6x - 5x = 17 - 2

x = 15

6 0
3 years ago
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What is the answer for number 74.
Alla [95]

Answer:

g

Step-by-step explanation:

8 0
2 years ago
In the vector space <img src="https://tex.z-dn.net/?f=P_2%28%5Cmathbb%7BR%7D%29" id="TexFormula1" title="P_2(\mathbb{R})" alt="P
tatuchka [14]

We could express any vector p\in P_2(\mathbb R) as

p=ax^2+bx+c

So any vector u\in U would have coefficients a,b,c that satisfy

16a+4b+c=0\implies c=-16a-4b

from which we can show that any such vector is a linear combination of some other vectors:

u=ax^2+bx+(-16a-4b)=a(x^2-16)+b(x-4)

More explicitly, we've shown that u is a linear combination of the vectors x^2-16 and x-4; in other words, u\in\mathrm{span}\{x^2-16,x-4\}, and in fact these two vectors form a basis for U. But this set does not span all of P_2(\mathbb R) because there's no combination of these two vectors that can be used to obtain a constant. We want the transformation to be usable for any vector in P_2(\mathbb R), so we need to add an additional vector extend the basis. We can do this simply by appending 1 into the spanning set. (Do check that the vectors remain linearly independent.)

Now we want the transformation to map those polynomials p(x) for which p(4)=0 to the zero vector. We know which vectors belong to the basis of U, so we need

f(x^2-16)=0

f(x-4)=0

f(1)=1

where the choice of the assignment for f(1) is arbitrary, so long as it's non-zero.

6 0
3 years ago
Figure ABCD is a rectangle find the value of X. X +11. 6X plusone
expeople1 [14]

Answer:

its 2

Step-by-step explanation:

because AE equals EC so x+11=6x+1

7 0
2 years ago
3u+7u+9u = show your work
Elenna [48]

Answer:

19u

Step-by-step explanation:

3u+7u+9u

4 0
3 years ago
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