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olga nikolaevna [1]

2 years ago

7

The ice cream Palace received 3 gallons of strawberry ice cream, 5 pints of mocha ice cream, and 1 quart of vanilla ice cream to

day. How many pints of ice cream did they receive?

1 answer:

Alina [70]2 years ago

6 0

They recived 31 pints of ice cream

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Brahmagupta solved a quadratic equation of the form ax2 + bx = c using the formula x =, which involved only one solution. Using

weeeeeb [17]

X2+7x-8=0
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x2-1x+8x-8=0
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1 year ago

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Gina has 5 2/6 feet of silver ribbon and 2 4/6 of gold ribbon. how much more silver ribbon does gina have than golden ribbon?

MrRissso [65]

To answer the question above, we are simply to subtract the length of the gold ribbon which is 2 4/6 ft from the length of the silver ribbon, 5 2/6 feet. Mathematically,
5 2/6 feet - 2 4/6 feet = 8/3 feet
Therefore, Gina has 8/3 feet more of the silver ribbon than the golden ribbon.

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1 year ago

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Which graph represents the function y = 2x - 4

WARRIOR [948]

Answer:

I will change my awnser when you tell me but what are the awnser choices

Step-by-step explanation:

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1 year ago

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Determine if each of the following sets is a subspace of ℙn, for an appropriate value of n. Type "yes" or "no" for each answer.

xxMikexx [17]

Answer:

1. Yes.

2. No.

3. Yes.

Step-by-step explanation:

Consider the following subsets of Pn given by

1.Let W1 be the set of all polynomials of the form $p(t)=at^2$ , where a is in ℝ.

2.Let W2 be the set of all polynomials of the form $p(t)=t^2+a$ , where a is in ℝ.

3. Let W3 be the set of all polynomials of the form $p(t)=at^2+at$ , where a is in ℝ.

Recall that given a vector space V, a subset W of V is a subspace if the following criteria hold:

- The 0 vector of V is in W.

- Given v,w in W then v+w is in W.

- Given v in W and a a real number, then av is in W.

So, for us to check if the three subsets are a subset of Pn, we must check the three criteria.

- First property:

Note that for W2, for any value of a, the polynomial we get is not the zero polynomial. Hence the first criteria is not met. Then, W2 is not a subspace of Pn.

For W1 and W3, note that if a= 0, then we have p(t) =0, so the zero polynomial is in W1 and W3.

- Second property:

W1. Consider two elements in W1, say, consider a,b different non-zero real numbers and consider the polynomials

$p_1 (t) = at^2, p_2(t)=bt^2$ .

We must check that p_1+p_2(t) is in W1.

Note that

$p_1(t)+p_2(t) = at^2+bt^2 = (a+b)t^2$

Since a+b is another real number, we have that p1(t)+p2(t) is in W1.

W3. Consider two elements in W3. Say $p_1(t) = a(t^2+t), p_2(t)= b(t^2+t)$ . Then

$p_1(t) + p_2(t) = a(t^2+t) + b(t^2+t) = (a+b) (t^2+t)$

So, again, p1(t)+p2(t) is in W3.

- Third property.

W1. Consider an element in W1 $p(t) = at^2$ and a real scalar b. Then

$bp(t) = b(at^2) = (ba)t^2)$ .

Since (ba) is another real scalar, we have that bp(t) is in W1.

W3. Consider an element in W3 $p(t) = a(t^2+t)$ and a real scalar b. Then

$bp(t) = b(a(t^2+t)) = (ba)(t^2+t)$ .

Since (ba) is another real scalar, we have that bp(t) is in W3.

After all,

W1 and W3 are subspaces of Pn for n= 2

and W2 is not a subspace of Pn.

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2 years ago

7. A hotelier conducted a survey of guests staying at her hotel. The table shows some

jonny [76]

Answer:

here's the solution :-

Step-by-step explanation:

i hope it helped...

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2 years ago