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spayn [35]
3 years ago
12

I need help figuring out the reason for the answers

Mathematics
2 answers:
Eddi Din [679]3 years ago
4 0
1. the sum of rational and irrational is always irrational2. the sum of 2 rationals is always rational3. the product of a non zero rational and an irrational is always irrational4. the product of 2 rationals is always rational
irina1246 [14]3 years ago
3 0
Result is correct
reason
1. the sum of rational and irrational is always irrational
2. the sum of 2 rationals is always rational
3. the product of a non zero rational and an irrational is always irrational
4. the product of 2 rationals is always rational
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John has 3 fewer dimes than quarters. He has twice as many nickels as quarters. Altogether he
sweet-ann [11.9K]
You gotta divide $6.45 and 3 together to find the answer
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3 years ago
CAN YOU PLEASE HELP ME WITH THIS PROBLEM???
ozzi

Answer:

Step-by-step explanation:

tan \ 50 =\dfrac{opposite \ side}{adjacent side}\\\\\\1.19=\dfrac{21}{MD}\\\\\\1.19*MD=21\\\\MD = \dfrac{21}{1.19}\\\\\\MD = 17.65

4 0
2 years ago
There are 15 pieces of fruit in a bowl and 6 of them are apples.what percent of the pieces of fruit in the bowl are apples?
Lelu [443]

Answer:

40%

Step-by-step explanation:

just devide the 4 by 15 and you get your answer

4 0
2 years ago
Calculate the following numbers into scientific notation (include units)
inna [77]
<span>7800nm + 95pm = (7800x 10^-9) + (95x10^-12) = 7.800095</span><span>x10^-6=</span><span>7.800095mm
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2500nm - 7pm    = (2500x 10^-9) + (7x10^-12) = 2.50007x10^-6 = 2.50007mm

(65x10^4m) x (4.5x10^-6s-1) = clarify this question

(24x10^5m) / (2x10^-8s) = clarify this question
                                                                              






7 0
3 years ago
Report Error Suppose $P(x)$ is a polynomial of smallest possible degree such that: $\bullet$ $P(x)$ has rational coefficients $\
motikmotik

Answer:

We want a polynomial of smallest degree with rational coefficients with zeros in \sqrt{7}, 1 - \sqrt{6} and -3. The last root gives us the factor (x+3). Hence, our polynomial is

P(x) =(x+3)q(x)

where q is a polynomial with rational coefficients and roots \sqrt{7} and 1 - \sqrt{6}. The root \sqrt{7} gives us a factor x-\sqrt{7}, but in order to obtain rational coefficients we must consider the factor x^2-7.

An analogue idea works with 1 - \sqrt{6}. For convenience write  x - 1 + \sqrt{6} = ( x - 1) + \sqrt{6}. This gives the factor (x-1)^2-6. Hence,

P(x) = (x+3)(x^2-7)((x-1)^2-6)=x^5+x^4-18x^3-22x^2+77x+105

Notice that P(-1)=24. So, in order to satisfy the last condition we divide by 3 the whole polynomial, without altering its roots. Finally, the wanted polynomial is

P(x) =(1/3)x^5+(1/3)x^4-6x^3-(22/3)x^2+(77/3)x+35

Step-by-step explanation:

We must have present that any polynomial it's determined by its roots up to a constant factor. But here we have irrational ones, in order to eliminate the irrational coefficients that a factor of the type x-\sqrt7 will introduce in the expression, we need to multiply by its conjugate x+\sqrt7. Hence, we will obtain x^2-7 that have rational coefficients. Finally, the last condition is given with the intention to fix the constant factor. Usually it is enough to evaluate in the point and obtain the necessary factor.

4 0
2 years ago
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