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lord [1]
2 years ago
12

WRITE 3/10 AS THE PRODUCT OF A WHOLE NUMBER AND A FRACTION. EXPLAIN YOUR ANSWER.

Mathematics
2 answers:
Darya [45]2 years ago
8 0
30 is the answer from 3/10
butalik [34]2 years ago
5 0
You divide 10 by 3 and the your remainder will be your whole number and the fraction will be your answer Will be 1 and3/3
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The actual distance between the two houses is 44.1ft. The contractor measures the distance as 46ft. Find the relative error of t
mixer [17]
The correct answer is: 44.1ft - 46ft = -1.9ft/44.1ft = .043% relative error. Please mark Brainliest if helpful
6 0
2 years ago
Anyone knows how to do questions 7 and 8? 15 pts!!
Oksi-84 [34.3K]
7) Certainly there is a typo in the statement, just see that the expression of item (ii) is different from that of item (i). Probably the correct expression is: 2x^2-4x+5. With this consideration, we can continue.

(i) Let E the expression that we are analyzing:

E=2x^2-4x+5\\\\ E=2x^2-4x+2-2+5\\\\ E=2(x^2-2x+1)-2+5\\\\ E=2(x-1)^2+3

Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.

(ii) Manipulating the expression:

2x^2+5=4x\\\\ 2x^2-4x+5=0

So, it's the case when E=0. However, E is always a positive number. Then, there is no real number x that satisfies the expression.

8) Let E the expression that we want to calculate:

E=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)+1\\\\ E-1=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)

Multiplying by (2-1) in the both sides:

(2-1)(E-1)=(2-1)(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2-1)(2+1)}_{2^2-1}(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2^2-1)(2^2+1)}_{2^4-1}(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ ... Repeating the process, we obtain: ...\\\\ E-1=(2^{32}-1)(2^{32}+1)\\\\ E-1=2^{64}-1\\\\ \boxed{E=2^{64}}
3 0
3 years ago
Read 2 more answers
12.73+0.06x=13.48+0.14x
Afina-wow [57]
X= -9.375. 0.06x-0.14x=13.48-12.73. -0.08x=0.75
3 0
3 years ago
Read 2 more answers
Which expressions are completely factored? Select each correct answer. 30a6−24a2=3a2(10a4−8) 16a5−20a3=4a3(4a2−5) 12a3+8a=4(3a3+
irinina [24]

When factoring an expression, we need to identify the Greatest Common Factor of the expression (GCF). By definition, GCF is the product of prime factors involved with its Lowest Exponent.

We need to divide each term by the GCF to get the expression inside the parenthesis.

30a^6-24a^2\\ GCF\; is \; 6a^2\\ \\ 30a^6-24a^2=6a^2(\frac{3a^6}{6a^2} -\frac{24a^2}{6a^2} )=6a^2(5a^4-4)\\ ---------------------\\ \\ 16a^5-20a^3\\ GCF \; is \; 4a^3\\ \\ 16a^5-20a^3=4a^3(\frac{16a^5}{4a^3} -\frac{20a^3}{4a^3})=4a^3(4a^2-5) \\ ---------------------\\ \\ 12a^3+8a\\ GCF \; is \; 4a\\ \\ 12a^3+8a=4a(\frac{12a^3}{4a}+\frac{8a}{4a})=4a(3a^2+2)\\   ---------------------\\ 24a^4+18\\GCF \; is \; 6\\\\24a^4+18=6(\frac{24a^4}{6} +\frac{18}{6})=6(4a^4+3) \\    ---------------------

Conclusion:

From above we can conclude that the below expressions are factored completely

16a^5-20a^3=4a^3(4a^2-5)\\ \\ 24a^4+18=6(4a^4+3)

6 0
2 years ago
Read 2 more answers
HELP PLZZZZZZ!!!!!!!!<br> F=fg/f+g-d<br> solve for f<br> F does not equal f<br> SHOW YOUR WORK
Flura [38]
Is this even a problem
3 0
2 years ago
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