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3 years ago

Solve for x in terms of a,b, and c: ax - 3b =c. . 1: a(c+3b). 2: a (c-3b). 3: c-3/a. 4: c+3b/a

1 answer:

8 0

We have to solve x in terms of a, b and c:
a x - 3 b = c
a x - 3 b + 3 b = c + 3 b
a x = c + 3 b
x = ( c + 3 b ) : a
Answer:
4 ) ( c + 3 b ) / a

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The line y=2/5x +7 is dilated by a scale factor of 4 centered at the point (5,9). What equation represents the image of the lone

Serjik [45]

Answer:

Step-by-step explanation:

I don’t understand

7 0

3 years ago

Please help me with this math question!

barxatty [35]

1) Change radical forms to fractional exponents using the rule:
The n<span>th root of "</span>a number" = "that number" raised to the<span> reciprocal of n.
For example </span> $\sqrt[n]{3} = 3^{ \frac{1}{n} }$ .

The square root of 3 ( $\sqrt{3}$ ) = 3 to the one-half power ( $3^{ \frac{1}{2} }$ ).
The 5th root of 3 ( $\sqrt[5]{3}$ ) = 3 to the one-fifth power ( $3^{ \frac{1}{5} }$ ).

2) Now use the product of powers exponent rule to simplify:
This rule says $a^{m} a^{n} = a^{m+n}
$ . When two expressions with the same base (a, in this example) are multiplied, you can add their exponents while keeping the same base.

You now have $(3^{ \frac{1}{2} })*(3^{ \frac{1}{5} })$ . These two expressions have the same base, 3. That means you can add their exponents:
$(3^{ \frac{1}{2} })(3^{ \frac{1}{5} })\\
= 3^{(\frac{1}{2} + \frac{1}{5}) }\\
= 3^{\frac{7}{10}}$

3) You can leave it in the form $3^{\frac{7}{10}}$ or change it back into a radical $\sqrt[10]{3^7}$

------

Answer: $3^{\frac{7}{10}}$ or $\sqrt[10]{3^7}$

6 0

3 years ago

Use distributive property to write an equivalent expression. Use paper and pencil to

just olya [345]

Answer:

A) 6Gn + 12 (or 6n+12, if that’s what you meant)

B) 4p²+9p

C) 5r²+20r

Step-by-step explanation:

A. If you meant 6(n+2)...

Multiply 6 by n and also 6 by 2:

6n + 12

B. Multiply p by 4p and also p by 9

4p²+9p

C. Multiply 5r by r and also 5r by 4

5r²+20r

Hope this helped!

8 0

3 years ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$