256% of 75 portion please help fast

The 3 sides of a triangle have lengths of (4x+3) cm, (2x+8) cm, and (3x-10) cm. What is the length of the shortest side if the p

galina1969 [7]

Answer:

The shortest sides is 3x-10

Step-by-step explanation:

This is because if you do the algebra which is 4x+3+2x+8+3x-10=136, x=15. Now that you have 15 you plug it into each sides x value. First, 4x+3= 4*15+3=63

Second, 2x+8=30+8=38

Lastly, 3x-10=45-10=35

So, as you can see here 35 is the lowest value which means the side (3x-10) is the shortest.

3 0

3 years ago

Please help I don’t understand

JulsSmile [24]

Answer:

Vertical: G an B. Adjacent: C and F. Supplementary: A and E. Straight: A, E. Complementary: G , B and E,A.

4 0

2 years ago

I'm very confused on this question and I don't know how to solve it. Any kind of help will be appreciated.

juin [17]

QP is the bisector so QPM is half NPM or

$m \angle NPM = 2 m \angle QPM$

6 0

3 years ago

An amusement park manager determined that about 23 of all customers would wait in long lines to ride the new roller coaster. Whi

laiz [17]

Answer:

Top Left in I.M.

Step-by-step explanation:

Trust me

4 0

2 years ago

Read 2 more answers

What is a quick and easy way to remember explicit and recursive formulas?

Oliga [24]

I always found derivation to be helpful in remembering. Since this question is tagged as at the middle school level, I assume you've only learned about arithmetic and geometric sequences.

First, remember what these names mean. An arithmetic sequence is a sequence in which consecutive terms are increased by a fixed amount; in other words, it is an additive sequence. If $a_n$ is the $n$ th term in the sequence, then the next term $a_{n+1}$ is a fixed constant (the common difference $d$ ) added to the previous term. As a recursive formula, that's

$a_{n+1}=a_n+d$

This is the part that's probably easier for you to remember. The explicit formula is easily derived from this definition. Since $a_{n+1}=a_n+d$ , this means that $a_n=a_{n-1}+d$ , so you plug this into the recursive formula and end up with

$a_{n+1}=(a_{n-1}+d)+d=a_{n-1}+2d$

You can continue in this pattern, since every term in the sequence follows this rule:

$a_{n+1}=a_{n-1}+2d$
$a_{n+1}=(a_{n-2}+d)+2d$
$a_{n+1}=a_{n-2}+3d$
$a_{n+1}=(a_{n-3}+d)+3d$
$a_{n+1}=a_{n-3}+4d$

and so on. You start to notice a pattern: the subscript of the earlier term in the sequence (on the right side) and the coefficient of the common difference always add up to $n+1$ . You have, for example, $(n-2)+3=n+1$ in the third equation above.

Continuing this pattern, you can write the formula in terms of a known number in the sequence, typically the first one $a_1$ . In order for the pattern mentioned above to hold, you would end up with

$a_{n+1}=a_1+nd$

or, shifting the index by one so that the formula gives the $n$ th term explicitly,

$a_n=a_1+(n-1)d$

Now, geometric sequences behave similarly, but instead of changing additively, the terms of the sequence are scaled or changed multiplicatively. In other words, there is some fixed common ratio $r$ between terms that scales the next term in the sequence relative to the previous one. As a recursive formula,

$a_{n+1}=ra_n$

Well, since $a_n$ is just the term after $a_{n-1}$ scaled by $r$ , you can write

$a_{n+1}=r(ra_{n-1})=r^2a_{n-1}$

Doing this again and again, you'll see a similar pattern emerge:

$a_{n+1}=r^2a_{n-1}$
$a_{n+1}=r^2(ra_{n-2})$
$a_{n+1}=r^3a_{n-2}$
$a_{n+1}=r^3(ra_{n-3})$
$a_{n+1}=r^4a_{n-3}$

and so on. Notice that the subscript and the exponent of the common ratio both add up to $n+1$ . For instance, in the third equation, $3+(n-2)=n+1$ . Extrapolating from this, you can write the explicit rule in terms of the first number in the sequence:

$a_{n+1}=r^na_1$

or, to give the formula for $a_n$ explicitly,

$a_n=r^{n-1}a_1$

6 0

3 years ago