A triangle has a base of 52/3 in. and a height of 3 in. what is the area of the triangle

dexar [7]

$\huge\boxed{x=25}$

The mistake is in Step 1. The solver should have also squared $9$ when squaring the other side of the equation.

<h3>Correctly solving</h3>

$\begin{aligned}\sqrt{3x+6}&=9\\(\sqrt{3x+6})^2&=9^2\\3x+6&=81\\3x+6-6&=81-6\\3x&=75\\\frac{3x}{3}&=\frac{75}{3}\\x&=25\end{aligned}$

4 0

1 year ago

Read 2 more answers

Mama had several notebooks. If she give 13 notebooks to each of her children, she will have 8 left. If she give 15 notebooks to

Goshia [24]

4 children
60 notebooks

3 0

3 years ago

Read 2 more answers

Help. Not good at word problems

earnstyle [38]

Answer:

Option A

Step-by-step explanation:

B U C = tax returns showing business income + tax filled in 2005

U represents Union relationship that means either B or C

5 0

3 years ago

Find a recurrence relation for the amount of money in a savings account after n years if the interest rate is 6 percent and $50

Korvikt [17]

Answer:

$a_n = a_{n-1} (1.06) + 50$

Step-by-step explanation:

Suppose, $a_0$ is initial amount in the saving account,

Here, the annual interest rate is 6% and additional amount in each year is $ 50,

So, the amount after one year,

$a_1 = a_0 + 6\%\text{ of }a_0 + 50 = a_0 + 0.06a_0 + 50 = a_0(1.06) + 50$

Amount after 2 years,

$a_2 = a_1 + 6\%\text{ of }a_1 + 50 = a_1(1.06) + 50$

Amount after 3 years,

$a_3 = a_2 + 6\%\text{ of }a_2 + 50 = a_2(1.06) + 50$

................................., so on....

Hence, by following the pattern,

The amount after n years,

$a_n = a_{n-1} (1.06) + 50$

Which is the required recurrence relation for the amount of money in a savings account

3 0

3 years ago

The sequence$$1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1,2,\dots$$consists of $1$'s separated by blocks of $2$'s with $n$ $2$'s i

kicyunya [14]

Consider the lengths of consecutive 1-2 blocks.

block 1 - 1, 2 - length 2

block 2 - 1, 2, 2 - length 3

block 3 - 1, 2, 2, 2 - length 4

block 4 - 1, 2, 2, 2, 2 - length 5

and so on.

Recall the formula for the sum of consecutive positive integers,

$\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2$

Now,

$1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016$

which means that the 1234th term in the sequence occurs somewhere about 1/5 of the way through the 49th 1-2 block.

In the first 48 blocks, the sequence contains 48 copies of 1 and 1 + 2 + 3 + ... + 47 copies of 2, hence they make up a total of

$\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224$

numbers, and their sum is

$\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400$

This leaves us with the contribution of the first 10 terms in the 49th block, which consist of one 1 and nine 2s with a sum of $1+9\times2=19$ .

So, the sum of the first 1234 terms in the sequence is 2419.

8 0

2 years ago