All categories

3 years ago

Find the side length of the square 16x2 +56x +49

Mathematics

1 answer:

Darya [45]3 years ago

8 0

Answer:

(4x+7)(4x+7)

Step-by-step explanation:

You might be interested in

16 POINTS

Marta_Voda [28]

Solve the following system using elimination:
{7 x + 2 y = -19 | (equation 1)
{2 y - x = 21 | (equation 2)

Add 1/7 × (equation 1) to equation 2:
{7 x + 2 y = -19 | (equation 1)
{0 x+(16 y)/7 = 128/7 | (equation 2)

Multiply equation 2 by 7/16:
{7 x + 2 y = -19 | (equation 1)
{0 x+y = 8 | (equation 2)

Subtract 2 × (equation 2) from equation 1:
{7 x+0 y = -35 | (equation 1)
{0 x+y = 8 | (equation 2)

Divide equation 1 by 7:
{x+0 y = -5 | (equation 1)
{0 x+y = 8 | (equation 2)

Collect results:
Answer: {x = -5, y = 8

8 0

3 years ago

Which box of crackers is a better deal?

a_sh-v [17]

Answer:

Since each cracker in the 190 gram box costs $ 0.0078 while the 850 gram box costs $ 0.0082 each, the 190 gram box is a better deal.

Step-by-step explanation:

Knowing that a 190-gram box of crackers costs $ 1.49 and an 850-gram box costs $ 6.99, to determine which of them is a better deal, the following calculation must be performed:

1.49 / 190 = 0.0078

6.99 / 850 = 0.0082

Therefore, since each cracker in the 190 gram box costs $ 0.0078 while the 850 gram box costs $ 0.0082 each, the 190 gram box is a better deal.

8 0

3 years ago

If $64,000 is invested in an IRA account with an annual interest rate of 8% compounded once a year, what is the value of the acc

Iteru [2.4K]

Answer: the value of the account after 6 years is $101559.96

Step-by-step explanation:

If $64,000 is invested in an IRA account, then

Principal = $64,000

So P = 64,000

The rate at which $64000 was compounded is 8%

So r = 8/100 = 0.08

If it is compounded once in a year, this means that it is compounded annually (and not semi annually, quarterly or others). So

n = 1

We want to determine the value of the account after 6 years, this means

time, t = 6

Applying the compound interest formula,

A = P(1 + r/n)^nt

A = amount after n number of years

A = 64000( 1 + 0.08/1)^1×6

A = 64000(1.08)^6

A= 64000×1.58687432294

A= 101559.956668416

Approximately $101559.96 to 2 decimal places

4 0

3 years ago

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

4 0

3 years ago

I need help on question 9 please!! Thank you!

jekas [21]

Answer:

Step-by-step explanation:

33000/0.27 =$8910

7 0

3 years ago