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3 years ago

Which shows the correct substitution of the values a,b,and c from the equation 0=4x^2+2x-1 into the quadratic formula be

1 answer:

6 0

It should be 4x^2 (x square) +2x-1
a=4x^2, b=2x, c=-1
Use the quadratic formula to solve it which is x=-b+- square root b^2-4ac/2a

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The composite figure is made up of a triangular prism and a pyramid. The two solids have congruent bases. A triangular pyramid i

Darina [25.2K]

Answer:

the answer is 3465

Step-by-step explanation:

got it right on ed

4 0

3 years ago

Read 2 more answers

Express the measurement 4.29478416 kg with 8, 6, 4, and 2 significant figures

ira [324]

Significant figures refers to digits within a number which must be included in other depict correct quantity of the figure.

4.29478416 = 4.2947842 ( 8 significant figures)
4.29478416 = 4.29478 (6 significant figures)
4.29478416 = 4.295 ( 4 significant figures)
4.29478416 = 4.3 (2 significant figures)

To round a number to a certain number of significant figures,

Only leading 0 which comes before the decimal Point are regarded as insignificant.
Once the number of significant figures have been identified, the next number after this is either rounded up to 1 and added to the last value(if number is ≥5) or rounded to 0 (if number is less than 5)

Learn. More : brainly.com/question/13386983?referrer=searchResults

3 0

2 years ago

If f(x) = 2x^4, then f'(3) =

valkas [14]

To find f'(3) (f prime of 3), you must find f' first. f' is the derivative of the function f(x).

Finding the derivative of f(x) = 2x⁴ requires the use of the power rule.

The power rule for derivatives is $\frac{d}{dx} [x^{n} ] =nx^{n-1}$ . In other words, you bring the exponent forward and multiply it by the coefficient of the term, and then you subtract 1 from the original exponent.

f'(x) = $\frac{d}{dx}$ (2x⁴)

f'(x) = 2(4)x³

f'(x) = 8x³

Now, to find f'(3), plug 3 into your derivative.

f'(3) = 8(3)³

f'(3) = 216

<h3>Answer:</h3>

f'(3) = 216

6 0

2 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

2 years ago

A car dealership's goal is to sell a mean of 25 cars per day. In one week, it sold the following numbers of cars each day:

LUCKY_DIMON [66]

Answer:

35 more cars

Step-by-step explanation:

17, 14, 19, 21, 27, 22, 20 add all of these numbers together

= 140

140 / 7 (number of days) = 20

25 x 7= 175

175 - 140 =35

8 0

2 years ago

Read 2 more answers