All categories

3 years ago

It has 5 questions so I'm giving more points ill also mark brainliest (or what ever its called) if its not a random answer :P

Mathematics

2 answers:

Neko [114]3 years ago

5 0

Answer:

Step-by-step explanation:

id like to help but i cant see the attachment :(

Maslowich3 years ago

5 0

1. The values for y in order are — 1, 2, 5, 7, 9
2. The values for y in order are— -11, -8, -5, -2, -1
3. There is not a consistent rate of change between x and y.
4. The line does not pass through the origin, (0,0)
5. The values for y in order are— -3, -2, -1, 0, 1

You might be interested in

The function f (x) = -4.922 +4.50 + 5 models the height of a ball thrown from five feet high with an initial vertical velocity o

solmaris [256]

Answer:

4.578? hope this helps

Step-by-step explanation:

6 0

3 years ago

| 1. Add the polynomials.<br> (-14x2 + 6x – 5) + (x3 + 2x2 + 5)

docker41 [41]

Answer:

So I'm not completely sure what the x is standing for in this. Is it standing at a multiplication sign like -14 TIMES 2? Or is it -14x TIMES 2?

Step-by-step explanation:

7 0

3 years ago

What is the square root of 30

MArishka [77]

5.47722557505 :) have a nice day hope I helped

4 0

3 years ago

$20000 is invested in an account that earned 6% p.A. Compounding yearly for 3 years. The interest rate then went up to 8% p.A. F

GuDViN [60]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the first 3 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$20000\\ r=rate\to 6\%\to \frac{6}{100}\dotfill &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases}$

$\bf A=20000\left(1+\frac{0.06}{1}\right)^{1\cdot 3}\implies A=20000(1.06)^3\implies \boxed{A=2382.032} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the next 4 years}}$

$\bf A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$2382.032\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}$

$\bf A=2382.032\left(1+\frac{0.08}{1}\right)^{1\cdot 4}\implies A=2382.032(1.08)^4\implies \boxed{A\approx 3240.73} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{amount for this period}}{2382.032+3240.73}\implies 5622.762$

4 0

4 years ago

Lsplslsplpsplslpsllspslpslpslpslpsplspslpslpslpslpslsplsplspls

Ivan

Answer:

257.61

Step-by-step explanation:

FORMULA= C=2πr=2·π·41≈257.6106

there is a screenshot included too

6 0

3 years ago

Read 2 more answers