All categories

4 years ago

Pls help me in this question

Mathematics

1 answer:

musickatia [10]4 years ago

8 0

I believe it's 7.5 but I could be wrong.

You might be interested in

What is the result when 3x^4+2x^3-5x^2-6x-2 is divided by x+1

SCORPION-xisa [38]

Answer:

The answer to your question is: 3x³ - x² - 4x - 2

Step-by-step explanation:

Data

3x⁴ + 2x³ - 5x² - 6x - 2

x + 1

Process

Synthetic division

3 2 -5 - 6 -2 -1

-3 1 4 2

3 -1 -4 -2 0

Result 3x³ - x² - 4x - 2

5 0

3 years ago

Express the given expanded numeral as a Hindu-Arabic numeral. (8x102) +(4x10)(2x1)

Marysya12 [62]

Answer:

The Hindu-Arabic numeral form of the given expanded numeral is 842.

Step-by-step explanation:

The given expanded numeral is

$(8\times 10^2)+(4\times 10)+(2\times 1)$

We need to express the given expanded numeral as a Hindu-Arabic numeral.

According to Hindu-Arabic numeral the given expanded numeral is written as

$(8\times 10^2)+(4\times 10)+(2\times 1)=(8\times 100)+(4\times 10)+(2\times 1)$

On simplification we get,

$(8\times 10^2)+(4\times 10)+(2\times 1)=(800)+(40)+(2)$

$(8\times 10^2)+(4\times 10)+(2\times 1)=842$

Therefore the Hindu-Arabic numeral form of the given expanded numeral is 842.

5 0

3 years ago

PLZ HELP I HAVE A MATH TEST FINAL AND THIS IS GOING TO BE ON THE TEST 0.3(6z+7.2)=0.5(8z+4)-0.94

SOVA2 [1]

Answer:

oh god let me try help

Step-by-step explanation:

hm, I seemed to have gotten 7.42 my deepest apologies if it's incorrect!! :(

7 0

3 years ago

George’s parents are saving for his college fund. they put $5,000 into an interest bearing account with a compound interest rate

sergey [27]

so hmmm is compound interest, now, we'll be assuming the compounding period is per year, or annually, so it happens once per year.

$~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 5.5\%\to \frac{5.5}{100}\dotfill &0.055\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &15 \end{cases} \\\\\\ A=5000\left(1+\frac{0.055}{1}\right)^{1\cdot 15}\implies A=5000(1.055)^{15}$

7 0

2 years ago

Paki answer po plsss

Otrada [13]

Answer:

what is the answer

Step-by-step explanation:

8 0

3 years ago