All categories

2 years ago

Unit 1 Review

Mathematics

1 answer:

Rzqust [24]2 years ago

3 0

Answer:

403

Step-by-step explanation:

divide 2.75 and 1,1109 and you will get your answer

You might be interested in

What is the percent of decrease from 100 to 49?<br><br> Write your answer using a percent sign (%).

slavikrds [6]

Answer:

Step-by-step explanation:

Decrease = 100 - 49 = 51

Percentage of decrease =

$=\frac{51}{100}*100$

= 51%

6 0

3 years ago

What is a situation in which you might locate points on a coordinate grid

777dan777 [17]

A math test is a situation in which you would need to do this :)

6 0

2 years ago

Read 2 more answers

Helppp can someone solve this please

Norma-Jean [14]

Answer:

$\displaystyle x=-15$

Step-by-step explanation:

<u>Solution Of A System Of Equations </u>

A system of linear equations is given as

$\displaystyle \left\{\begin{matrix}ax+by=c\\ dx+ey=f\end{matrix}\right.$

There are many methods to solve them. We will use the method of reduction

The given system is

$\displaystyle \left\{\begin{matrix}2x+3y=45\\ x+y=10\end{matrix}\right.$

Multiplying the second equation by -3

$\displaystyle \left\{\begin{matrix}2x+3y=45\\ -3x-3y=-30\end{matrix}\right.$

Adding the resulting equations

$\displaystyle -x=15$

$\displaystyle x=-15$

5 0

2 years ago

Can someone help me please thank you

icang [17]

Answer:

c and d

Step-by-step explanation:

a slae price cant be one whole so those are the only reasonable options.

3 0

2 years ago

Read 2 more answers

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