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2 years ago

Factorise. Help would be massively appreciated:)

Mathematics

1 answer:

kenny6666 [7]2 years ago

4 0

Answer:

A. 2x(x+3)

B. 2y(y-4)

C. 5p(p+2)

D. 7c(c-3)

E. 3x(2x+3)

I hope this helps

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The half-life of a radioactive kind of tellurium is 25 minutes. If you start with 88 grams of it, how much will be left after 50

LUCKY_DIMON [66]

Notice that 50 minutes is twice the half-life.

After 25 minutes, half of the starting amount remains, so that 88g decays to 44g.

After another 25 minutes (a total of 50 minutes), the 44g will decay to 22g.

5 0

3 years ago

I need help! Math isn’t my best subject

Dmitriy789 [7]

Answer:

A and c

Step-by-step explanation:

multiply the 3 to both parts of equation

6 0

2 years ago

Read 2 more answers

The population of a town in 2007 is 113, 505

ELEN [110]

from 2007 to 2012 is only 5 years, so we can see this as a compound interest with a rate of 1.2% per annum for 5 years, so

$~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$113505\\ r=rate\to 1.2\%\to \frac{1.2}{100}\dotfill &0.012\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per year, thus once} \end{array}\dotfill &1\\ t=years\dotfill &5 \end{cases} \\\\\\ A=113505\left(1+\frac{0.012}{1}\right)^{1\cdot 5}\implies A=113505(1.012)^5\implies A\approx 120481$

7 0

1 year ago

Let p = the product of all the odd integers between 500 and 598, and let q = the product of all the odd integers between 500 and

tia_tia [17]

$p=501\cdot503\cdot\cdots\cdot597$

$q=\underbrace{501\cdot503\cdot\cdots\cdot597}_p\cdot599\cdot601$

So we have $q=359,999p$ . Then

$\dfrac1p+\dfrac1q=\dfrac{359,999}{359,999p}+\dfrac1q=\dfrac{359,999+1}q=\dfrac{360,000}q$

and the answer is D.

4 0

3 years ago

Find all possible values of x in the similar triangles pqr stu

antoniya [11.8K]

If triangles PQR and STU are similar then PQ corresponds to ST and PR corresponds to SU. Therefore, PQ/ST=PR/SU
Considering that, PQ= 7-x, ST= 13-x, PR= x²+5 and SU= x² +20
therefore, (7-x)/(13-x)= (x²+5)/(x²+20)
cross multiplying,
7x² +140-x³+20x =13x²+65-x³-5x
combining the like terms,
6x² +15x -75=0
solving for x,
x = 5/2 or -5

5 0

2 years ago