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4 years ago

Solve by factorising the equation 8x^2-30x-27=0

Mathematics

1 answer:

IrinaK [193]4 years ago

6 0

I got 9/2,-3/4. hope it helps

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What’s the answer?????????????

Natali [406]

Answer:

1) binomial

2) monomial

Step-by-step explanation:

mono-think one

bi-think two

tri-think thee

poly-1 or more

So 1 goes with binomial

and 2 goes with monomial

See if you can figure out 3 and 4

I can check them.

6 0

4 years ago

You invest $15,000 in a savings account with an annual interest rate of 2.5% in

almond37 [142]

Answer:

$\$16,990.62$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=5\ years\\ P=\$15,000\\ r=2.5\%=2.5/100=0.025\\n=4$

substitute in the formula above

$A=15,000(1+\frac{0.025}{4})^{4*5}$

$A=15,000(1.00625)^{20}$

$A=\$16,990.62$

5 0

3 years ago

What is the answer to -6(a+8)

Dafna11 [192]

Answer: -6a-48

Step-by-step explanation:

First -6 goes into a and that turns into -6a and then -6 goes into 8 and that turns into -48 so ur answer would be: -6a-48

ur welcome

8 0

3 years ago

What is the value of x 3x 2x 55

Juli2301 [7.4K]

Answer:

Step-by-step explanation:

3x+2x+55=180

5x+55=180

5x=180-55

5x=125

X=125/5

X=25

So the value of X is 25

8 0

3 years ago

The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence where an i

Anit [1.1K]

Answer:

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The $n^{th}$ term of geometric sequence is given by:

$a_n = ar^{n-1}$

$a_4 = 16 = ar^3\\a_6 = 4 = ar^5$

Dividing the two equations, we get,

$\dfrac{16}{4} = \dfrac{ar^3}{ar^5}\\\\4}=\dfrac{1}{r^2}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}$

the first term can be calculated as:

$16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128$

Thus, the required geometric sequence is

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

4 0

3 years ago