All categories

3 years ago

Given the quadratic equation below, determine which situation it could represent? x2 -4x = 60

Mathematics

1 answer:

yarga [219]3 years ago

4 0

Answer:

The solution is x = 10 and x = -6. This quadratic could be represented in factored form as (x - 10)(x + 6) or on a graph with x-intercepts at (10,0) and (-6,0).

Step-by-step explanation:

To solve the quadratic, write the quadratic in standard form and factor the equation.

x² - 4x = 60

x² - 4x - 60 = 0

(x - 10)(x + 6) = 0

x = 10 and x = -6

You might be interested in

Fifty-four and three hundredths written in standard form is 54.3

jasenka [17]

False, should be 54.03

3 0

2 years ago

15 times as much as 1 fith of 12

Mila [183]

The answer is 36.

12 / 5 = 2.4

2.4 • 15 = 36

I hope this helps!

8 0

3 years ago

Help asap !! will get branliest.

kompoz [17]

Answer:

0.8

Step-by-step explanation:

(2 x 10^-4)

= 0.0002

(4x10^3)

= 4000

0.0002 x 4000 = 0.8

7 0

3 years ago

What is the solution of n/6 = 36

igor_vitrenko [27]

Answer:

n=216.

Step-by-step explanation:

Very simple problem - we just have to multiply both sides by 6, which will get rid of the 6 on the left.

36 x 6 = 216

8 0

2 years ago

Find the sum of the geometric series 40 + 40(1.005) + 40(1.005)^2 + ⋯ + 40(1.005)^11.

KiRa [710]

Answer:

The sum is 493.4

Step-by-step explanation:

In order to find the value of the sum, you have to apply the geometric series formula, which is:

$\sum_{i=1}^{n} ar^{i-1} = \frac{a(1-r^{n})}{1-r}$

where i is the starting point, n is the number of terms, a is the first term and r is the common ratio.

The finite geometric series converges to the expression in the right side of the equation. Therefore, you don't need to calculate all the terms. You can use the expression directly.

In this case:

a=40

b= 1.005

n=12 (because the first term is 40 and the last term is 40(1.005)^11 )

Replacing in the formula:

$\frac{a(1-r^{n})}{1-r} = \frac{40(1-1.005^{12})}{1-1.005}$

Solving it:

The sum is 493.4

4 0

3 years ago