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3 years ago

Order the following numbers from least to greatest: 105, 10−99, 10−17, 1014, 10−5, 1030

Mathematics

2 answers:

neonofarm [45]3 years ago

8 0

If I understood your question correctly, it should be...
-89, -7, 5, 105, 1014, 1030

JulijaS [17]3 years ago

6 0

Nice mdbdhidjdbdjdhdjsiwj

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Factor the quadratic equation below to reveal the solutions. X^2+4x-21=-9

a_sh-v [17]

Answer:

x = 3 and x = -7

Step-by-step explanation:

The given quadratic equation is $x^2+4x-21=0$ . We need to find the solution of this equation.

If the equation is in the form of $ax^2+bx+c=0$ , then its solutions are given by :

$x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}$

Here, a = 1, b = 4 and c = -21

Plugging all the values in the value of x, such that :

$x=\dfrac{-b+ \sqrt{b^2-4ac} }{2a},\dfrac{-b- \sqrt{b^2-4ac} }{2a}\\\\x=\dfrac{-4+ \sqrt{(4)^2-4\times 1\times (-21)} }{2(1)},\dfrac{-4- \sqrt{(4)^2-4\times 1\times (-21)} }{2(1)}\\\\x=3, -7$

So, the solutions of the quadratic equation are 3 and -7.

6 0

3 years ago

A species of beetles grows 32% every year. Suppose 100 beetles are released into a field. How many beetles will there be in 10 y

siniylev [52]

Given that a species of beetles grows 32% every year.

So growth rate is given by

r=32%= 0.32

Given that 100 beetles are released into a field.

So that means initial number of beetles P=100

Now we have to find about how many beetles will there be in 10 years.

To find that we need to setup growth formula which is given by

$A=P(1+r)^n$ where A is number of beetles at any year n.

Plug the given values into above formula we get:

$A=100(1+0.32)^n$

$A=100(1.32)^n$

now plug n=10 years

$A=100(1.32)^{10}=100(16.0597696605)=1605.97696605$

Hence answer is approx 1606 beetles will be there after 20 years.

Now we have to find about how many beetles will there be in 20 years.

To find that we plug n=20 years

$A=100(1.32)^{20}=100(257.916201549)=25791.6201549$

Hence answer is approx 25791 beetles will be there after 20 years.

Now we have to find time for 100000 beetles so plug A=100000

$A=100(1.32)^n$

$100000=100(1.32)^n$

$1000=(1.32)^n$

$log(1000)=n*log(1.32)$

$\frac{\log\left(10000\right)}{\log\left(1.32\right)}=n$

33.174666862=n

Hence answer is approx 33 years.

4 0

3 years ago

Read 2 more answers

2200 dollars is placed in an account with an annual interest rate of 7.25%. How much

SCORPION-xisa [38]

Answer:

$3190.00

Step-by-step explanation:

7 0

3 years ago

Find the sum: 5 + 6. A. -11 B. -1 C. 1 D. 11

uranmaximum [27]

The answer would be 11

4 0

3 years ago

Read 2 more answers

(a2 − 6) and (a2 + 4) are the factors of

morpeh [17]

Just multiply them together and get a4-2a2-24.
(a2-6)(a2+4)
F: a4
O:4a2
I:-6a2
L:-24
Add together
a4-2a2-24

6 0

3 years ago