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3 years ago

Round 12,309,453 to the nearest million

Mathematics

2 answers:

Oliga [24]3 years ago

7 0

Answer:

12,000,000

Step-by-step explanation:

Mnenie [13.5K]3 years ago

4 0

So 12,309,453 rounded to the nearest millions

is 2,000,000 because

2 is less and 3 is also less

so it's only 2

I hope this helps

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Tatiana [17]

It should be the second one

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2 years ago

$4000 is invested in a mutual fund that pays annual interest compounded quarterly for 15 years. What annual interest is needed t

Neporo4naja [7]

Answer:

The interest needed to reach that value of $6000 is 2.75%

Step-by-step explanation:

Given as :

The principal that invested in mutual fund = p = $4000

The time period = t = 15 years

The Amount after 15 years = A = $6000

Let The interest needed to reach that value = r%

Now, According to question

From Compound Interest method

Amount = principal × $(1+\dfrac{\textrm rate}{100})^{\textrm time}$

Or, A = p × $(1+\dfrac{\textrm r}{100})^{\textrm t}$

Or, $6000 = $4000 × $(1+\dfrac{\textrm r}{100})^{\textrm 15}$

or, $\dfrac{6000}{4000}$ = $(1+\dfrac{\textrm r}{100})^{\textrm 15}$

Or, $\frac{3}{2}$ = $(1+\dfrac{\textrm r}{100})^{\textrm 15}$

or, 1.5 = $(1+\dfrac{\textrm r}{100})^{\textrm 15}$

or, $1.5^{\frac{1}{15}}$ = $(1+\dfrac{\textrm r}{100})$

or, 1.0275 = $(1+\dfrac{\textrm r}{100})$

or, 1.0275 - 1 = $\dfrac{r}{100}$

or, 0.0275 = $\dfrac{r}{100}$

∴ r = 0.0275 × 100

I.e r = 2.75

So, The interest needed to reach that value = r = 2.75%

Hence,The interest needed to reach that value of $6000 is 2.75% Answer

3 0

2 years ago

Solve.

Ksenya-84 [330]

Solve.

x² + 4x + 4 = 18

A x=−4±3√ 2

B x=2±3√ 2

C x=4±9√ 2

D x=−2±3√2

 $x^2 + 4x + 4 = 18 \\ \\ x^2+4x+4-18= 0 \\ \\ x^2+4x-14=0 \\ \\ x_1_y_2= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} \qquad a= 1\qquad b= 4\qquad c= -14 \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{4^2-4(1)(-14)} }{2(1)} \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{16-(-56)} }{2} \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{72} }{2} \\ \\ \\ x_1= \dfrac{-4+ \sqrt{72} }{2} \qquad\qquad x_2= \dfrac{-4+ \sqrt{72} }{2}\\ \\ \\ x_1= \dfrac{-4+ \sqrt{2^3*3^2} }{2} \qquad\qquad x_2= \dfrac{-4- \sqrt{2^3*3^2} }{2}$

 $x_1= \dfrac{-4+2*3 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4- 2*3\sqrt{2} }{2} \\ \\ \\ x_1= \dfrac{-4+6 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4- 6\sqrt{2} }{2} \\ \\ \\ x_1= \dfrac{-4}{2} + \dfrac{6 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4}{2}- \dfrac{ 6\sqrt{2} }{2} \\ \\ \\ x_1= -2 + 3 \sqrt{2} \qquad\qquad\quad x_2= -2- 3\sqrt{2} \\ \\ \\ \boxed{x= -2\pm 3\sqrt{2} } \to D)$

7 0

3 years ago

Read 2 more answers

Can someone please help me with this? I’ll give brainiest to the best answer

Olin [163]

What’s the problem ?

7 0

2 years ago

Question 2 of 10

attashe74 [19]

Answer:

This is true

LMN is a right triangle

7 0

3 years ago

Round 12,309,453 to the nearest million​

Round 12,309,453 to the nearest million