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2 years ago

If I make $1,000 per minute how much time will it take to make $1,000,000?

Mathematics

2 answers:

Dominik [7]2 years ago

8 0

Answer: I think it’ll take 1,000 minutes

Step-by-step explanation:

If you multiply 1,000 and 1,000 it’ll equal 1 million.

Hoped this helps.

ryzh [129]2 years ago

4 0

Answer:

it would take 1,000 minutes

Step-by-step explanation:

1,000,000/1,000=1,000

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ared has taken out a loan for his car. His monthly payments amount to $650. If his loan is for 36 months, and he paid a down pay

LekaFEV [45]

36*650 $ +400 $ =23800 $ is the total cost

5 0

3 years ago

The dimension of a mural pointed on the wall of a school is 10 ft by 22 ft. the art club wants to paint a new mural that is x fe

Westkost [7]

Answer:

18 ft and 30 ft

Step-by-step explanation:

The new dimensions will be:

10 + x and 22 + x

The area is 540

Using area equation we get:

(10 + x)(22 + x) = 540
x² + 32x + 220 = 540
x² + 32x = 320
x² + 2x*16 + 256 = 320 + 256
(x + 16)² = 576
x + 16 = √576 (Ignoring the negative root as the length is never negative)
x + 16 = 24
x = 8

The new dimensions are:

10 + 8 = 18 ft and
22 + 8 = 30 ft

6 0

2 years ago

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kiruha [24]

Answer:

i did this before xD

6 0

2 years ago

(X^2+ x– 12) (x^2+ 10x + 25)

yulyashka [42]

Solution:

(x² + x – 12)(x² + 10x + 25)
=> (x⁴ + 10x³ + 25x²) + (x³ + 10x² + 25x) + (-12x² - 120x - 300)
=> x⁴ + 10x³ + 25x² + x³ + 10x² + 25x - 12x² - 120x - 300
=> x⁴ + (10x³ + x³) + (25x² + 10x² - 12x²) + (25x - 120x) - 300
=> x⁴ + (11x³) + (23x²) + (-95x) - 300
=> x⁴ + 11x³ + 23x² - 95x - 300

The only term that has a x-variable is "-95x".

The coefficient of x is -95.

6 0

1 year ago

Read 2 more answers

Help math question derivative!

atroni [7]

Let $f(x)=\sec^{-1}x$ . Then $\sec f(x)=x$ , and differentiating both sides with respect to $x$ gives

$(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1$
$f'(x)=\dfrac1{\sec f(x)\tan f(x)}$

Now, when $x=\sqrt2$ , you get

$(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}$

You have $\sec^{-1}\sqrt2=\dfrac\pi4$ , so $\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2$ and $\tan\left(\sec^{-1}\sqrt2\right)=1$ . So $(\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}$

5 0

3 years ago