All categories

2 years ago

An area of 3.5m² is the same as 350000mm². True or false. Justify your answer

Mathematics

2 answers:

boyakko [2]2 years ago

5 0

no because 3.5 times 1000= 3500

hope this helps

Vsevolod [243]2 years ago

3 0

No, because to convert meter to millimeter you'll have to multiply the length value by 1000 which will give you 3500.

3.5 × 1000

= 3500.

have a great day!

You might be interested in

Ellie borrows money at a year simple rate of 6 1/2% after 4 years ellie owes 39 interest

kupik [55]

Answer: Ellie borrowed $150.

Step-by-step explanation:

I = PRT

39 = P x 0.065 x 4

39 = P x 0.26

Divide both sides by 0.26

150 = P

6 0

2 years ago

Riley and Erik have earned a total of 135 tokens to buy items in the school store. The ratio of the number of tokens that Riley

lisabon 2012 [21]

Answer:

Riley has 72 tokens

Step-by-step explanation:

System of Equations

We have two conditions for the tokens Riley and Erik have earned. Let's call x and y to the number of tokens of Riley and Erik respectively. The first condition states that

$x+y=135$

Solving for y

$y=135-x$

The second condition is that the ratio of the number of tokens that Riley had to the number of tokens that Erik has is 8 to 7. It's written as

$\displaystyle \frac{x}{y}=\frac{8}{7}$

Or equivalently

$7x=8y$

Replacing y from the first equation

$7x=8(135-x)$

Operating

$7x=1080-8x$

Simplifying

$15x=1080$

$\boxed{x=72}$

Riley has 72 tokens

3 0

3 years ago

Percentage change 42 increased to 145

dedylja [7]

The answer should be 103? because when you add 103 to 42 u should get 103%.

3 0

3 years ago

A circle cuts the x-axis at (4,0) and (14,0), and cuts the y-axis at (0,6) and (0,8). Find it's equation

gayaneshka [121]

It is not a circle but you can find the equation of the ellipse.
To do this we need to work out the major and minor radii and the centre
The centre is at (9, 7)
The major (y) radius is 1 and the minor (x) radius is 5
Therefore the equation is (x-9)/5 + (y-7)/1 = 1

5 0

3 years ago

Am i correct? calculus

Advocard [28]

Check the picture below.

now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer. That matters because the derivative of a constant is zero.

now, those are the values when the rocket is 30 feet up above.

$\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{y}{15}\implies tan(\theta )=\cfrac{1}{15}\cdot y
\\\\\\
\stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot\cfrac{d\theta }{dt}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}
\\\\\\
\boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\
-------------------------------\\\\
$

$\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\
-------------------------------\\\\
\cfrac{d\theta }{dt}=\cfrac{\left( \frac{1}{\sqrt{5}} \right)^2\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{1}{5}\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{11}{5}}{15}\implies \cfrac{d\theta }{dt}=\cfrac{11}{5}\cdot \cfrac{1}{15}
\\\\\\
\cfrac{d\theta }{dt}=\cfrac{11}{75}$

8 0

3 years ago