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3 years ago

7

Which ratio is equivalent to 3/7

1 answer:

Alex3 years ago

8 0

That is the simplest ratio because 3 and 7 do not have common factors.

But some similar ratios are

6/14

9/21

12/28

15/35

And so on

Hope I helped!

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At the fair, Kate found a strange machine with a sign on it labeled, "Enter a number." When she pushed the

Elanso [62]

Answer:

it reduce the number by 6 mark me brainlist

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3 years ago

If you raise 4 to the 214th power, what is the units digit of the result

Len [333]

$\begin{array}{lrlll}
4^1&\boxed{4}\\
4^2&1\boxed{6}\\
4^3&6\boxed{4}\\
4^4&25\boxed{6}\\
4^5&102\boxed{4}\\
4^6&409\boxed{6}
\end{array}\impliedby \textit{see the pattern for 4?}$

notice, it goes 4,6 then 4, 6 again, and so on and so on
based on whatever exponent it may have

now, you're being asked, what 214 will give? 4 or 6?

4 0

3 years ago

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katovenus [111]

Answer:

13.4

Step-by-step explanation:

7 0

3 years ago

Y=11 + x<br> 5x+2y= -6<br> What is the answer

Serjik [45]

X=3.1:

Step-by-step explanation:

5 0

3 years ago

Find a cubic function with the given zeros.

Fed [463]

Answer:

The correct option is D) $f(x) = x^3 + 2x^2 - 2x - 4$ .

Step-by-step explanation:

Consider the provided cubic function.

We need to find the equation having zeros: Square root of two, negative Square root of two, and -2.

A "zero" of a given function is an input value that produces an output of 0.

Substitute the value of zeros in the provided options to check.

Substitute x=-2 in $f(x) = x^3 + 2x^2 - 2x + 4$ .

$f(x) = x^3 + 2x^2 - 2x + 4\\f(x) = (-2)^3 + 2(-2)^2 - 2(-2) + 4\\f(x) =-8 + 2(4)+4 + 4\\f(x) =8$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 + 2x^2 + 2x - 4$ .

$f(x) = x^3 + 2x^2 + 2x - 4\\f(x) = (-2)^3 + 2(-2)^2 + 2(-2) - 4\\f(x) =-8+2(4)-4-4\\f(x) =-8$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 - 2x^2 - 2x - 4$ .

$f(x) = x^3 - 2x^2 - 2x - 4\\f(x) = (-2)^3 - 2(-2)^2 - 2(-2) - 4\\f(x) =-8-8+4-4\\f(x) =-16$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-2)^3+2(-2)^2 - 2(-2) - 4\\f(x) =-8+8+4-4\\f(x) =0$

Now check for other roots as well.

Substitute x=√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (\sqrt{2})^3+2(\sqrt{2})^2 - 2(\sqrt{2}) - 4\\f(x) =2\sqrt{2}+4-2\sqrt{2}-4\\f(x) =0$

Substitute x=-√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-\sqrt{2})^3+2(-\sqrt{2})^2 - 2(-\sqrt{2}) - 4\\f(x) =-2\sqrt{2}+4+2\sqrt{2}-4\\f(x) =0$

Therefore, the option is correct.

8 0

3 years ago