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3 years ago

What is 5/16 expressed as a percent? Enter your answer in the box.

Mathematics

1 answer:

kenny6666 [7]3 years ago

4 0

Answer:

31.25 %

Step-by-step explanation:

The fraction bar also means divide:

5 / 16

5 Divided By 16

0.3125

0.3125 * 100

= 31.25

- I.A -

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Rationalize the denominator of 4/14 root 2?

Katarina [22]

First of all, you can simplify the 4 at the numerator and the 14 at the denominator (they're both multiple of 2):

$\dfrac{4}{14\sqrt{2}} = \dfrac{2\cdot 2}{2\cdot 7\cdot\sqrt{2}} = \dfrac{2}{7\sqrt{2}}$

Now, rationalize a denominator means that you have to get rid of the square root, in order to have an integer denominator.

To do so, remember that you can always multiply any number by 1 without changing its value, and you can always think of 1 as a fraction where numerator and denominator are equal:

$\dfrac{2}{7\sqrt{2}} = \dfrac{2}{7\sqrt{2}}\cdot 1 = \dfrac{2}{7\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{2\sqrt{2}}{7\cdot 2} = \dfrac{\sqrt{2}}{7}$

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3 years ago

‼️HELP FAST‼️<br> What are the zeros of this function?

nika2105 [10]

Answer:

Step-by-step explanation:

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3 years ago

How much money will you have if you started with $45 and put it in an account that earned 10% every year for 20 years

Veronika [31]

Probably it's all wrong
4.5 = 10 %If is the same then 4.5*20+45=135
If not then
FV=P(1+r)n45* (1+0.1)20 = 45*6.727=302.737

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2 years ago

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

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3 years ago

You and a friend go to the local taco joint for lunch. You order three tacos and three burritos and your bill totals $11.25. You

anzhelika [568]

Answer:

Step-by-step explanation:

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2 years ago