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3 years ago

Plz help me I will be giving a brainliest to the first person to answer

Mathematics

2 answers:

pishuonlain [190]3 years ago

8 0

15 is a positive value, -3 is a negative value.

A positive value is always greater than a negative value.

The answer is: 15 > -3

jeyben [28]3 years ago

8 0

Answer:

15 > -3

Step-by-step explanation:

15 is greater than -3 because 15 is greater than 0 and -3 is less than 0 this means 15 is greater than -3.

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12 decreased by some number h

Nimfa-mama [501]

Answer:

12 - h

Step-by-step explanation:

5 0

2 years ago

What is 1/4 X 5/16 equal

VashaNatasha [74]

$\frac{1}{4}$ x $\frac{5}{16}$

1 x 5 = 5
16 x 4 = 64

$\frac{5}{64}$ is the answer

In simplest form the answer would be $\frac{5}{64}$ because it cant be divided any farther.

6 0

3 years ago

Read 2 more answers

ipn [44]

Answer:

50 soldiers must be transferred elsewhere.

Step-by-step explanation:

We solve this question by proportions, using a rule of three.

As the number of soldiers decrease, the provisions last for more time. This means that the measures are inversely proportional, and we have an inverse rule of three, using line multiplication, instead of cross.

30 days of provisions - 200 soldiers

40 days of provisions - x soldiers

$40x = 200*30$

Simplifying by 40

$x = 5*30 = 150$

The provisions will last for 40 days with 150 soldiers, which means that 200 - 150 = 50 soldiers must be transferred elsewhere.

4 0

2 years ago

How do you solve the problem 2p = 60

d1i1m1o1n [39]

You have to separate the 2 and p so divide 60 by 2 which will leave p=30 and now if you plug it in you see 2 times 30 equals 60 :)

4 0

3 years ago

Find the derivative of <br> |x|/(x-1)

ki77a [65]

$\bf \cfrac{|x|}{x-1}\iff \cfrac{\sqrt{x^2}}{x-1}\iff \cfrac{(x^2)^{\frac{1}{2}}}{x-1}
\\\\\\
\textit{using the quotient rule}
\\\\\\
\cfrac{dy}{dx}=\cfrac{\frac{1}{2}(x^2)^{-\frac{1}{2}}\cdot 2x(x-1)-(x^2)^{\frac{1}{2}}\cdot 1}{(x-1)^2}$

$\bf \cfrac{dy}{dx}=\cfrac{\frac{x(x-1)-(x^2)^{\frac{1}{2}}}{(x^2)^{\frac{1}{2}}}}{(x-1)^2}\implies \cfrac{dy}{dx}=\cfrac{x(x-1)-(x^2)^{\frac{1}{2}}}{(x^2)^{\frac{1}{2}}(x-1)^2}
\\\\\\\cfrac{dy}{dx}=\cfrac{x(x-1)-|x|}{|x|(x-1)^2}$

4 0

3 years ago

Plz help me I will be giving a brainliest to the first person to answer​

Plz help me I will be giving a brainliest to the first person to answer