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3 years ago

Computer hardware is best described as

Mathematics

1 answer:

likoan [24]3 years ago

8 0

The physical parts of a computer. *Software* refers to programs that actually run on the hardware.

You might be interested in

Which is the distance from point A to B?

Aleks04 [339]

Answer:

D) radical 26

Step-by-step explanation:

So this is a question where you need to find the hypotenuse of a triangle

the sides are 1 and 5

you can use Pythagoras (a^2+b^2=c^2)

substitute the values in

1^2+5^2=c^2

simplify the exponents

1+25=c^2

add them together

26=c^2

take the square root of 26

the answer is radical 26

5 0

2 years ago

The pattern for the kite calls for 169 square inches of fabric. What is the

lys-0071 [83]

Answer:

13 inches.

Step-by-step explanation:

Square root 169 and you will get the answer 13. Hope this helped.

3 0

2 years ago

Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.

Let's start with setting the $y$ 's equal to find those $x$ 's for which the $y$ 's are the same.

$\log(x)=\frac{1}{2}\log(x+1)$

By power rule:

$\log(x)=\log((x+1)^\frac{1}{2})$

Since $\log(u)=\log(v)$ implies $u=v$ :

$x=(x+1)^\frac{1}{2}$

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

This is a quadratic equation.

Subtract $(x+1)$ on both sides:

$x^2-(x+1)=0$

$x^2-x-1=0$

Comparing this to $ax^2+bx+c=0$ we see the following:

$a=1$

$b=-1$

$c=-1$

Let's plug them into the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$x=\frac{1 \pm \sqrt{1+4}}{2}$

$x=\frac{1 \pm \sqrt{5}}{2}$

So we have the solutions to the quadratic equation are:

$x=\frac{1+\sqrt{5}}{2}$ or $x=\frac{1-\sqrt{5}}{2}$ .

The second solution definitely gives at least one of the logarithm equation problems.

Example: $\log(x)$ has problems when $x \le 0$ and so the second solution is a problem.

So the $x$ where the equations intersect is at $x=\frac{1+\sqrt{5}}{2}$ .

Let's find the $y$ -coordinate.

You may use either equation.

I choose $y=\log(x)$ .

$y=\log(\frac{1+\sqrt{5}}{2})$

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

6 0

2 years ago

Which statement is true?

diamong [38]

3 is correct , base is never equal to 1 cause all of solves of logarithm will be 0

7 0

3 years ago

Read 2 more answers

Guy's y'all plzz answer this who knows about it...and also exam is near

Yanka [14]

Answer:

Step-by-step explanation:

Okay so I will explain the first one.

2 to the power of 4 = 2x2x2x2

2 to the power of 6 = 2x2x2x2x2x2

2 to the power of 4 = 2x2x2x2

2^4 = 16

2^6 = 64

When you are multiplying powers you add.

When you subtract you takeaway.

So you would do

4+6+4

= 14

Meaning 2 to the power of 14

2 $x^{14}$

So what is 2 to the power of 14.

You would do

2x2x2x2x2x2x2x2x2x2x2x2x2x2

The answer is 16384

You can also use a calculator

GOOD LUCK FOR YOUR TEST!

7 0

3 years ago