All categories

3 years ago

Write 4 tens and 3 hundredths as a decimal

Mathematics

2 answers:

Rina8888 [55]3 years ago

8 0

This would be 40.03. The 4 is in the tens place and the 3 is in the hundredths place.

mamaluj [8]3 years ago

5 0

I believe it is written as 0.43

You might be interested in

In the diagram below, the radii of the two concentric circles are 3 centimeters and 7 centimeters, respectively.

True [87]

Answer:

125.6 cm²

Step-by-step explanation:

Area of the shaded region = area of larger circle - area of the smaller circle

Area of the smaller circle = πr²

π = 3.14, r = 3 cm

Area of smaller circle = 3.14*3² = 3.14*9 = 28.26 cm²

Area of larger circle = πr²

π = 3.14, r = 7

Area of larger circle = 3.14*7² = 3.14*49 = 153.86 cm²

Area of the shaded region = 153.86 - 28.26 = 125.6 cm²

5 0

3 years ago

Solve for t and find the value for t if I= 115, P=650, r=5%

Brrunno [24]

P=650
R= 5% annually= 0.05
T=1 month=1/12 year

Sl=Prt=650*0.05*1/12= $2.70
So the answer would be 2.7 or 2.70

5 0

3 years ago

Given: 3x - 5 = 4 (2x – 15) Prove: x = 11 Statement Reason

Shalnov [3]

Step-by-step explanation:

3x-5=8x-60

55=5x

x=11

6 0

2 years ago

1. What is the center and radius of the circle? (x-4)² + (x-7) ² - 49

tester [92]

Answer:

Center at (4, 7) and radius is √49, or 7

Step-by-step explanation:

Didn't you mean (x-4)² + (y-7) ² = 49?

Comparing (x-4)² + (y-7) ² = 49

to (x - h)^2 + (y - k)^2 = r^2, we see that the center is at (h, k) => (4, 7) and that the radius is √49, or 7.

4 0

3 years ago

*Asymptotes* g(x) =2x+1/x-3 Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .

Or!

I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).

$y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}$

$y=2+\frac{7}{x-3}$

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

6 0

3 years ago