All categories

3 years ago

What is 46.08 divided by 0.8

Mathematics

2 answers:

TiliK225 [7]3 years ago

6 0

The answer is 5.76...

professor190 [17]3 years ago

4 0

The answer is fifty-seven point six.... 57.6

You might be interested in

A triangle has side lengths of (q+r)(q+r) centimeters, (5q-10s)(5q−10s) centimeters, and (5s-7r)(5s−7r) centimeters. Which expre

Stolb23 [73]

The expression represents the perimeter, in centimeters, of the triangle is 6q - 6r - 5s

<h3>What is the perimeter?</h3>

The formula for perimeter of a triangle is expressed as;

Perimeter = a + b + c

Where a , b and c are the lengths of its side

(q+r)
(5q-10s)
(5s-7r)

Now, let's substitute the values

Perimeter = (q + r) + (5q - 10s) + (5s - 7r)

expand the bracket

Perimeter = q + r + 5q - 10s + 5s - 7r

collect like terms

Perimeter = q + 5q + r - 7r -10s + 5s

Add like terms

Perimeter = 6q - 6r - 5s

Thus, the expression represents the perimeter, in centimeters, of the triangle is 6q - 6r - 5s

Learn more about perimeter here:

brainly.com/question/24571594

#SPJ1

8 0

1 year ago

Solve the equation. 2x + 12 = 2(x + 6)

Vilka [71]

Answer:

Infinitely many solutions

Step-by-step explanation:

Step 1- Distribute into the parenthesis.

2x+12= 2(x+6)

2x+12= (2)x+(2)6

Step 2- Multiply

2x+12= 2x+12

Since they are equal, they are infinite.

5 0

3 years ago

Find the area of each sector.

sergey [27]

Answer:

D. $\frac{525\pi}{8}$ m²

Step-by-step explanation:

Area of the sector = ½*r²∅

Where,

Radius (r) = 15 m

Angle in radians (∅) = $\frac{7\pi}{12}$

Plug in the values into the equation

Area of sector = $\frac{1}{2} * 15^2 * \frac{7\pi}{12}$

Area of sector = $\frac{1}{2} * 225 * \frac{7\pi}{12}$

Area of sector = $\frac{1*225*7\pi}{2*12}$

Area of sector = $\frac{1,575\pi}{24}$

Simplify

Area of sector = $\frac{525\pi}{8}$

5 0

3 years ago

What is x in 1/150/1/200x2=x

choli [55]

Answer:

6.66666667E-5

hope this helped a bit!

5 0

3 years ago

Suppose x is a real number and epsilon > 0. Prove that (x - epsilon, x epsilon) is a neighborhood of each of its members; in

kaheart [24]

Answer:

See proof below

Step-by-step explanation:

We will use properties of inequalities during the proof.

Let $y\in (x-\epsilon,x+\epsilon)$ . then we have that $x-\epsilon$ . Hence, it makes sense to define the positive number delta as $\delta=\min\{x+\epsilon-y,y-(x-\epsilon)\}$ (the inequality guarantees that these numbers are positive).

Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ , and if we prove this, we are done. To prove it, let $z\in (y-\delta,y+\delta)$ , then $y-\delta$ . First, $\delta \leq y-(x-\epsilon)$ then $-\delta \geq -y+x-\epsilon$ hence $z>y-\delta \geq x-\epsilon$

On the other hand, $\delta \leq x+\epsilon-y$ then $z$ hence $z$ . Combining the inequalities, we have that $x-\epsilon$ , therefore $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ as required.

3 0

4 years ago