All categories

3 years ago

The scale used on a map is 5 millimeters represents 32 kilometers. How many millimeters represent 160 kilometers?

Mathematics

1 answer:

goblinko [34]3 years ago

7 0

160/32= 5

5*5= 25 mm

25 mm is how many millimeters represents the kilometers.

I hope this helps!
~kaikers

You might be interested in

Find the least common denominator of -7/x and -5/6x^2

Jlenok [28]

That is 6x^2

Both x and 6x^2 will divide into 6x^2

7 0

3 years ago

S а о -3 S - 14 2 -S # -3 -2 1+ A + Lovezujo edos e seu molaguude 16 avuq UUDIUM

julsineya [31]

Answer:

Line Q is the line with the slope of 0.

Step-by-step explanation:

It does not increase or decrease.

6 0

3 years ago

A store sells mugs. Each mug costs the same amount. During a sale, the store reduces the price of each mug by $1.75. Wen spends

Ostrovityanka [42]

Answer:

$7.39

Step-by-step explanation:

Let the initial price be x

Then the reduced price is

x - 1.75

If 5 mugs cost $28.20, we have equation:

(x - 1.75)*5 = 28.20
x - 1.75 = 28.20/5
x - 1.75 = 5.64
x = 1.75 + 5.64
x = 7.39

Initial price for 1 mug was $7.39

4 0

3 years ago

Read 2 more answers

What is the number gonna be if you divide a negative by a negative?

mina [271]

If you divide a negative by a negative, your result will be positive.
For example, -5/-1=5.

8 0

3 years ago

Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)

viktelen [127]

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}$

The first line segment can be parameterized by $\mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle$ with $0\le t\le1$ . Denote this first segment by $C_1$ . Then

$\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt$
$=108$

The second line segment ( $C_2$ ) can be described by $\mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle$ , again with $0\le t\le1$ . Then

$\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt$
$=-\dfrac{229}3$

Finally,

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3$

5 0

3 years ago