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2 years ago

What % of 40 is 15? Please put how u do it

Mathematics

2 answers:

andriy [413]2 years ago

6 0

Answer:

37.5

Step-by-step explanation:

so x% of 40 is 15

the equation for that is

x/100 * 40 = 15

40x/100 = 15

multiply 100 on both sides

40x = 1500

divide 40 on both sides

37.5

37.5% of 40 is 15

telo118 [61]2 years ago

6 0

Answer:

37.5

Step-by-step explanation:

Just do 15 out of 40 and divide 15 by 40 for the answer.

40/15

add a zero to the 15

40 goes into 150 3 times (40*3= 120)

150-120=30

40 doesn't go into 30 so you add another 0 which makes it 300

40 goes into 300 7 times

300-280= 20

40 doesn't go into 20 so add another zero (200)

40 goes into 200 5 times (40*5=200)

37.5

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3 years ago

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What is the area of this triangle ?

oee [108]

Answer:

Area of triangle is 9.88 units^2

Step-by-step explanation:

We need to find the area of triangle

Given E(5,1), F(0,4), D(0,8)

We will use formula:

$Area\,\,of\,\,triangle =\sqrt{s(s-a)(s-b)s-c)} \\where\,\, s = \frac{a+b+c}{2}$

We need to find the lengths of side DE, EF and FD

Length of side DE = a = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of side DE = a = $=\sqrt{(5-0)^2+(1-8)^2}\\=\sqrt{(5)^2+(-7)^2}\\=\sqrt{25+49}\\=\sqrt{74}\\=8.60$

Length of side EF = b = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of side EF = b = $=\sqrt{(0-5)^2+(4-1)^2}\\=\sqrt{(-5)^2+(3)^2}\\=\sqrt{25+9}\\=\sqrt{34}\\=5.8$

Length of side FD = c = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of side FD = c = $=\sqrt{(0-0)^2+(8-4)^2}\\=\sqrt{(0)^2+(4)^2}\\=\sqrt{0+16}\\=\sqrt{16}\\=4$

so, a= 8.60, b= 5.8 and c = 4

s = a+b+c/2

s= 8.6+5.8+4/2

s= 9.2

Area of triangle= $=\sqrt{s(s-a)(s-b)s-c)}\\=\sqrt{9.2(9.2-8.6)(9.2-5.8)(9.2-4)}\\=\sqrt{9.2(0.6)(3.4)(5.2)}\\=\sqrt{97.5936}\\=9.88$

So, area of triangle is 9.88 units^2

4 0

3 years ago

10/18=c/4.5<br> what is c

zhannawk [14.2K]

$\frac{10}{18}=\frac{c}{4,5}\\\\cross\ multiplication\\\\
4,5*10=18c\\\\
45=18c\ \ \ \ | divide\ by\ 18\\\\
c=\frac{45}{18}=2\frac{1}{2}\\\\\
\boxed{c=2\frac{1}{2}}$

7 0

3 years ago