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3 years ago

Choose the fraction that has not been reduced to simplest form.

Mathematics

2 answers:

Thepotemich [5.8K]3 years ago

8 0

first one since it’s 1/16 simplest form or 11/17 cuz it’s obv

denis23 [38]3 years ago

8 0

Answer:

3/57

Step-by-step explanation:

3/57 because 3/3*19=1/19

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If 8 apples can make 3 pies, then how many pies can you make with 24 apples?

serious [3.7K]

D answer is 9 , 24/8 = 3
3x3=9

4 0

3 years ago

Read 2 more answers

SELECT ALL OF THE EQUATIONS THAT

anygoal [31]

Answer:

x+9= 12 yes x = 3

4 + X= 9 no x = 5

X x 5 = 18 no x = 3.6

21 / x= 7 yes x= 3

hope this helps

4 0

3 years ago

After reading 80% of her e-mails in her inbox, Danette still has M unread e-mails. Which of the following expressions could repr

sergey [27]

Answer:

(Choice A) 5M

Step-by-step explanation:

Let total no. of e-mails in inbox be x

Since she read 80 % of mails i.e 80 % of x

⇒ $80\% \times x$

⇒ $\frac{80}{100}\times x$

⇒ $0.8 x$

So, she read 0.8x e-mails.

So, remaining email = total e-mails - read e-mails

$=x-0.8x$

Since we are given that Danette still has M unread e-mails.

⇒ $M=x-0.8x$

⇒ $M=x(1-0.8)$

⇒ $M=x\times 0.2$

⇒ $\frac{M}{0.2}=x$

⇒ $\frac{10M}{2}=x$

⇒ $5M=x$

Thus the expression represent the number of e-mails Danette had in her inbox before she started reading is 5M.

Hence Choice A is correct.

7 0

3 years ago

Due in an hour please help!!! :(

Anton [14]

Answer:

Step-by-step explanation:

1,7 are same side exterior

1,8 alternate exterior

2,4 none

3,5 same side int

2,6 corrosponding

4,6 same side int

4,7 is none

5,4 alternate int

8,5 vertical

8,7 alternate exterior

7 0

2 years ago

Use a surface integral to find the general formula for the surface area of a cone with height latex: h and base radius latex: a(

BlackZzzverrR [31]

We can parameterize this part of a cone by

$\mathbf s(u,v)=\left\langle u\cos v,u\sin v,\dfrac hau\right\rangle$

with $0\le u\le a$ and $0\le v\le2\pi$ . Then

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=\sqrt{1+\dfrac{h^2}{a^2}}u\,\mathrm du\,\mathrm dv$

The area of this surface (call it $\mathcal S$ ) is then

$\displaystyle\iint_{\mathcal S}\mathrm dS=\sqrt{1+\frac{h^2}{a^2}}\int_{v=0}^{v=2\pi}\int_{u=0}^{u=a}u\,\mathrm du\,\mathrm dv=a^2\sqrt{1+\frac{h^2}{a^2}}\pi=a\sqrt{a^2+h^2}\pi$

6 0

3 years ago