PLS help,269 divided by 4

Ann took a taxi home from the airport. The taxi fare was

Zinaida [17]

Answer:

She would have traveled 21 miles, because it you multiply 2.1 by 21 you get 44.1 and then Ann gave a $5 tip so that would bring it up to $49.10.

Step-by-step explanation:

6 0

2 years ago

What is the slope of the function?<br> -10<br> -5<br> 5<br> 10

bixtya [17]

Answer:

m= 1/2

Step-by-step explanation:

3 0

2 years ago

April worked 1 1/2 times as long on her math project as did Carl. Debbie worked 1 1/4 times as long as Sonia. Richard worked 1 3

vlada-n [284]

Answer:

Student Hours worked

April. $7\frac{7}{8} \ hrs$

Debbie. $8\frac{1}{8}\ hrs$

Richard. $7\frac{19}{24}\ hrs$

Step-by-step explanation:

Some data's were missing so we have attached the complete information in the attachment.

Given:

Number of Hours Carl worked on Math project = $5\frac{1}{4}\ hrs$

$5\frac{1}{4}\ hrs$ can be Rewritten as $\frac{21}{4}\ hrs$

Number of Hours Carl worked on Math project = $\frac{21}{4}\ hrs$

Number of Hours Sonia worked on Math project = $6\frac{1}{2}\ hrs$

$6\frac{1}{2}\ hrs$ can be rewritten as $\frac{13}{2}\ hrs$

Number of Hours Sonia worked on Math project = $\frac{13}{2}\ hrs$

Number of Hours Tony worked on Math project = $5\frac{2}{3}\ hrs$

$5\frac{2}{3}\ hrs$ can be rewritten as $\frac{17}{3}\ hrs.$

Number of Hours Tony worked on Math project = $\frac{17}{3}\ hrs.$

Now Given:

April worked $1\frac{1}{2}$ times as long on her math project as did Carl.

$1\frac{1}{2}$ can be Rewritten as $\frac{3}{2}$

Number of Hours April worked on math project = $\frac{3}{2}$ $\times$ Number of Hours Carl worked on Math project

Number of Hours April worked on math project = $\frac{3}{2}\times \frac{21}{4} = \frac{63}{8}\ hrs \ \ Or \ \ 7\frac{7}{8} \ hrs$

Also Given:

Debbie worked $1\frac{1}{4}$ times as long as Sonia.

$1\frac{1}{4}$ can be Rewritten as $\frac{5}{4}$ .

Number of Hours Debbie worked on math project = $\frac{5}{4}$ $\times$ Number of Hours Sonia worked on Math project

Number of Hours Debbie worked on math project = $\frac{5}{4}\times \frac{13}{2}= \frac{65}{8}\ hrs \ \ Or \ \ 8\frac{1}{8}\ hrs$

Also Given:

Richard worked $1\frac{3}{8}$ times as long as tony.

$1\frac{3}{8}$ can be Rewritten as $\frac{11}{8}$

Number of Hours Richard worked on math project = $\frac{11}{8}$ $\times$ Number of Hours Tony worked on Math project

Number of Hours Debbie worked on math project = $\frac{11}{8}\times \frac{17}{3}= \frac{187}{24}\ hrs \ \ Or \ \ 7\frac{19}{24}\ hrs$

Hence We will match each student with number of hours she worked.

Student Hours worked

April. $7\frac{7}{8} \ hrs$

Debbie. $8\frac{1}{8}\ hrs$

Richard. $7\frac{19}{24}\ hrs$

5 0

3 years ago

Read 2 more answers

1 Hector puts a fence around his rectangular

prisoha [69]

Answer:

48 feet of fence

Step-by-step explanation:

A rectangle has 4 sides, the parallel sides are equal to each other in length.

If Hector used 14 feet of fence on the long side, he used 28 feet in total for both long sides.

If Hector used 10 feet for the width/short side, he used 20 feet in total for both short sides.

Put simply: (14 x 2) + (10 x 2) = 48 feet of fence in total

4 0

2 years ago

) The density of oil in a circular oil slick on the surface of the ocean at a distance of r meters from the center of the slick

Feliz [49]

Answer:

Therefore the mass of the of the oil is 409.59 kg.

Step-by-step explanation:

Let us consider a circular disk. The inner radius of the disk be r and the outer diameter of the disk be (r+Δr).

The area of the disk

=The area of the outer circle - The area of the inner circle

= $\pi (r+\triangle r)^2- \pi r^2$

$=\pi [r^2+2r\triangle r+(\triangle r)^2-r^2]$

$=\pi [2r\triangle r+(\triangle r)^2]$

Since (Δr)² is very small, So it is ignorable.

∴ $A=2\pi r\triangle r$

The density $\delta (r)= \frac{40}{1+r^2}$

We know,

Mass= Area× density

$=(2r \pi \triangle r)(\frac{40}{1+r^2}})$

Total mass $M=\sum_{i=1}^n \frac{80r_i\pi }{1+r^2}\triangle r_i$

Therefore

$\sum_{i=1}^n \frac{80r_i\pi }{1+r^2}\triangle r_i=\int_0^5 \frac{80r\pi }{1+r^2}dr$

$=40\pi[ln(1+r^2)]_0^5$

$=40\pi [ln(1+5^2)-ln(1+0^2)]$

$=40\pi ln(26)$

= 409.59 kg (approx)

Therefore the mass of the of the oil is 409.59 kg.

3 0

2 years ago

PLS help,269 divided by 4​

PLS help,269 divided by 4