Ask question

Ask question

All categories

3 years ago

15

a darts competition starts at 6:25, the competition lasts for three and a quarter hours. which watch face displays the time that

competition endes ?

1 answer:

algol133 years ago

3 0

They’re aren’t any pictures you provided but I can still assist you a little. It’s starts at 6:25 so you add 3 hours then add 15 minutes.

So add 6 hours + 3 hours = 9 hours, then do 25+15 = 40

So the competition ends at 9:40

You might be interested in

The graph of an exponential function is given. Which of the following is the correct equation of the function?

Nataliya [291]

Answer:

$y = 2.4^x \\ from \: intercepts \: using \: natural \: log( ln)$

6 0

3 years ago

Read 2 more answers

What is the perimeter of rhombus WXYZ?

oksian1 [2.3K]

<u>Answer</u>:

The perimeter of rhombus WXYZ is $4 \sqrt{13}$

<u>Step-by-step explanation:</u>

Step 1 :Finding length of XY

Distance formula = $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

here

$x_1$ = 5

$x_2$ =3

$y_1$ = -1

$y_2$ =2

XY = $\sqrt{(3-5)^2 +(2 -(-1))^2}$

XY = $\sqrt{(3-5)^2 +(2 +1))^2}$

XY = $\sqrt{(-2)^2 +(3))^2}$

XY = $\sqrt{4 +9}$

XY = $\sqrt{(13)}$

Step 2 :Finding length of YZ

Distance formula = $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

here

$x_1$ = 3

$x_2$ =5

$y_1$ = 2

$y_2$ =5

YZ = $\sqrt{(5-3)^2 +(5-2)^2}$

YZ = $\sqrt{(2)^2 +(3)^2}$

YZ = $\sqrt{4 +9}$

YZ = $\sqrt{(13)}$

Step 3 : :Finding length of ZW

Distance formula = $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

here

$x_1$ = 5

$x_2$ =7

$y_1$ = 5

$y_2$ =2

ZW = $\sqrt{(7-5)^2 +(5-2)^2}$

ZW = $\sqrt{(2)^2 +(3)^2}$

ZW = $\sqrt{4 +9}$

ZW = $\sqrt{(13)}$

Step 4 :Finding length of WX

Distance formula = $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

here

$x_1$ = 7

$x_2$ =5

$y_1$ = 2

$y_2$ = -1

WX = $\sqrt{(7-5)^2 +((-1)-2)^2}$

WX = $\sqrt{(2)^2 +(-3)^2}$

WX = $\sqrt{4 +9}$

WX = $\sqrt{(13)}$

Step 5: finding the perimeter of the rhombus

Perimeter= 4 X side

=> $4 \times \sqrt{13}$

=> $4 \sqrt{13}$

5 0

3 years ago

Read 2 more answers

At the annual music festival, the number of Bayview Middle School students competing in the band, chorus, and orchestra competit

Ghella [55]

Answer:

30 students

Step-by-step explanation:

The problem above can be solved using a Venn diagram :

Using the information given :

The number of students competing :

Band and chorus only (BnC) only = 5 - 4 = 1

Band and Orchestra (BnO) only = 6 - 4 = 2

Chorus and Orchestra (CnO) only = 7-4 = 3

Chorus only (C) = 13 - (1+4+3) = 5

Band only (B) = 14 - (1+4+2) = 7

Orchestra only (O) = 17 - (2+4+3) = 8

The Total Number of Bayview students who competed :

(8+7+5+3+2+1+4) = 30

6 0

3 years ago

Joyce baked a cake using 2 3/4 cups of brown sugar and 1 3/4 cups of powdered sugar. How many cups of sugar did she use in her c

BartSMP [9]

So she has used two kinds of sugar, brown and powdered.

She used $2 \frac{3}{4}$ cups of brown sugar and $1 \frac{3}{4}$ cups of powdered sugar: we need to add the two numbers:

$2 \frac{3}{4}$ + $1 \frac{3}{4}$ = 2+1+ $\frac{3}{4}+ \frac{3}{4}$ =3 $\frac{6}{4}=3+1+ \frac{2}{4} =4 \frac{2}{4}=4 \frac{1}{2}=4.5$

so the correct answer is 4.5

4 0

4 years ago

How do you work these out im confused

Vitek1552 [10]

$\boxed {45)}$ Multiply 520 students by 20% (0.20).

520 * 0.20 = 104

Multiply 104 students by 12.5% (0.125).

104 * 0.125 = 13

13 students were on the wrestling team.

$\boxed {46)}$ dimes (d) = n + .80, nickels (n) = n

Make the dimes and nickels add up to $1.40 in an equation.

(n + .80) + n = 1.40

Combine like terms.

2n + .80 = 1.40

Subtract .80 from both sides.

2n = 0.60

Divide both sides by 2.

n = 0.30

Divide this by 0.05 (how many nickels are in a dollar).

0.30/0.05 = 6

d = n + 0.80 ⇒ 0.30 + 0.80 = 1.10

Divide this by 0.10 (how many dimes are in a dollar).

1.10/0.10 = 11

There are 6 nickels and 11 dimes in the bank.

4 0

3 years ago