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PtichkaEL [24]
3 years ago
13

A school contains 357 boys and 323 girls.

Mathematics
1 answer:
PolarNik [594]3 years ago
3 0

Answer:

19/40

Step-by-step explanation:

just divide the number of girls by the total number of students.

323/680 = 19/40 or 0.48

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In J, K is the midpoint. The coordinates of ) are (2, 2), and the coordinates of Kare (10, 11). What are the coordinates of L?
DiKsa [7]

Answer:

L(18, 20)

Step-by-step explanation:

In JL, K is the midpoint. The coordinates of J are (2, 2), and the

coordinates of K are (10, 11). What are the coordinates of L?

Solution:

If O(x, y) is the midpoint between two points A(x_1,y_1) and B(x_2,y_2). The equation to determine the location of O is given by:

x=\frac{x_1+x_2}{2} \\\\y=\frac{y_1+y_2}{2}

Since JL is a line segment and K is the midpoint. Given the location of J as (2, 2) and K as (10, 11). Let (x_2,y_2) be the coordinate of L. Therefore:

10=\frac{2+x_2}{2} \\\\20=2+x_2\\\\x_2=18

11=\frac{2+y_2}{2} \\\\22=2+y_2\\\\y_2=20

Therefore L = (18, 20)

4 0
3 years ago
Simplified form of 12x-1 = 2(6+5x) +7
horrorfan [7]
X=10



12x-1=12+10x+7
2x=20
X=10
4 0
3 years ago
You can buy 5 cans for green beans at the village market for 3.70. You can buy 10 of the same cans of beans at sams club for 4.1
il63 [147K]

the rate for 5 cans for $3.70 is 0.74

the rate for 10 cans for $4.10 is 0.41

so the better buy is 10 cans for $4.10 at sams

4 0
3 years ago
please help expert help please! giving 16 points please help! will mark brainliest! Please do question 3!!! love y’all!
likoan [24]

Answer:

12 meters i think but I may be wrong

5 0
2 years ago
Find all solutions to
BARSIC [14]

Answer:

x= 0 , \frac{1}{14} , \frac{-1}{12}

Step-by-step explanation:

Given, equation is \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x}. →→→ (1)

Now, by cubing the equation on both sides, we get

( \sqrt[3]{15x-1} + \sqrt[3]{13x+1} )³ = (4\sqrt[3]{x})³

⇒ (15x-1) + (13x+1) + 3×\sqrt[3]{15x-1}×\sqrt[3]{13x+1} (\sqrt[3]{15x-1} + \sqrt[3]{13x+1}) = 64 x.

⇒ 28x + 3×\sqrt[3]{15x-1}×\sqrt[3]{13x+1} (4\sqrt[3]{x}) = 64x.        

(since from (1),  \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x})

⇒ 12× \sqrt[3]{15x-1}×\sqrt[3]{13x+1} (\sqrt[3]{x})= 36x.

⇒ 3x = \sqrt[3]{(15x-1)(13+1)(x)}.

Now, once again cubing on both sides, we get

(3x)³ = (\sqrt[3]{(15x-1)(13+1)(x)})³.

⇒ 27x³ = (15x-1)(13x+1)(x).

⇒ 27x³ = 195x³ + 2x² - x

⇒ 168x³ + 2x² - x = 0

⇒ x(168x² + 2x -1) = 0

⇒ by, solving the equation we get ,

x = 0 ; x = \frac{1}{14} ; x = \frac{-1}{12}

therefore, solution is x= 0 , \frac{1}{14} , \frac{-1}{12}

7 0
3 years ago
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