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3 years ago

Can someone please be nice enough and give me all the answer for these questions.

Mathematics

1 answer:

alexandr402 [8]3 years ago

5 0

1. 1/9 since you take all the "a"s in the words which is 2/18
2.2/9 same thing as above
3.18/18 since there is no "O" it's a trick question
4.3/9
5.5/6 since you don't count 6
6.1/2 and 1/2 coins are always 1/2 unless you flip twice
7.5/13
8.1/2 because 1,3,5 odd 2,4,6 even
Hope that helps

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3 years ago

Which ordered pair (x, y) satisfies the system of equations? 1/2 x − 1/4 y = 10 1/8 x − 1/8 y = 19

zlopas [31]

Answer:

-36,-192

Step-by-step explanation:

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3 years ago

Plz help me well mark brainliest if you are correct!.....

Alex17521 [72]

Answer:

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3 years ago

John sells hamburgers ($3) and cheeseburgers ($3.50). One afternoon he sells a total of 24 burger for $79. How many of these wer

ankoles [38]

John sold 10 hamburgers and 14 cheeseburgers.

Step-by-step explanation:

Given,

Cost of one hamburger = $3

Cost of one cheeseburger = $3.50

Total burgers sold = 24

Total revenue earned = $79

Let,

Number of hamburgers sold = x

Number of cheeseburgers sold = y

According to given statement;

x+y=24 Eqn 1

3x+3.50y=79 Eqn 2

Multiplying Eqn 1 by 3

$3(x+y=24)\\3x+3y=72\ \ \ Eqn\ 3$

Subtracting Eqn 3 from Eqn 2

$(3x+3.50y)-(3x+3y)=79-72\\3x+3.50y-3x-3y=7\\0.50y=7$

Dividing both sides by 0.50

$\frac{0.50y}{0.50}=\frac{7}{0.50}\\y=14$

Putting y=14 in Eqn 1

$x+14=24\\x=24-14\\x=10$

John sold 10 hamburgers and 14 cheeseburgers.

Keywords: linear equation, elimination method

Learn more about elimination method at:

#LearnwithBrainly

3 0

3 years ago

Es 18 de junio de 1815, las tropas napoleónicas se encuentran justo adelante, tu eres el ingeniero en balística y el general Wel

Airida [17]

Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

$(3) \, t =\dfrac{2}{\cos \theta}$

Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

3 0

3 years ago