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2 years ago

You are given two algebraic equations, y = 3x, and 2y = x + 20. Solve for x and y.

Mathematics

1 answer:

Yuri [45]2 years ago

7 0

Answer:

x= -4 & y=-12

Step-by-step explanation:

multiply y=3x by 2

giving you 2y=6x

subtracting 2y=x+20 from 2y =6x

=> 5x=-20

dividing through by 5

=> x=-4

put x=-4 into y=3x

=> y=3(-4)

y=-12

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The temperature at noon on Monday

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Answer:

There was a 5% change

Step-by-step explanation:

On Monday, the temperature was 80.

On Tuesday, the temp was 77

On Wednesday, the temp was 84

there was a difference of 4

4 is 5% of 80

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2 years ago

Lines k and p are perpendicular, neither is vertical and p passes through the origin. Which is greater? A. The product of the sl

AlladinOne [14]

What do we know about those two lines?

They are perpendicular, meaning they have the same slope.
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Therefore either
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Answer A

4 0

3 years ago

A bad punter on a football team kicks a football approximately straight upward with an initial velocity of 89 ft/sec.

Andrej [43]

Answer:

See below

Step-by-step explanation:

Vertical velocity will be affected by gravity in this scenario

df = do + vo t + 1/2 a t^2 do = original height = 4 ft a = -32.2 ft/s^2

df = 4 + 89 t - 1/2 (32.2) t^2 df = height

On the way up and the way down, the ball may reach height of 102.2125 ft :

102.2125 = 4 + 89 t - 1/2 (32.2) t2 re-arrange to:

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6 0

2 years ago

2 u + 12 < 23 0,1 d + 8 > 0 3 - 4 r > 7 13 < 2 z + 3 6 ≥ 9 - 0, 15 i

BartSMP [9]

Answer:

Part 1) $u< 5.5$

Part 2) $d > -80$

Part 3) $r < -1$

Part 4) $z>5$

Part 5) $i\geq 20$

Step-by-step explanation:

The question is

Solve each inequality for the indicated variable

Part 1) we have

$2u+12$

subtract 12 both sides

$2u< 23-12\\2u$

Divide by 2 both sides

$u< 5.5$

The solution is the interval (-∞,5.5)

In a number line the solution is the shaded area at left of u=5.5 (open circle)

The number 5.5 is not included in the solution

Part 2) we have

$0.1d+8 >0$

subtract 8 both sides

$0.1d > -8$

Divide by 0.1 both sides

$d > -80$

The solution is the interval (-80,∞)

In a number line the solution is the shaded area at right of d=-80 (open circle)

The number -80 is not included in the solution

Part 3) we have

$3-4r> 7$

Subtract 3 both sides

$-4r>7-3\\-4r>4$

Divide by -4 both sides

Remember that, when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

$r < -1$

The solution is the interval (-∞,-1)

In a number line the solution is the shaded area at left of r=-1.1 (open circle)

The number -1.1 is not included in the solution

Part 4) we have

$13< 2z+3$

Subtract 3 both sides

$13-3$

Divide by 2 both sides

$5$

Rewrite

$z>5$

The solution is the interval (5,∞)

In a number line the solution is the shaded area at right of z=5 (open circle)

The number 5 is not included in the solution

Part 5) we have

$6\geq 9-0.15i$

Subtract 9 both sides

$6-9\geq -0.15i\\-3\geq -0.15i$

Divide by -0.15 both sides

Remember that, when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

$20\leq i$

Rewrite

$i\geq 20$

The solution is the interval [20,∞)

In a number line the solution is the shaded area at right of i=20 (closed circle)

The number 20 is included in the solution

6 0

3 years ago

Csc A - sin A = A. (cos^2A) / (sinA) B. (1 - sinA) / (sinA) C. (1 - sinA) / (cosA)

julsineya [31]

Answer:

A) $\frac{cos^2A}{sinA}$

Step-by-step explanation:

$cscA-sinA\\\\\frac{1}{sinA}-sinA\\ \\\frac{1}{sinA}-\frac{sin^2A}{sinA}\\ \\ \frac{1-sin^2A}{sinA}\\ \\\frac{cos^2A}{sinA}$

4 0

2 years ago

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