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3 years ago

Solve for y 2x+y=3 thank you.

Mathematics

2 answers:

xxMikexx [17]3 years ago

8 0

Y = -2x + 3
Hope that helps.

Alexxx [7]3 years ago

4 0

Hello there!

$\boxed{ 2*x+y-(3)=0 } \\ \\ \boxed{ Solve \ : 2x+y-3 = 0 } \\ \\ y=? \\ \\ \boxed{ y-intercept = -2x+3} \\ \\ y= \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Uparrow\ \ \ \ \ \ \ \ \ \ \Uparrow$

I hope this helps you!

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Find AB А. B N M -3. finele ooie 7 35. D 22 uz to с 29

gizmo_the_mogwai [7]

Answer: 15

Step-by-step explanation:

By the trapezoid midsegment theorem,

$\frac{AB+CD}{2}=MN\\\frac{29+AD}{2}=22\\29+AD=44\\AD=15$

7 0

2 years ago

Please I need help on this question, A line passes through (6,3), (8,4), and (n,-2). Find the value of n.

Vika [28.1K]

Answer:

n= -4

Step-by-step explanation:

First we have to find the equation of line which passes through (6,3) and (8,4).

Then we have to pass this line through the point (n , -2).

we know, equation of line passing through (x1,y1) and (x2,y2) is

$\frac{x-x1}{x1-x2}=\frac{y-y1}{y1-y2}$

So, the equation of line passing through (6,3) and (8,4) is

$\frac{x-6}{6-8}=\frac{y-3}{3-4}$

or, $\frac{x-6}{-2} =\frac{y-3}{-1}$

multiplying both sides by -2

x-6=2(y-3)

or, x-6=2y-6

or, x-2y=6-6

or, x-2y=0

Now, we have to pass this line through (n,-2)

So, using the point in the line, that is using x=n , y=-2

we get

n-2(-2)=0

or, n+4=0

or, n=-4

6 0

3 years ago

Read 2 more answers

PLEASE HELP 30 POINTS!

tamaranim1 [39]

Answer:

The pentagon.

Step-by-step explanation:

Mark me Brainliest.

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2 years ago

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Choose a justification for each step in the derivation of the sine difference identity.

Natali [406]

Answer:

Step 1:

✔ Cofunction identity

Step 5:

✔ Sine difference identity

Step 6:

✔ Cofunction Identity

Step 7:

✔ Cosine function is even, sine function is odd.

Step-by-step explanation:

8 0

2 years ago

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1. The height of a triangle is 6 m more than its base. The area of the triangle is 56 m². What is the length of the base? Enter

Elodia [21]

Answers:
1. 8 m
2. 17 m
3. 7 cm
4. 2 s

Explanations:

1. Let x = length of the base
x + 6 = height of the base

Then, the area of the triangle is given by

(Area) = (1/2)(base)(height)
56 = (1/2)(x)(x + 6)
56 = (1/2)(x²  + 6x)

Using the symmetric property of equations, we can interchange both sides of equations so that

(1/2)(x²  + 6x) = 56

Multiplying both sides by 2, we have

x² + 6x = 112

The right side should be 0. So, by subtracting both sides by 112, we have

x² + 6x - 112 = 112 - 112
x² + 6x - 112 = 0

By factoring, x² + 6x - 112 = (x - 8)(x + 14). So, the previous equation becomes

(x - 8)(x +14) = 0

So, either

x - 8 = 0 or x + 14 = 0

Thus, x = 8 or x = -14. However, since x represents the length of the base and the length is always positive, it cannot be negative. Hence, x = 8. Therefore, the length of the base is 8 cm.

2. Let x = length of increase in both length and width of the rectangular garden

Then,

14 + x = length of the new rectangular garden
12 + x = width of the new rectangular garden

So,

(Area of the new garden) = (length of the new garden)(width of the new garden)

255 = (14 + x)(12 + x) (1)

Note that

(14 + x)(12 + x) = (x + 14)(x + 12)
= x(x + 14) + 12(x + 14)
= x² + 14x + 12x + 168
= x² + 26x + 168

So, the equation (1) becomes

255 = x² + 26x + 168

By symmetric property of equations, we can interchange the side of the previous equation so that

x² + 26x + 168 = 255

To make the right side becomes 0, we subtract both sides by 255:

x² + 26x + 168 - 255 = 255 - 255
x² + 26x - 87 = 0

To solve the preceding equation, we use the quadratic formula.

First, we let

a = numerical coefficient of x² = 1

Note: if the numerical coefficient is hidden, it is automatically = 1.

b = numerical coefficient of x = 26
c = constant term = - 87

Then, using the quadratic formula

$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} = \frac{-26 \pm \sqrt{26^2 - 4(1)(-87)} }{2(1)} 
\newline x = \frac{-26 \pm \sqrt{1,024} }{2}
\newline
\newline x = \frac{-26 \pm 32 }{2}$

So,

$x = \frac{-26 + 32 }{2} \text{ or } x = \frac{-26 - 32 }{2}
\newline x = \frac{6 }{2} \text{ or } x = \frac{-58 }{2}
\newline \boxed{ x = 3 \text{ or } x = -29}$

Since x represents the amount of increase, x should be positive.

Hence x = 3.

Therefore, the length of the new garden is given by

14 + x = 14 + 3 = 17 m.

3. The area of the shaded region is given by

(Area of shaded region) = π(outer radius)² - π(inner radius)²
= π(2x)² - π6²
= π(4x² - 36)

Since the area of the shaded region is 160π square centimeters,

π(4x² - 36) = 160π

Dividing both sides by π, we have

4x² - 36 = 160

Note that this equation involves only x² and constants. In these types of equation we get rid of the constant term so that one side of the equation involves only x² so that we can solve the equation by getting the square root of both sides of the equation.

Adding both sides of the equation by 36, we have

4x² - 36 + 36 = 160 + 36
4x² = 196

Then, we divide both sides by 4 so that

x² = 49

Taking the square root of both sides, we have

$x = \pm 7$

Note: If we take the square root of both sides, we need to add the plus minus sign $(\pm)$ because equations involving x² always have 2 solutions.

So, x = 7 or x = -7.

But, x cannot be -7 because 2x represents the length of the outer radius and so x should be positive.

Hence x = 7 cm

4. At time t, h(t) represents the height of the object when it hits the ground. When the object hits the ground, its height is 0. So,

h(t) = 0 (1)

Moreover, since $v_0 = 27$ and $h_0 = 10$ ,

h(t) = -16t² + 27t + 10 (2)

Since the right side of the equations (1) and (2) are both equal to h(t), we can have

-16t² + 27t + 10 = 0

To solve this equation, we'll use the quadratic formula.

Note: If the right side of a quadratic equation is hard to factor into binomials, it is practical to solve the equation by quadratic formula.

First, we let

a = numerical coefficient of t² = -16
b = numerical coefficient of t = 27
c = constant term = 10

Then, using the quadratic formula

$t = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} = \frac{-27 \pm \sqrt{27^2 - 4(-16)(10)} }{2(-16)} \newline t = \frac{-27 \pm \sqrt{1,369} }{-32} \newline \newline t = \frac{-27 \pm 37 }{32}$

So,

$t = \frac{-27 + 37 }{-32} \text{ or } t = \frac{-27 - 37 }{-32} \newline t = \frac{-10}{32} \text{ or } t = \frac{-64 }{-32} \newline \boxed{ t = -0.3125 \text{ or } t = 2}$

Since t represents the amount of time, t should be positive.

Hence t = 2. Therefore, it takes 2 seconds for the object to hit the ground.

3 0

3 years ago

Read 2 more answers