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3 years ago

Find the equation of the axis of symmetry of the graph of y = (x + 3)^2 - 5

1 answer:

3 0

Equation of the parabola: y = ax^2 + bx + c. Find a, b, and c.
x of axis of symmetry:
x
=
−
b
2
a
=
3
-> b = -6a
Writing that the graph passing at point (1, 0) and point (4, -3):
(1) 0 = a + b + c -> c = - a - b = - a + 6a = 5a
(2) -3 = 16a + 4b + c --> -3 = 16a - 24a + 5a = -3a --> a = 1
b = -6a = -6; and c = 5a = 5
y
=
x
2
−
6
x
+
5

Check with x = 1: -> y = 1 - 6 + 5 = 0. OK

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If 3x-2+5x=2,find the value of 3x

BlackZzzverrR [31]

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3 years ago

Find UV (Pic Stated Below)

natta225 [31]

Answer:

UV=29

Step-by-step explanation:

In right triangles AQB and AVB,

∠AQB = ∠AVB ...(i) {Right angles}

∠QBA = ∠VBA ...(ii) {Given that they are equal}

We know that sum of all three angles in a triangle is equal to 180 degree. So wee can write sum equation for each triangle

∠AQB+∠QBA+∠BAQ=180 ...(iii)

∠AVB+∠VBA+∠BAV=180 ...(iv)

using (iii) and (iv)

∠AQB+∠QBA+∠BAQ=∠AVB+∠VBA+∠BAV

∠AVB+∠VBA+∠BAQ=∠AVB+∠VBA+∠BAV (using (i) and (ii))

∠BAQ=∠BAV...(v)

Now consider triangles AQB and AVB;

∠BAQ=∠BAV {from (v)}

∠QBA = ∠VBA {from (ii)}

AB=AB {common side}

So using ASA, triangles AQB and AVB are congruent.

We know that corresponding sides of congruent triangles are equal.

Hence

AQ=AV

5x+9=7x+1

9-1=7x-5x

8=2x

divide both sides by 2

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3 years ago

-8z+(4.5)+3.5z+7y-1.5

bogdanovich [222]

Answer:

$\boxed{\sf{7y+3-4.5z}}$

Step-by-step explanation:

In order to combine like terms, you have to isolate x and y from one side of the equation.

$\sf{-8z+\left(4.5\right)+3.5z+7y-1.5}$

First, thing you do is remove parentheses.

$\Longrightarrow: \sf{-8z+4.5+3.5z+7y-1.5}$

Solve.

Then, you combine like terms.

$\Longrightarrow:\sf{-8z+3.5z+7y+4.5-1.5}$

Add/subtract the numbers from left to right.

-8z+3.5z=-4.5z

Rewrite the problem down.

$\Longrightarrow: \sf{-4.5z+7y+4.5-1.5}$

Solve.

4.5-1.5=3

$\Longrightarrow: \boxed{\sf{7y+3-4.5z}}$

Therefore, the correct answer is 7y+3-4.5z.

I hope this helps, let me know if you have any questions.

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2 years ago

67. The line contains the point (4,0) and is parallel to the line defined by 3x = 2y.

olganol [36]

Answer:

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

Step-by-step explanation:

Hi there!

What we need to know:

Linear equations are typically organized in slope-intercept form:

$y=mx+b$ where m is the slope of the line and b is the y-intercept (the value of y when the line crosses the y-axis)

Parallel lines will always have the same slope but different y-intercepts.

1) Determine the slope of the parallel line

Organize 3x = 2y into slope-intercept form. Why? So we can easily identify the slope, m.

$3x = 2y$

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$2y=3x$

Divide both sides by 2 to isolate y

$\frac{2y}{2} = \frac{3}{2} x\\y=\frac{3}{2} x$

Now that this equation is in slope-intercept form, we can easily identify that $\frac{3}{2}$ is in the place of m. Therefore, because parallel lines have the same slope, the parallel line we're solving for now will also have the slope $\frac{3}{2}$ . Plug this into $y=mx+b$ :

$y=\frac{3}{2} x+b$

2) Determine the y-intercept

$y=\frac{3}{2} x+b$

Plug in the given point, (4,0)

$0=\frac{3}{2} (4)+b\\0=6+b$

Subtract both sides by 6

$0-6=6+b-6\\-6=b$

Therefore, -6 is the y-intercept of the line. Plug this into $y=\frac{3}{2} x+b$ as b:

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

I hope this helps!

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3 years ago

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balu736 [363]

Step-by-step explanation:

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3 years ago