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masya89 [10]
3 years ago
10

What's the answer for this ?

Mathematics
1 answer:
7nadin3 [17]3 years ago
3 0
B is your answer. Goodluck
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How to find the x-intercepts of a parabola from the vertex, (-1,-108) and the y intercept (0,-105)?
Ulleksa [173]

Answer:

* The x-intercepts are -7 and 5

Step-by-step explanation:

* At first lets revise the standard and general forms of the

 quadratic function which represented graphically by the parabola

- f(x) = a(x - h)² + k ⇒ standard form

- Where point (h , k) is the vertex of the parabola

- f(x) = ax² + bx + c ⇒ general form

- Where a, b, c are constant

- c is the y-intercept ⇒ means x = 0

- h = -b/2a

- k = f(h)

* Lets solve the problem

- We will find the equation of the parabola

∵ The vertex is (-1 , -108)

∴ h = -1 and k = -108

∵ y-intercept = -105

- Equate the two forms

∵ ax² + bx + c = a(x - h)² + k ⇒ solve the (   )²

∴ ax² + bx + c = a(x² - 2hx + h²) + k ⇒ open the bracket

∴ ax² + bx + c = ax² - 2ahx + ah² + k ⇒ by comparing the two sides

∴ c = ah² + k

- Substitute the value of c , h , k in it

∴ -105 = a(-1)² + -108

∴ -105 = a - 108 ⇒ add 108 to the both sides

∴ 3 = a

- Lets write the equation in the standard form

∴ y = 3(x - -1)² + -108

∴ y = 3(x + 1)² - 108

* To find the x-intercepts means the parabola intersects the x-axis,

  then put y = 0

∴ 3(x + 1)² - 108 = 0 ⇒ Add 108 to the both sides

∴ 3(x + 1)² = 108 ⇒ divide the both sides by 3

∴ (x + 1)² = 36 ⇒ take square root for both sides

∴ (x + 1) = ± 6

# x + 1 = 6  OR x + 1 = -6

∵ x + 1 = 6 ⇒ subtract 1 from both sides

∴ x = 5

∵ x + 1 = -6 ⇒ subtract 1 from both sides

∴ x = -7

* The x-intercepts are -7 and 5

∴

3 0
3 years ago
Activating Schema: Draw the figure described below. Then solve for length PR. Right Triangle PQR with the following characterist
Marina CMI [18]

The given triangle is

To solve for PR, we have to use the sine function which is equivalent to the ratio between the opposite leg and the hypothenuse.

\begin{gathered} \sin 52=\frac{PR}{47} \\ PR=47\cdot\sin 52 \\ PR\approx37.04 \end{gathered}<h2>Hence, PR is 37.04 mm long, approximately.</h2>

6 0
2 years ago
If $7500 is invested at an interest rate of 5% each year, what is the value of the investment in 4 years? Write an exponential f
Vanyuwa [196]
The answer is 9,116.3
7500*(1+0.05)^4 
3 0
3 years ago
The function f(x)=x√3 has a domain of (−∞,∞) and a range of [0,∞)<br><br> True<br><br> False
Vlad [161]
True I’m not sure I’m just guessing cause I gotta answer Two questions before mine
7 0
3 years ago
The equation y = \large 1\frac{1}{2}x represents the number of cups of dried fruit, y, needed to make x pounds of granola. Deter
Natasha2012 [34]

Answer:

(1\frac{1}{2},1) - False

(4,6) - True

(18,12) -- False

(0,0) -- True

(2\frac{1}{2},3\frac{3}{4}) -- True

Step-by-step explanation:

The points are

(1\frac{1}{2},1) , (4,6), (18,12), (0,0) and (2\frac{1}{2},3\frac{3}{4}) ---- missing from the question

Given

y = 1\frac{1}{2}x

Required

Determine if each of the points would be on y = 1\frac{1}{2}x

To do this, we simply substitute the value of x and of each point in y = 1\frac{1}{2}x.

(a) (1\frac{1}{2},1)

In this case;

x = 1\frac{1}{2} and y = 1

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 1\frac{1}{2}

y = \frac{3}{2} * \frac{3}{2}

y = \frac{9}{4}

y = 2\frac{1}{4}

<em>The point </em>(1\frac{1}{2},1)<em>  won't be on the graph because the corresponding value of y for </em>x = 1\frac{1}{2}<em> is </em>y = 2\frac{1}{4}<em></em>

(b) (4,6)

In this case;

x = 4

y = 6

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 4

y = \frac{3}{2} * 4

y = \frac{3* 4}{2}

y = \frac{12}{2}

y = 6

<em>The point </em>(4,6)<em>  would be on the graph because the corresponding value of y for </em>x = 4 is y = 6

(c) (18,12)

In this case:

x = 18;y = 12

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 18

y = \frac{3}{2} * 18

y = \frac{3* 18}{2}

y = \frac{54}{2}

y = 27

<em>The point </em>(18,12)<em>  wouldn't be on the graph because the corresponding value of y for </em>x = 18<em> is </em>y = 12<em></em>

(d) (0,0)

In this case;

x =0; y = 0

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 0

y = 0

<em>The point </em>(0,0)<em>  would be on the graph because the corresponding value of y for </em>x = 0 is y = 0

(e) (2\frac{1}{2},3\frac{3}{4})

In this case:

x = 2\frac{1}{2}; y = 3\frac{3}{4}

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 2\frac{1}{2}

y = \frac{3}{2} * \frac{5}{2}

y = \frac{15}{4}

y = 3\frac{3}{4}

<em>The point </em>(2\frac{1}{2},3\frac{3}{4}) <em>  would be on the graph because the corresponding value of y for </em>x = 2\frac{1}{2} is y = 3\frac{3}{4}

3 0
3 years ago
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