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levacccp [35]
3 years ago
9

I need help from question 11- 16! Please help!

Mathematics
1 answer:
insens350 [35]3 years ago
3 0
<h2>11. Find discriminant.</h2>

Answer: D) 0, one real solution

A quadratic function is given of the form:

ax^2+bx+c=

We can find the roots of this equation using the quadratic formula:

x_{12}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

Where \Delta=b^2-4ac is named the discriminant. This gives us information about the roots without computing them. So, arranging our equation we have:

4a^2-4a-6=-7 \\ \\ Adding \ 7 \ to \ both \ sides \ of \ the \ equation: \\ \\ 4a^2-4a-6+7=-7+7 \\ \\ 4a^2-4a+1=0 \\ \\ Then \ the \ discriminant: \\ \\ \Delta=(-4)^2-4(4)(1) \\ \\ \Delta=16-16 \\ \\ \boxed{Delta=0}

<em>Since the discriminant equals zero, then we just have one real solution.</em>

<h2>12. Find discriminant.</h2>

Answer: D) -220, no real solution

In this exercise, we have the following equation:

-r^2-2r+14=-8r^2+6

So we need to arrange this equation in the form:

ax^2+bx+c=

Thus:

-r^2-2r+14=-8r^2+6 \\ \\ Adding \ 8r^2 \ to \ both \ sides \ of \ the \ equation: \\ \\ -r^2-2r+14+8r^2=-8r^2+6+8r^2 \\ \\ Associative \ Property: \\ \\ (-r^2+8r^2)-2r+14=(-8r^2+8r^2)+6 \\ \\ 7r^2-2r+14=6 \\ \\ Subtracting \ 6 \ from \ both \ sides: \\ \\ 7r^2-2r+14-6=6-6 \\ \\ 7r^2-2r+8=0

So the discriminant is:

\Delta=(-2)^2-4(7)(8) \\ \\ \Delta=4-224 \\ \\ \boxed{\Delta=-220}

<em>Since the discriminant is less than one, then there is no any real solution</em>

<h2>13. Value that completes the squares</h2>

Answer: C) 144

What we need to find is the value of c such that:

x^2+24x+c=0

is a perfect square trinomial, that are given of the form:

a^2x^2\pm 2axb+b^2

and can be expressed in squared-binomial form as:

(ax\pm b)^2

So we can write our quadratic equation as follows:

x^2+2(12)x+c \\ \\ So: \\ \\ a=1 \\ \\ b=12 \\ \\ c=b^2 \therefore c=12^2 \therefore \boxed{c=144}

Finally, the value of c that completes the square is 144 because:

x^2+24x+144=(x+12)^2

<h2>14. Value that completes the square.</h2>

Answer: C) \frac{121}{4}

What we need to find is the value of c such that:

z^2+11z+c=0

So we can write our quadratic equation as follows:

z^2+2\frac{11}{2}z+c \\ \\ So: \\ \\ a=1 \\ \\ b=\frac{11}{2} \\ \\ c=b^2 \therefore c=\left(\frac{11}{2}\left)^2 \therefore \boxed{c=\frac{121}{4}}

Finally, the value of c that completes the square is \frac{121}{4} because:

z^2+11z+\frac{121}{4}=(x+\frac{11}{2})^2

<h2 /><h2>15. Rectangle.</h2>

In this problem, we need to find the length and width of a rectangle. We are given the area of the rectangle, which is 45 square inches. We know that the formula of the area of a rectangle is:

A=L\times W

From the statement we know that the length of the rectangle is is one inch less than twice the width, this can be written as:

L=2W-1

So we can introduce this into the equation of the area, hence:

A=L\times W \\ \\ \\ Where: \\ \\ W:Width \\ \\ L:Length

A=(2W-1)(W) \\ \\ But \ A=45: \\ \\ 45=(2W-1)(W) \\ \\ Distributive \ Property:\\ \\ 45=2W^2-W \\ \\ 2W^2-W-45=0 \\ \\ Quadratic \ Formula: \\ \\ x_{12}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\ W_{1}=\frac{-(-1)+ \sqrt{(-1)^2-4(2)(-45)}}{2(2)} \\ \\ W_{1}=\frac{1+ \sqrt{1+360}}{4} \therefore W_{1}=5 \\ \\ W_{2}=\frac{-(-1)- \sqrt{(-1)^2-4(2)(-45)}}{2(2)} \\ \\ W_{2}=\frac{1- \sqrt{1+360}}{4} \therefore W_{2}=-\frac{9}{2}

The only valid option is W_{1} because is greater than zero. Recall that we can't have a negative value of the width. For the length we have:

L=2(5)-1 \\ \\ L=9

Finally:

The \ length \ is \ 9 \ inches \\ \\ The \ width \ is \ 5 \ inches

<h2>16. Satellite</h2>

The distance in miles between mars and a satellite is given by the equation:

d=-9t^2+776

where t is the number of hours it has fallen. So we need to find when the satellite will be 452 miles away from mars, that is, d=452:

d=-9t^2+776 \\ \\ 452=-9t^2+776 \\ \\ 9t^2=776-452 \\ \\ 9t^2=324 \\ \\ t^2=\frac{324}{9} \\ \\ t^2=36 \\ \\ t=\sqrt{36} \\ \\ \boxed{t=6h}

Finally, <em>the satellite will be 452 miles away from mars in 6 hours.</em>

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PLEASEEE EXPLAIN AND HELP
Firdavs [7]

The length of the unknown sides of the triangles are as follows:

CD = 10√2

AC = 10√2

BC =  10

AB = 10

<h3>Triangle ACD</h3>

ΔACD is a right angle triangle. Therefore, Pythagoras theorem can be used to find the sides of the triangle.

  • c² = a² + b²

where

c = hypotenuse side = AD = 20

a and b are the other 2 legs

lets use trigonometric ratio to find CD,

cos 45 = adjacent / hypotenuse

cos 45 = CD / 20

CD = 1 / √2 × 20

CD = 20 / √2 = 20√2 / 2 = 10√2

20² - (10√2)² = AC²

400 - 100(2)  = AC²

AC² = 200

AC = √200 = 10√2

<h3>Triangle ABC</h3>

ΔABC is a right angle triangle too. Therefore,

  • AB² + BC² = AC²

Using trigonometric ratio,

cos 45 = BC / 10√2

BC = 10√2 × cos 45

BC = 10√2 × 1 / √2

BC = 10√2 / √2 =  10

(10√2)² - 10² = AB²

200 - 100 = AB²

AB² = 100

AB = 10

learn more on triangles here: brainly.com/question/24304623?referrer=searchResults

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