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DerKrebs [107]
2 years ago
7

Help please I don’t understand it What is the sum of these questions?

Mathematics
1 answer:
weqwewe [10]2 years ago
5 0

I need some help since I don't comprehend what you're asking.

1. A(x) = (f(x)-f(a)) / (x-a) This average rate of change function is called A. The function f's input is represented by x minus a, while the function f's response to the input changing from a to x is represented by f(x) minus f(a).

f(5)=-2 f(9)=14 (f(5)-f(9))/(5-9)=(-2-14)/-4=4 from the table.

Over the range of x = 5 to x = 9, the average rate of change is 4.

2. g(x)=2x2+13x+1, f(x)=4x2+6x

(f/g)(x)=(4x²+6x)/(2x²+13x+1)

Using the numbers shown in the table, x = 4,5,6,7,8,9 and f(x) = 88,85,155,116,180,151,238.190,304,233,378.280

3. f(x)=x2-6x+8, g(x)=x-2, x2-6x-x+8+2=0, x2-7x+10=0 f(x)=g(x)x2-6x+8=x-2x2-6x-x+8+2=0

x₁,₂=(7⁺₋√(7²-4*10))/2=(7⁺₋3)/2 x₁=5, x₂=2

Because 2 doesn't appear in the table of values, the answer using the table of values is x=5.

To learn more about average rate visit : brainly.com/question/8061202

#SPJ9

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Greatest common factor <br>3a^3+9a^2​
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Answer:

27A+81A =108

Step-by-step explanation:

7 0
3 years ago
find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM​
Snezhnost [94]

\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}

  • Given - <u>A </u><u>trapezium</u><u> </u><u>ABCD </u><u>with </u><u>non </u><u>parallel </u><u>sides </u><u>of </u><u>measure </u><u>1</u><u>5</u><u> </u><u>cm </u><u>each </u><u>!</u><u> </u><u>along </u><u>,</u><u> </u><u>the </u><u>parallel </u><u>sides </u><u>are </u><u>of </u><u>measure </u><u>1</u><u>3</u><u> </u><u>cm </u><u>and </u><u>2</u><u>5</u><u> </u><u>cm</u>

  • To find - <u>Area </u><u>of </u><u>trapezium</u>

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

<u>Construction</u><u> </u><u>-</u>

draw \: CE \: \parallel \: AD \:  \\ and \: CD \: \perp \: AE

Now , we can clearly see that AECD is a parallelogram !

\therefore AE = CD = 13 cm

Now ,

AB = AE + BE \\\implies \: BE =AB -  AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm

Now , In ∆ BCE ,

semi \: perimeter \: (s) =  \frac{15 + 15 + 12}{2}  \\  \\ \implies \: s =  \frac{42}{2}  = 21 \: cm

Now , by Heron's formula

area \: of \: \triangle \: BCE =  \sqrt{s(s - a)(s - b)(s - c)}  \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)}  \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21}  \: cm {}^{2}

Also ,

area \: of \: \triangle \:  =  \frac{1}{2}  \times base \times height \\  \\\implies 18 \sqrt{21} =  \: \frac{1}{\cancel2}  \times \cancel12  \times height \\  \\ \implies \: 18 \sqrt{21}  = 6 \times height \\  \\ \implies \: height =  \frac{\cancel{18} \sqrt{21} }{ \cancel 6}  \\  \\ \implies \: height = 3 \sqrt{21}  \: cm {}^{2}

<u>Since </u><u>we've </u><u>obtained </u><u>the </u><u>height </u><u>now </u><u>,</u><u> </u><u>we </u><u>can </u><u>easily </u><u>find </u><u>out </u><u>the </u><u>area </u><u>of </u><u>trapezium </u><u>!</u>

Area \: of \: trapezium =  \frac{1}{2}  \times(sum \: of \:parallel \: sides) \times height \\  \\ \implies \:  \frac{1}{2}  \times (25 + 13) \times 3 \sqrt{21}  \\  \\ \implies \:  \frac{1}{\cancel2}  \times \cancel{38 }\times 3 \sqrt{21}  \\  \\ \implies \: 19 \times 3 \sqrt{21}  \: cm {}^{2}  \\  \\ \implies \: 57 \sqrt{21}  \: cm {}^{2}

hope helpful :D

6 0
2 years ago
In triangle ABC, angle = 90 degrees. AC = 7, BC = 12. AB = _______ Measure of angle A = _______ Measure of angle B = _______
Ulleksa [173]

Answer:

AB = 13.89

Measure of angle A = 59.74°

Measure of angle B = 30.26°

Step-by-step explanation:

The given parameters are;

∠C = 90°

AC = 7

BC = 12

Part 1

Hence, the question has the dimensions of the two adjacent sides of the right angle (angle 90°)

From Pythagoras theorem, we have;

A² = B² + C²

Where, A is the opposite side to the right angle, hence;

In the ΔABC,

AB ≡ A

Therefore;

AB² = AC² + BC² = 7² + 12² = 193

∴ AB = √193 = 13.89

Part 2

∠A is the side opposite side BC such that by trigonometric ratios

tan \angle A = \frac{Opposite \, side \,  to \,  angle \,  A}{Adjacent  \,  side  \, to  \, angle \,  A}  = \frac{BC}{AC} = \frac{12}{7} = 1.714

∴ ∠A = Arctan(1.714) or tan⁻¹(1.714) = 59.74°

Part 3

∠B is found from knowing that the sum of the angles in a triangle = 180°

∴ ∠A + ∠B + ∠C = 180° which gives

59.74° + 90° + ∠B = 180°

Hence, ∠B = 180° - (59.74° + 90°) = 180° - 149.74° = 30.26°.

7 0
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no

Step-by-step explanation:

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