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2 years ago

Assume that triangle GHI is congruent to LMN. Which of the following congruence statement are correct? check all that apply

Mathematics

2 answers:

Vsevolod [243]2 years ago

6 0

Answer.

GH = LM

<N = <I

<M = <H

IH = NM

Just got it wrong on APEX

Airida [17]2 years ago

5 0

If two triangles are congruent, then they have equal corresponding angles and also the sides.
Therefore, if GHI is congruent to LMN, then GH =LM, HI=MN and GI=LN, and also angle G=angle L, Angle H=angle M, while angle I = angle N, therefore the correct answers is a) ∠M= ∠H, c) ∠L=∠G. and e)IH=NM.

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Solve x2 - 8x - 9 = 0. Rewrite the equation so that it is of the form x2 + bx = c.

barxatty [35]

Answer:

x=9,-1 and $x^2+(-8)x=9$

Explanation:

we have been given with the quadratic equation $x^2-8x-9$

we compare the given quadratic equation with general quadratic equation

general quadratic is $ax^2+bx+c=0$

from given quadratic equation a=1,b= -8,c= -9

substituting these values in the formula for discriminant $D=b^{2}-4ac$

$D=(-8)^2-4(1)(-9)=100$

Now, to find the value of x

Formula is $x=\frac{-b\pm\sqrt{D}}{2a}$

Now, substituting the values we will get

$x=\frac{-(-8)\pm\sqrt{100}} {2}= \frac{8\pm10}{2}=9,-1$

And rewritting the given equation by shifting 9 to right hand side of the given equation and taking minus inside the bracket so as to convert it in the form of

$x^2+bx=c$

4 0

2 years ago

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The equation y=−0.0088x2+0.79x+15 models the speed x (in miles per hour) and average gas mileage y (in miles per gallon) for a v

rewona [7]

Answer: 30.72 miles per gallon

To find <span>best approximate for the average gas mileage at a speed of 60 miles per hour you need to replace the variable x with 60. The calculation would be:

</span><span> y=−0.0088x2+0.79x+15
y= -31.68 +47.4 + 15
y= 30.72
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2 years ago

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Rounded to the nearest hundredth, what is the positive solution to the quadratic equation 0=2x2+3x-8?

Basile [38]

Answer:

Step-by-step explanation:

The given quadratic equation is

2x^2+3x-8 = 0

To find the roots of the equation. We will apply the general formula for quadratic equations

x = -b ± √b^2 - 4ac]/2a

from the equation,

a = 2

b = 3

c = -8

It becomes

x = [- 3 ± √3^2 - 4(2 × -8)]/2×2

x = - 3 ± √9 - 4(- 16)]/2×2

x = [- 3 ± √9 + 64]/2×2

x = [- 3 ± √73]/4

x = [- 3 ± 8.544]/4

x = (-3 + 8.544) /4 or x = (-3 - 8.544) / 4

x = 5.544/4 or - 11.544/4

x = 1.386 or x = - 2.886

The positive solution is 1.39 rounded up to the nearest hundredth

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2 years ago

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6 people sharing 40 lbs of gold evenly. How many do they get? Then subtract 1 2/3 from the result of the problem above & sim

r-ruslan [8.4K]

Answer: 6

Step-by-step explanation:

Given Data:

Number of people = 6

Amount of Gold to be shared = 40lbs

Subtract 2/3 from your answer.

Therefore

If six people are to share 40lbs of gold evenly

= 40lbs / 6

Divide through by 2

= 20lbs / 3

= 20lbs / 3 - 2/3

= 18/3

= 6

Each person would get approximately 6lbs of gold

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1 year ago

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A random draw is being designed for 210 participants. A single winner is to be chosen, and all the participants must have an equ

Over [174]

Answer: The correct number of balls is (b) 4.

Step-by-step explanation: Given that a single winner is to be chosen in a random draw designed for 210 participants. Also, there is an equal probability of winning for each participant.

We are using 10 balls, numbered through 0 to 9. We are to find the number of balls which needs to be picked up, regardless of order, so that each of the 210 participants can be assigned a unique set of numbers.

Let 'r' represents the number of balls to be picked up.

Since we are choosing from 10 balls, so we must have

$^{10}C_r=210.$

The value of 'r' can be any one of 0, 1, 2, . . , 10.

Now,

if r = 1, then

$^{10}C_1=\dfrac{10!}{1!(10-1)!}=\dfrac{10!}{1!9!}=\dfrac{10\times 9!}{1\times 9!}=10$

If r = 2, then

$^{10}C_2=\dfrac{10!}{2!(10-2)!}=\dfrac{10!}{2!8!}=\dfrac{10\times 9\times 8!}{2\times 1\times 8!}=45$

If r = 3, then

$^{10}C_3=\dfrac{10!}{3!(10-3)!}=\dfrac{10!}{3!7!}=\dfrac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}=120$

If r = 4, then

$^{10}C_4=\dfrac{10!}{4!(10-4)!}=\dfrac{10!}{4!6!}=\dfrac{10\times 9\times 8\times\times 7\times 6!}{4\times 3\times 2\times 1\times 6!}=210.$

Therefore, we need to pick 4 balls so that each participant can be assigned a unique set of numbers.

Thus, (b) is the correct option.

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2 years ago

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