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2 years ago

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The image of ΔABC after a reflection across Line E G is ΔA'B'C'. 2 triangles are shown. A line of reflection is between the 2 tr

iangles. Line segment B B prime has a midpoint at point E. Line segment A A prime has a midpoint at point F. Line segment C C prime has a midpoint at point G. Which statement is true about point F? F is the midpoint of AA' because Line E G bisects AA'. F is the midpoint of EG because AA' bisects EG. F is the midpoint of AA' because AA' bisects EG. F is the midpoint of EG because Line E G bisects AA'.

2 answers:

nirvana33 [79]2 years ago

6 0

Answer:

It's A - F is the midpoint of AA' because Line E G bisects AA'

Step-by-step explanation: :D

True [87]2 years ago

3 0

Answer:

A

Step-by-step explanation:

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1. Mac is about to sue his contractor who promised to install a water tank that holds 1880 gallons of water. Mac knows that 1880

zzz [600]

Answer:

option C.

Yes, the water tank is about 245 cubic feet too small

Step-by-step explanation:

step 1

Determine the volume of the cylindrical tank

we know that

The volume of a cylinder is equal to

$V=\pi r^{2} h$

Remember that

$1\ ft=12\ in$

we have

$r=6\ ft\\h=2\ ft\ 2\ in=2+(2/12)=\frac{13}{6}\ ft$

assume

$\pi =3.14$

substitute

$V=(3.14)(6)^{2}(\frac{13}{6})$

$V=244.92\ ft^3$

Compare with 251 cubic feet

$244.92\ ft^3 < 251\ ft^3$

therefore

Yes, the water tank is about 245 cubic feet too small

8 0

2 years ago

Please answer all of them need this

VikaD [51]

First Question

For a better understanding of the solution provided here please find the first attached file which has the diagram of the the isosceles trapezoid.

We dropped perpendiculars from C and D to intersect AB at Q and P respectively.

As can be seen in $\Delta BCQ$ , we can easily find the values of CQ and BQ.

Since, $Sin(75^0)=\frac{CQ}{8}$

$\therefore CQ=8\times Sin(75^0)\approx 7.73$ ft

In a similar manner we can find BQ as:

$Cos(75^0)=\frac{BQ}{8}$

$BQ\approx2.07$ ft

All these values can be found in the diagram attached.

Thus, because of the inherent symmetry of the isosceles trapezoid, PQ can be found as:

$PQ=22-(AP+QB)=22-(2.07+2.07)=17.86$

Let us now consider $\Delta AQC$

We can apply the Pythagorean Theorem here to find the length of the diagonal AC which is the hypotenuse of $\Delta AQC$ .

$AC=\sqrt{(AQ)^2+(QC)^2}=\sqrt{(AP+PQ)^2+(QC)^2}=\sqrt{(2.07+17.86)^2+(7.73)^2}\approx21.38$ feet.

Thus, out of the given options, Option B is the closest and hence is the answer.

Second Question

For this question we can directly apply the formula for the area of a triangle using sines which is as:

Area= $\frac{1}{2}$ (First Side)(Second Side)(Sine of the angle between the two sides)

Thus, from the given data,

Area= $\frac{1}{2}\times 218.5\times 224.5\times sin(58.2^0)\approx20845$ $m^2$

Therefore, Option D is the correct option.

Third Question

For this question we will apply the Sine Rule to the $\Delta ABC$ given to us.

Thus, from the triangle we will have:

$\frac{AB}{Sin(\angle C)}=\frac{BC}{Sin(\angle A)}$

$\frac{c}{Sin(\angle C)}=\frac{a}{Sin(\angle A)}$

$\frac{17}{Sin(25^0)}=\frac{a}{Sin(45^0)}$

This gives $a$ to be:

$a\approx28.44$

Which is not close to any of the given options.

Fourth Question

Please find the second attachment for a better understanding of the solution provided her.

As can be clearly seen from the attached diagram, we can apply the Cosine Rule here to find the return distance of the plane which is CA.

$AC=\sqrt{(AB)^2+(BC)^2-2(AB)(BC)\times Cos(\angle B)}$

$\therefore AC=\sqrt{(172.20)^2+(111.64)^2-2(172.20)(111.64)\times Cos(177.29^0)}\approx283.8$ miles.

Thus, Option D is the answer.

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2 years ago

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2 years ago

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Alexander's dividing oranges into eighths he has 5 oranges.how many eights will be have

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Ther will be 40 eights. Hope this helps!

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2 years ago

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A quality control engineer tests the quality of produced computers. suppose that 5% of computers have defects, and defects occur

11111nata11111 [884]

(a) 0.059582148 probability of exactly 3 defective out of 20

(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.

(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So

0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)

= 0.05^3 * 0.95^17 * 20! / (3!17!)

= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)

= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)

= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)

= 0.05^3 * 0.95^17 * 20*19*3

= 0.000125* 0.418120335 * 1140

= 0.059582148

(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.

The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)

= 0.05*0.95^3*24/(1!3!)

= 0.05*0.857375*24/6

= 0.171475

So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.

3 0

2 years ago