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Usimov [2.4K]
3 years ago
8

Geometry Homework Help Please?

Mathematics
1 answer:
kolbaska11 [484]3 years ago
3 0
Ok what is the geometry u need help with
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In a camp ,140 students from school A,70 students from school B, 105 students from school C, and 35 students from school D, part
Schach [20]
Total no. of students - 140 students
Total no. of students from all schools= 140+70+35+105=350
school A= 140÷350×100=40./. (multiplying by 100 because it's the total percentage )
School B = 70÷350×100=20./.
School C = 105÷350×100=30./.
School D = 105÷350×100= 10./.
Consisting all the schools percentage it gives 100./.

Hope this helps
7 0
3 years ago
Given the equation 4x2 − 8x 20 = 0, what are the values of h and k when the equation is written in vertex form a(x − h)2 k = 0?
Westkost [7]

The vertex form of the given quadratic equation is

4(x-1)^{2} + 16 =0.

According to the given question.

We have a quadratic equation

4x^{2} -8x + 20 = 0

Since, for the standard quadratic form is ax^{2} +bx + c = y, the vertex form of a quadratic equation is y = a(x -h)^{2} + k where (h, k) is the vertex.    

And h and k can be calculated as

h = -b/2a and y = k

So, for the given equation 4x^{2} -8x + 20 =0 the vertex (h, k) is given by

h = -(-8)/2(4) = 8/8 = 1   (X coordinate of vertex)

and,

y =4x^{2} -8x+ 20

substitute x = 1 in the above equation for the value of k

Y = 4(1)(1) - 8(1) + 20

⇒ Y = 4 - 8 + 20

⇒ Y = -4 + 20

⇒ Y = 16

so, k = 16          (Y coordinate of vertex)

Now, substitute the value of a, k and h in y = a(x-h)^{2} + k.

⇒  0 =4(x-1)^{2} + 16

Therefore, the vertex form of the given quadratic equation is

4(x-1)^{2} + 16 =0.

Find out more information about vertex form of a quadratic equation here:

brainly.com/question/12223454

#SPJ4

6 0
2 years ago
Two boats are equidistant from a lighthouse. The boats are 30 miles apart. The angle formed between the two boats, with the ligh
anastassius [24]
Using the Law of Sines  (sina/A=sinb/B=sinc/C for any triangle)

(sin40)/30=(sin(180-40)/2)/d

d=(30sin70)/sin40

d≈43.857mi

d≈44 mi to the nearest whole mile...
3 0
3 years ago
Read 2 more answers
Bob is driving along a straight and level road toward a mountain. At some point on his trip, he measures the angle of elevation
DerKrebs [107]

Answer:

Option A.

Step-by-step explanation:

Distance between Bob and the center of the mountain at the base = 19,427.5 ft.

Angle of elevation to the top of the mountain = 25°11'

We know that

1 degree = 60 minutes

1/60 degree = 1 minute

25^{\circ}11'=25^{\circ}+\frac{11}{60}{\circ}\Rightarrow 25^{\circ}+0.183^{\circ}=25.183^{\circ}

Let the height of the mountain be h.

In a right angled triangle

\tan\theta = \frac{opposite}{adjacent}

\tan(25.183^{\circ}) = \frac{h}{19427.5}

\tan(25.183^{\circ})\times 19427.5 = h

h=0.4702\times 19427.5

h=9134.8105

h\approx 9135

Therefore, the height of the mountain to the nearest foot 9135. Option A is correct.

5 0
3 years ago
Which expression is equivalent to <img src="https://tex.z-dn.net/?f=%5Csqrt128x%5E%7B8%7D%20y%5E%7B3%7D%20z%5E%7B9%7D" id="TexFo
Elis [28]

Answer:

(C)8x^4z^4\sqrt{2yz}

Step-by-step explanation:

We want to determine an expression equivalent to: \sqrt{128x^8y^3z^9}

\sqrt{ab} =\sqrt{a}*\sqrt{b}

Therefore:

\sqrt{128x^8y^3z^9}=\sqrt{128}*\sqrt{x^8}*\sqrt{y^3}*\sqrt{z^9}

=\sqrt{64*2}*\sqrt{x^{4*2}}*\sqrt{y^{2+1}}*\sqrt{z^{8+1}}\\=8\sqrt{2}*\sqrt{x^{4*2}}*\sqrt{y^2*y}*\sqrt{z^{8}*z}}

=8\sqrt{2}*x^4*y \sqrt{y}*z^4\sqrt{z}\\=8*x^4*z^4*\sqrt{2}*\sqrt{y}*\sqrt{z}\\=8x^4z^4\sqrt{2yz}

4 0
3 years ago
Read 2 more answers
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