MATHEMATICS HIGH SCHOOL

Twice the difference of a number and ten is fifty-four. Find the number,

Answers

Answer 1
Answer:

Answer:

37

Step-by-step explanation:

2(x-10)=54

2x-20=54

2x=74

x=37

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Given the information below, QRS is UVT similar to ? If so, give the scale factor.QR= 8 RS = 16 UV = 18 VT= 34 M<S=36 M<T=36

A.No, the triangles are not similar.
B.Yes, the scale factor is 1:2.
C.Yes, the scale factor is 4:9.
D.Yes, the scale factor is 8:17.

Answers

Look\ at\ the\ picture.\n\n\Delta QRS\sim\Delta UVT\ if\ (|QR|)/(|UV|)=(|RS|)/(|VT|)\n\n(|QR|)/(|UV|)=(8)/(18)=(4)/(9)\n\n(|RS|)/(|VT|)=(16)/(34)=(8)/(17)\n\nAnswer:A.\ No,\ the\ triangles\ are\ not\ similar.

An engineer on the ground is looking at the top of a building the angle of elevation to the top of the building is 22 degrees the engineer knows the building is 450 ft tall. What is the distance from the engineer to the base of the building to the nearest whole foot?

Answers

Answer:

1114 foot

Step-by-step explanation:

Let the engineer is at a distance of x feet from the base of the building.

Please see the attached image.

In the triangle, we have

\tan22^(\circ)=(450)/(x)

Solving the equation for x, we get

x=(450)/(\tan22^(\circ))\n\nx\approx1114

Thus, the engineer is 1114 foot from the base of the building.
Hello,

tan 22°=450/d
==>d=450/tan 22°=1113.7890... ≈1114 (ft)

Rationalise:
(1)              4/(2+root3+root7)
(2)              4/(2root3+root5)

Answers

(4)/(2+\sqrt3+\sqrt7)\cdot(2-(\sqrt3+\sqrt7))/(2-(\sqrt3+\sqrt7))=(8-4\sqrt3-4\sqrt7)/(2^2-(\sqrt3+\sqrt7)^2)=(8-4\sqrt3-4\sqrt7)/(4-3-2√(3\cdot7)-7)\n\n=(8-4\sqrt3-4\sqrt7)/(-6-2√(21))=(-2(2\sqrt3+2\sqrt7-4))/(-2(3+√(21)))=(2\sqrt3+2\sqrt7-4)/(3+√(21))\cdot(3-√(21))/(3-√(21))\n\n=(6\sqrt3-2√(63)+6\sqrt7-2√(147)-12+4√(21))/(3^2-(√(21))^2)=(6\sqrt3-2√(9\cdot7)+6\sqrt7-2√(49\cdot3)-12+4√(21))/(9-21)

=(6\sqrt3-6\sqrt7+6\sqrt7-14\sqrt3-12+4√(21))/(-12)=(-8\sqrt3+4√(21)-12)/(-12)=(-4(2\sqrt3-√(21)+3))/(-12)\n\n=(2\sqrt3-√(21)+3)/(3)

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(4)/(2\sqrt3+\sqrt5)\cdot(2\sqrt3-\sqrt5)/(2\sqrt3-\sqrt5)=(8\sqrt3-4\sqrt5)/((2\sqrt3)^2-(\sqrt5)^2)=(8\sqrt3-4\sqrt5)/(4\cdot3-5)=(8\sqrt3-4\sqrt5)/(12-5)\n\n=(8\sqrt3-4\sqrt5)/(7)
(1) (4)/(2+√(3) +√(7)) \n \n or, (4)/(2+√(3) +√(7)) * (2 - √(3) -√(7))/(2-√(3)-√(7)) \n \n => \frac{ \sqrt[2]{3} - √(21)+3}{3} \n \n \n (2) \frac{4}{\sqrt[2]{3} + √(5)} \n \n or, \frac{4}{\sqrt[2]{3} + √(5)} * \frac{\sqrt[2]{3}-√(5)}{\sqrt[2]{3}-√(5)} \n \n => \frac{\sqrt[8]{3}-\sqrt[4]{5}}{7}

How to solve this system of equation 3x + y= 17 and 4x + 2 y = 20

Answers

3x + y= 17        (a)
4x + 2 y = 20    (b)

Multiply (a) by 2:

6x + 2y = 34   (a)
4x + 2y = 20   (b)

make:   (a) - (b)

2x = 14 -----> x =7

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Replace the value of x = 7 in one of anterior equation:

3x + y= 17  
3 x 7 + y = 17
21 + y = 17
y = 17 - 21
y = -4

Orígenes de los números naturales

Answers

The natural numbers are presumed to have started before recorded history when humans began to count things. The Babylonians developed a place-value system based on the numerals for 1 (one) and 10 (ten). The ancient Egyptians added to this system to include all the powers of 10 up to one million.

Which is another way to check the sum of 52+23+10+78

Answers

well first off those numbers add up to 163.

A way to check this is to add 52 with 23 which is 75 then add 78 by 10 which is 88 now add 75 with 88 which is 163
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