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Elanso [62]
3 years ago
8

iddle" class="latex-formula">

Mathematics
2 answers:
4vir4ik [10]3 years ago
6 0

x= -1

your solve X that is it

anzhelika [568]3 years ago
5 0

a) y-intercept = 5

b) (x-1)(x-5)

      step 1: find factors of 1st and last terms

      step 2: multiply the factors "rainbow" style (like in the picture) and add          the products to make sure they equal the middle term

x^{2} -6x+5\\1,1-1,-5

(-1)+(-5)=-6\\=(x-1)(x-5)

c) x=1,5

(x-1)(x-5)\\x-1=0x-5=0

x=1             x=5

d) AoS= x=3

e) vertex=(3,-4)

(h,k)\\h=-\frac{b}{2a} \\h=-\frac{(-6)}{2(1)} \\h=-(-3)\\h=3\\k=f(3)=(3)^2-6(3)+5\\k=9-18+5\\k=-4

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Need help with pre-cal
timurjin [86]

A plot of the points A(-2, 11), B(5, 7), C(1, 0) is given by the option;

  • D

  • The triangle is a right triangle given that (AC)² = (AB)² + (BC)²

  • The area of the ∆ABC is 32.5 square units.

  • Sum of the squares of the lengths of the legs of the triangle = <u>130</u>

  • Square of the length of the hypotenuse of the triangle = <u>130</u>

<h3>How can ∆ABC be proven to be a right triangle from its dimensions?</h3>

The coordinates of the vertices of the triangle are;

A = (-2, 11), B = (5, 7), C = (1, 0)

Therefore, on the coordinate plane, we have;

The highest and leftmost point of the triangle is the vertex, <em>A</em>

The second highest and rightmost point of the triangle is the vertex, <em>B</em>

The<em> </em>vertex of the triangle that is midway between <em>A </em>and <em>B </em>and the lowest vertex of the triangle is the vertex <em>C</em>

  • The<em> </em>correct<em> </em>option that shows the points ABC and triangle ABC is the option <em>D</em>.

The lengths of the sides of the triangle are;

AB = √((-2 - 5)² + (11 - 7)²) = √(65)

BC = √((5 - 1)² + (7 - 0)²) = √(65)

AC = √((-2 - 1)² + (11 - 0)²) = √(130)

Therefore;

(AC)² = 130 = (AB)² + (BC)² = 65 + 65

Which gives;

  • (AC)² = (AB)² + (BC)²

Therefore;

  • ∆ABC is a right triangle, from the definition of a right triangle.

The legs of ∆ABC are AB and BC

AC is the hypotenuse of ∆ABC

The area of ∆ABC is therefore;

Area = (1/2) × AB × BC

Which gives;

Area of ∆ABC = (1/2) × √(65) × √(65)

√(65) × √(65) = 65

  • Area of ∆ABC = (1/2) × 65 = 32.5

(AB)² + (BC)² = 65 + 65 = 130

Therefore;

  • Sum of the squares of the lengths of the legs of the triangle = 130

(AC)² = 130

  • Square of the length of the hypotenuse of the triangle = 130

Learn more about right triangles here:

brainly.com/question/2284306

#SPJ1

3 0
2 years ago
A box has a volume of 192 cubic inches, a length that is twice as long as its width, and a height that is 2 inches greater than
Volgvan

Answer:

Length = 8 inches

Width = 4 inches

Height = 6 inches

Step-by-step explanation:

Volume of the box = 192 cubic inches

Volume of a box = length × width × height

Let

Width = x

Length = 2x

Height = x + 2

Volume of a box = length × width × height

192 = 2x * x * (x + 2)

192 = 2x^2 (x + 2)

192 = 2x^3 + 4x^2

Divide through by 2

96 = x^3 + 2x^2

Subtract 96 from both sides

x ^3 + 2x^2 - 96 = 0

Factorise

(x - 4) (x^2 + 6x + 24) = 0

x^2 + 6x + 24 has no real x-value

So,

Divide both sides by (x^2 + 6x + 24).

0 ÷ (x^2 + 6x + 24) = 0

So,

(x - 4) = 0

x = 4

Width = x = 4 inches

Length = 2x

= 2(4)

= 8 inches

Height = x + 2

= 4 + 2

= 6 inches

Check:

Volume of a box = length × width × height

= (8 * 4 * 6) inches

= 192 inches

4 0
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liubo4ka [24]

Answer:

It would be A.10

Step-by-step explanation:

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9/2.25= 4

40/10 = 4

6 0
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Anastasy [175]
3/5 of 180 is 108
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Answer: 7 to the 12 power

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