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Svetlanka [38]
3 years ago
11

Two boxes have the same volume. One box has a base that is 5cm by 5 The other box has a base that is 10cm by 10cm

Mathematics
2 answers:
Nadusha1986 [10]3 years ago
7 0

Answer:2

Step-by-step explanation:

Lorico [155]3 years ago
5 0
The answers is 2 do you need explain
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Find the inverse function F -1 (X) for the given function F (X)
charle [14.2K]

<u>Solution for question 5:</u>

f(x) = 4x - 8

To find f^-1(x) write it as an equation:

y = 4x - 8

Then take x to left and y to right;

x = 4y - 8

Then solve for y;

y = (x + 8)/4 = x/4 + 2

Your y is the inverse of f(x) = 4x - 8

f^-1(x) = x/4 + 2


8 0
3 years ago
1/2(12•x)=276 solve for x
tatiyna
The answer is 11.5 (please mark me as brainliest if true)
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Someone please help this is on a test
ElenaW [278]

Answer:

The angle 7 is equal to angle 1. They are vertically opposite angles.

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Find the area of the trapezoid. helppppopop thank you
Arada [10]

Answer:

\frac{4}{3}\:\mathrm{cm^2}

Step-by-step explanation:

The area of a trapezoid can be found by multiplying the average of its bases and its height.

We're given:

  • One base of 4 cm
  • One base of 12 cm
  • Height of 1/6 cm

To find the average of a set of n values, add all the values in the set and divide by n. Therefore, to find the average of the two bases, we add 4 to 12 and divide by 2.

The average of the bases is therefore \frac{4+12}{2}=\frac{16}{2}=8

Thus, the area of the trapezoid is 8\cdot \frac{1}{6}=\frac{8}{6}=\boxed{\frac{4}{3}\:\mathrm{cm^2}}

6 0
3 years ago
Triangle $ABC$ is situated within an ellipse whose major and minor axes have lengths 10 and 8, respectively. Point $A$ is locate
mina [271]

Answer:

The coordinates of the incenter ≈ (0.0563, 0.136)

The length of the inradius ≈ 2.3634

Step-by-step explanation:

The lengths of the major and minor axis of the ellipse in which ΔABC is situated are;

Major axis = 10, minor axis = 8

The location of point A = The focus of the ellipse

The location of point B = An endpoint of the minor axis

The location of point C =  On the ellipse such that the other focus lies on \overline{BC}

B(0, 4), P(3, 0)

The equation of the line BPC y - 0 = ((-4)/3)·(x - 3)

y = -4·x/3 + 4

Therefore, at 'C', we have;

1 = x²/5² + y²/4² =  x²/5² + (-4·x/3 + 4)²/4²

Using an online tool, we get;

1184·x²/49 = 400

x = ± √(400×49/1184)

At C, x = √(400×49/1184) ≈ 4.07

y = -4 × (√(400×49/1184))/3 + 4 ≈ -1.425

The coordinates of point C = (4.07, -1.425)

The coordinates of point A = (-3, 0)

The coordinates of point B = (0, 4)

The length of AB = √(4² + (0 - 3)²) = 5

The length of AC = √((-3 - 4.07)² + (0 - (-1.425))²) ≈ 7.2122

The length of BC = √((0 - 4.07)² + (4 - (-1.425))²) ≈ 6.782

The coordinates of the incenter = (4.07 + (-3) + 0)/(5 + 7.2122 + 6.782), (0 + 4 + (-1.425))/(5 + 7.2122 + 6.782)) ≈ (0.0563, 0.136)

The perpendicular to the line AB (y = (4/3)·x + 4) from the incenter

The slope equation of the line perpendicular to AB = -3/4

The equation of the line perpendicular to AB = y - 0.136 = (-3/4)·(x - 0.0563)

y = (-3/4)·(x - 0.0563) + 0.136

∴ At the point of intersection, (-3/4)·(x - 0.0563) + 0.136 = (4/3)·x + 4)

Solving gives, x = -(46.368 - 2.0268/4)/25 ≈ -1.83442

y = (-3/4)×((-1.83442) - 0.0563) + 0.136 ≈ 1.55404

The radius ≈ √((1.55404 - 0.136)² + ((-1.83442) - 0.0563)²) ≈ 2.3634

The length of the inradius ≈ 2.3634

4 0
3 years ago
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