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3 years ago

Please help me with this Math

Mathematics

2 answers:

Sliva [168]3 years ago

7 0

Last ! I think sorry if it’s wrong

spayn [35]3 years ago

6 0

The last one I think

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Using symbols write the product of 18 and 7+5

Mashutka [201]

Eighteen times twelve is 216... if that's what you're asking...

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4 years ago

Please help this is due tomorrow

tia_tia [17]

Danielle had a percent error of 13.6 (farthest)
Iris had a percent error of 6.8
HB has a percent error of -2.3 (closest)

So the order is HB, Iris, and then Danielle

7 0

3 years ago

PLEASE ANSWERRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR

nadezda [96]

Hey mate. Here is your answer.

Answer: D. The highlighted land is perfect for growing crops due to abundance of water.

Hope this helps!

4 0

4 years ago

a. For each of the Five Platonic Solids, count the number V of vertices, the number F of faces, and the number E of edges. Check

natita [175]

Answer:

1. The tetrahedron has 4 vertices, 6 edges and 4 faces. Then V-E+F=4-6+4=2

2. The cube has 8 vertices, 12 edges and 6 faces. Then V-E+F=8-12+6=2

3. The octahedron has 6 vertices, 12 edges and 8 faces. Then V-E+F=6-12+8=2

4. The icosahedron has 12 vertices, 30 edges and 20 faces. Then V-E+F=12-30+20=2

5. The dodecahedron has 20 vertices, 30 edges and 12 faces. Then V-E+F=20-30+12=2.

8 0

3 years ago

LORAN is a long range hyperbolic navigation system. Suppose two LORAN transmitters are located at the coordinates (−60,0) and (6

USPshnik [31]

LORAN follows an hyperbolic path.

The equation of the hyperbola is: $\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}$

The coordinates are given as:

$\mathbf{(x,y) = (-60,0)\ (60,0)}$

The center of the hyperbola is

$\mathbf{(h,k) = (0,0)}$

The distance from the center to the focal points is given as:

$\mathbf{c = 60}$

Square both sides

$\mathbf{c^2 = 3600}$

The distance from the receiver to the transmitters is given as:

$\mathbf{2a = 100}$

Divide both sides by 2

$\mathbf{a = 50}$

Square both sides

$\mathbf{a^2 = 2500}$

We have:

$\mathbf{b^2 = c^2 - a^2}$

This gives

$\mathbf{b^2 = 3600 - 2500}$

$\mathbf{b^2 = 1100}$

The equation of an hyperbola is:

$\mathbf{\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1}$

So, we have:

$\mathbf{\frac{(x - 0)^2}{2500} + \frac{(y - 0)^2}{1100} = 1}$

$\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}$

Hence, the equation of the hyperbola is: $\mathbf{\frac{x^2}{2500} + \frac{y^2}{1100} = 1}$