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2 years ago

Greg's home is 1.8 miles from his school.

Mathematics

1 answer:

stiv31 [10]2 years ago

4 0

Answer:

a. 0.9 miles

b. 0.54 miles

c. 1.44 miles

Step-by-step explanation:

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Z2 + (2j – 3)z + (5 – j) = 0

BARSIC [14]

Answer:

2z+(2j-3)z+(5-j)=0

2z +2jz-3z+5-j=0

2z-3z+2jz-j=-5

-z+2jz-j=-5

2jz-z-j=-5

I hope this help

4 0

3 years ago

The area of a trapezoid is 39 square millimeters. The height of the trapezoid is 6 millimeters. One of the base lengths of the t

krok68 [10]

39×2÷6=13
13-5=8 mm. The other base of the trapezoid is 8 mm. Let check it:
1/2(8+5)×6
=1/2×13×6
=39 square mm. Hope it help!

5 0

4 years ago

The equation t^3=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

ankoles [38]

Answer:

2√2

Step-by-step explanation:

We can find the relationship of interest by solving the given equation for A, the mean distance.

<h3>Solve for A</h3>

$T^3=A^2\\\\A=\sqrt{T^3}=T\sqrt{T}\qquad\text{take the square root}$

<h3>Substitute values</h3>

The mean distance of planet X is found in terms of its period to be ...

$D_x=T_x\sqrt{T_x}$

The mean distance of planet Y can be found using the given relation ...

$T_y=2T_x\\\\D_y=T_y\sqrt{T_y}=2T_x\sqrt{2T_x}=(2\sqrt{2})T_x\sqrt{T_x}\\\\D_y=2\sqrt{2}\cdot D_x$

The mean distance of planet Y is increased from that of planet X by the factor ...

2√2

8 0

2 years ago

Can someone please help me fill in the answer in the bottom.

Julli [10]

Answer:

Step-by-step explanation:

1) Initial water level = 30 feet

2) Rate of change = (30 - 25) / (0 - 1) = 5/(-1) = -5

5ft per hour

3) y-intercept = 30

6 0

3 years ago

Read 2 more answers

High Points!!!

Karo-lina-s [1.5K]

Answer:

Step-by-step explanation:

5 0

3 years ago