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4 years ago

A train travels 558 miles in 3 hours at this rate how far does the train travel per hour

Mathematics

2 answers:

nikitadnepr [17]4 years ago

5 0

558/3=186 ;P so it is 186

adell [148]4 years ago

4 0

Divide 558 by 3 for your answer bud!!!

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Evaluate and answer in standard form:<br><br>(3^2)^2

algol13

${ ({3}^{2}) }^{2}$

${3}^{2 \times 2}$

${3}^{4}$

$= 81$

4 0

3 years ago

10. The area of a square frame is 55 square inches. Find the length of one side of the

Amanda [17]

Answer:

see explanation

Step-by-step explanation:

The area (A) of a square is calculated as

A = s² ( s is the side length )

Given A = 55 , then

s² = 55 ( take the square root of both sides )

s = $\sqrt{55}$ ≈ 7 in ( to the nearest whole inch )

s ≈ 7.4 in ( to the nearest tenth of an inch )

6 0

3 years ago

Read 2 more answers

What is the area of the circle?

dmitriy555 [2]

We know that the area of a circle has the formula A=πr²

We are given the diameter, and as we know that the diameter is twice the radius, we can divide the given diameter by 2.

27 ÷ 2= 13.5

Now, plug-in the number into the formula.

A=πr²

A=π×13.5²

A=π×182.25

(Assume π=3.14)

A=3.14 × 182.25

A=572.265

A=572.3 square meters.

Therefore, the area of the circle is 572.3 square meters.

8 0

3 years ago

Simplify the complex numbers using de Moivre's Theorem and match them with their solutions.

alexandr1967 [171]

For the numbers in $a+bi$ form, convert to polar form:

$1+i=\sqrt2\dfrac{1+i}{\sqrt2}=\sqrt2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)$

By DeMoivre's theorem,

$(1+i)^5=(\sqrt2)^5\left(\cos\dfrac{5\pi}4+i\sin\dfrac{5\pi}4\right)=4\sqrt2\dfrac{-1-i}{\sqrt2}=-4-4i$

$-1+i=\sqrt2\dfrac{-1+i}{\sqrt2}=\sqrt2\left(\cos\dfrac{3\pi}4+i\sin\dfrac{3\pi}4}\right)$

$\implies(-1+i)^6=(\sqrt2)^6\left(\cos\dfrac{18\pi}4+i\sin\dfrac{18\pi}4\right)=8i$

$\sqrt3+i=2\dfrac{\sqrt3+i}2=2\left(\cos\dfrac\pi6+i\sin\dfrac\pi6\right)$

$\implies2(\sqrt3+i)^{10}=2^{11}\left(\cos\dfrac{10\pi}6+i\sin\dfrac{10\pi}6\right)=2^{11}\dfrac{1-i\sqrt3}2=2^{10}(1-i\sqrt3)$

For the numbers already in polar form, DeMoivre's theorem can be applied directly:

$2\left(\cos20^\circ+i\sin20^\circ\right)^3=2\left(\cos60^\circ+i\sin60^\circ\right)=2\dfrac{1+i\sqrt3}2=1+i\sqrt3$

$2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)^4=2(\cos\pi+i\sin\pi)=-2$

At second glance, I think the 2s in the last two numbers should also be getting raised to the 3rd and 4th powers:

$\left(2(\cos20^\circ+i\sin20^\circ)\right)^3=8\left(\cos60^\circ+i\sin60^\circ\right)=4+4\sqrt3$

$\left(2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)\right)^4=16(\cos\pi+i\sin\pi)=-16$

4 0

3 years ago

The line through (1,6) and (0,3) passes through every quadrant expect one. Which one?

KATRIN_1 [288]

4 quadrant

plot the graph of the line

6 0

3 years ago