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2 years ago

What does 6/4 x 2/1 equals?

Mathematics

2 answers:

Lynna [10]2 years ago

6 0

Answer:

Step-by-step explanation:

steposvetlana [31]2 years ago

5 0

Answer:

Step-by-step explanation:

6/4 .2/1 divide the numbers 6/4 .2/1 3/2 .2/1 then multiply the numbers 3/2 .2/1/3/1 so it's 3

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Write as many equations as possible that could represent the relationship between

IgorC [24]

Answer:

a) Here we have:

"In Family A, the youngest child is 7 years younger than the oldest, who is 18"

Let's define:

Y = age of the youngest child.

O = age of the oldest child.

Then we know that:

Y = O - 7

O = 18

Then we can replace the second equation into the first one:

Y = 18 - 7 = 11.

b) Here we have:

"In Family B, the middle child is 5 years older than the youngest child."

Let's define:

Y = age of the youngest child.

M = age of the middle child.

Here we have only one equation:

M = Y + 5.

5 0

3 years ago

Zebras migrate in a circle with a round trip of about 300 miles. They travel at 64 km/h. How many hours does it take zebra to do

Tresset [83]

Answer:

7.5 hours (7 hours and 30 minutes)

Step-by-step explanation:

ok Let's start with 64km/h.

1.6 km=1 mile,

64÷1.6=40miles

so 64km/h changes into 40miles/hour.

Zebras's round trip =300miles,

300/40=7.5

so answer is 7.5 hours.

8 0

3 years ago

11 3/4-7 1/4 evaluate answer need asap

Debora [2.8K]

Answer:

−6

Step-by-step explanation:

8 0

2 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$