All categories

2 years ago

The length of the midsegment of a triangle is always half the length of the side it is parallel to true or false

1 answer:

3 0

True. The midsegment of a triangle is halfway up the triangle, and since the top half and the full triangle are similar triangles, its dimensions are proportional.

This means half the height will give half the base (midsegment). So the midsegment is always half the length of the side it's parallel to.

You might be interested in

Marley has read 112.5 pages of a book. By the end of today, she plans to have read at least a total of 360 pages. If she reads 4

Vinvika [58]

5.5 hours. 5 hours and 30 minutes

The total pages she wants to finish (360) minus the pages she has read (112.5) equals to 247.5
Then divide it by the 45 (the rate of pages she reads per hour)
5.5 hours.

6 0

2 years ago

Really need help please

Anettt [7]

I am too lazy I'll explain so you find 688 minus 267 then you divide the difference by 6 and then u subtract the difference from 267x100 because that will be the total. Then u find the mean of the 94 numbers u get!

3 0

2 years ago

Simplify 9/4 using fractions.

seropon [69]

First to Answer:

The answer is 2 1/4

Have a great day!

Work:

9 divided by 4 is 2.25.

0.25 is equivalent to 1/4 and 2.00 is equivalent to 2.

put those together and you get 2 1/4.

8 0

2 years ago

Let R be the region in the first quadrant of the xy-plane bounded by the hyperbolas xyequals1, xyequals9, and the lines yequa

Tema [17]

Answer:

The area can be written as

$\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = 0.2274$

And the value of it is approximately 1.8117

Step-by-step explanation:

x = u/v

y = uv

Lets analyze the lines bordering R replacing x and y by their respective expressions with u and v.

x*y = u/v * uv = u², therefore, x*y = 1 when u² = 1. Also x*y = 9 if and only if u² = 9

x=y only if u/v = uv, And that only holds if u = 0 or 1/v = v, and 1/v = v if and only if v² = 1. Similarly y = 4x if and only if 4u/v = uv if and only if v² = 4

Therefore, u² should range between 1 and 9 and v² ranges between 1 and 4. This means that u is between 1 and 3 and v is between 1 and 2 (we are not taking negative values).

Lets compute the partial derivates of x and y over u and v

$x_u = 1/v$

$x_v = u*ln(v)$

$y_u = v$

$y_v = u$

Therefore, the Jacobian matrix is

$\left[\begin{array}{ccc}\frac{1}{v}&u \, ln(v)\\v&u\end{array}\right]$

and its determinant is u/v - uv * ln(v) = u * (1/v - v ln(v))

In order to compute the integral, we can find primitives for u and (1/v-v ln(v)) (which can be separated in 1/v and -vln(v) ). For u it is u²/2. For 1/v it is ln(v), and for -vln(v) , we can solve it by using integration by parts:

$\int -v \, ln(v) \, dv = - (\frac{v^2 \, ln(v)}{2} - \int \frac{v^2}{2v} \, dv) = \frac{v^2}{4} - \frac{v^2 \, ln(v)}{2}$

Therefore,

$\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = \int\limits_1^2 (\frac{1}{v} - v \, ln(v) ) (\frac{u^2}{2}\, |_{u=1}^{u=3}) \, dv= \\4* \int\limits_1^2 (\frac{1}{v} - v\,ln(v)) \, dv = 4*(ln(v) + \frac{v^2}{4} - \frac{v^2\,ln(v)}{2} \, |_{v=1}^{v=2}) = 0.2274$

4 0

3 years ago

Evaluate. 4⋅(4+(26⋅2−3)) 48 24 12 2

RSB [31]

I got the answer 212

4*(4+(26*2-3))

4*(4+52-3)

4*(56-3)

4*53

212

Hope this helps, have a BLESSED day! :-)

8 0

2 years ago