In the rectangle LOBA, if m∠1 = 36°. Find m∠OBA and m∠2.<br> m∠OBA = <br> m∠2 =

A millimetre is ______meter(s)

Dmitriy789 [7]

Answer:

1000 mm

Step-by-step explanation:

5 0

3 years ago

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Whick scenario is most likely the one shown on the graph?

serg [7]

Answer:

yes

Step-by-step explanation:

5 0

3 years ago

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If triangle FGH, FG=8.7, FH=4.5 and GH=6.1, Which is the smallest angle of the triangle

jolli1 [7]

∠G is the shortest

Step-by-step explanation:

Given is a ΔFGH

FG = 8.7

FH = 4.5

GH = 6.1

The triangle will be formed with three angles

F, G and H

"The side opposite to the largest angle of the triangle is the largest"

Similarly, The side opposite to the shortest angle is the shortest.

Converse of this postulate will be:

The interior angle opposite to the shortest side of the triangle is the smallest.

The triangle formed is attached in the form of a picture.

The angle opposite to FH will be the shortest.

Hence,

∠G is the shortest

Keywords: Angles, triangles

Learn more about angles at:

#LearnwithBrainly

7 0

3 years ago

A deli sells a sandwich bread for $6.40 per pound . Much do you have to pay for 24 ounces of the spread?

anygoal [31]

I think it is $2.40 (but I am not too sure) try to get some more people to answer.
Good Luck!!!!!☺☻☺☻☺☻

3 0

3 years ago

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Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,

sweet [91]

Answer:

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

$f'' = x - \frac{3}{2}$

$f' = \int {\left(x-\frac{3}{2}\right) } \, dx$

$f' = \int {x} \, dx -\frac{3}{2}\int \, dx$

$f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C$ , where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative ( $f'(4) = 1$ ):

$1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C$

$C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)$

$C = -1$

The first derivative is $y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1$ , and the particular solution is found by integrating one more time and using the initial condition ( $f(0) = 0$ ):

$y = \int {\left(\frac{1}{2}\cdot x^{2}-\frac{3}{2}\cdot x -1 \right)} \, dx$

$y = \frac{1}{2}\int {x^{2}} \, dx - \frac{3}{2}\int {x} \, dx - \int \, dx$

$y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x + C$

$C = 0 - \frac{1}{6}\cdot 0^{3} + \frac{3}{4}\cdot 0^{2} + 0$

$C = 0$

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

5 0

4 years ago

In the rectangle LOBA, if m∠1 = 36°. Find m∠OBA and m∠2. m∠OBA = m∠2 =