Determine the value of y, if x is -2. y=

F = Gm^2/r^2 solve for g

TEA [102]

$F=\frac{Gm^{2}}{r^{2}}$ $\rightarrow Fr^{2}=Gm^{2}$ $\rightarrow G=\frac{Fr^{2}}{m^{2}}$

5 0

3 years ago

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Alex777 [14]

Answer:

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3 0

2 years ago

Simplify -1.5(-3.6+2) Pleaseee people

Temka [501]

Answer:

Your answer is 2.4

Step-by-step explanation:

7 0

3 years ago

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Prove diagonals of a rhombus are perpendicular

Lisa [10]

Answer:

A rhombus is a parallelogram with four congruent sides.

So, all sides of rhombus ABCD are congruent.

i.e, $\overline{JM} \cong \overline{JK} \cong \overline{KL}\cong \overline{LM}$

Also, we know that the diagonals of a parallelogram bisect each other.

Since a rhombus is a parallelogram.

By property of rhombus , if point N is the intersection of the diagonals as shown in the figure, then

$\overline{MN} \cong \overline{NK}$ .....[1]

$\overline{JN} \cong \overline{NL}$

In ΔJNM and ΔJNK

$\overline{MN} \cong \overline{NK}$ [side] [by (1)]

$\overline{JM} \cong \overline{JK}$ [side] [Given]

By reflexive property states that a segment is congruent to itself:

$\overline{JN} \cong \overline{JN}$ [Side] [Reflexive Property]

SSS(Side-Side-Side) postulates states that if three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent.

then by SSS congruence,

$\triangle JNM \cong \triangle JNK$

By CPCT [Corresponding Part of congruent triangles are congruent]

Since, JNM and JNK are corresponding angles therefore,

$\angle JNM \cong JNK$

Linear pair theorem states that two angles that form a linear pair are supplementary.

By linear pair theorem, JNM and JNK are supplementary

this mean:

$m\angle JNM +m\angle JNK =180^{\circ}$

Since, the angles are congruent i.e, $\angle JNM \cong JNK$

so;

$m\angle JNK +m\angle JNK =180^{\circ}$

or

$m\angle JNK +m\angle JNK =180^{\circ}$

2 $m\angle JNK=180^{\circ}$

Simplify:

$m\angle JNK =90^{\circ}$

also; $m\angle JNM =90^{\circ}$

therefore, the diagonals of JKLM are perpendicular to each other i,e $\overline{JL} \perp \overline{MK}$

4 0

3 years ago

2. Kelsey has a list of possible functions. Pick one of the g(x) functions below and then describe to Kelsey the key features of

Gemiola [76]

Answer:

g(x) = (x + 2)(x − 1)(x − 2)

I picked the first one, because why not. :)

End behavior. This is a positive function; there are no negatives next to any of the xs. There are 3 xs, so it's a cubic; a positive cubic has a down-up end behavior.

Y-intercept. We can either graph this on the calculator/online or work it out algebraically. Multiply together the constants: +2 * -1 = -2, -2 * -2 = 4. So the y-intercept is at y = 4. (0,4). I've also attached the graph.

Zeros / x-intercept. This is really simple: set g(x) equal to 0, separate these three parenthesis (x+2) (x-1) and (x-2) into three equations by the Zero Product Property.

x+2 = 0

x = -2

x-1 = 0

x = 1

x-2 = 0

x = 2

So the function has three zeros at (-2,0) (1,0) and (2,0).

3 0

3 years ago

Determine the value of y, if x is -2. y= | x |– 9 Answer: