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3 years ago

A number cube has faces numbered 1 to 6.

Mathematics

2 answers:

Arturiano [62]3 years ago

7 0

Answer:

Hmmmm, what I think is you are 100% right if I'm wrong put it in the comment please, Thanks!

nikdorinn [45]3 years ago

5 0

Answer:

yes

Step-by-step explanation:

your right

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Jodie bought some shirts for $6 each. Marge bought some shirts for $8 each. The girls spent the same amount of money on shirts.

ELEN [110]

$24.00 i hope this helped you

3 0

3 years ago

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Solve pls brainliest

Oksana_A [137]

Answer:

y = 13x

Step-by-step explanation:

Substitute 2 in x and you'll get 26:

y = 13•2

y = 26

4 0

3 years ago

Jane draws a line in the coordinate plane that passes through the points (2,3) and (6,9)

timofeeve [1]

Answer:

y=x+1

Step-by-step explanation:

To find the line of the equation, use the slope formula to find the slope.

Slope formula is: y2-y1/x2-x1

9-3/6-2= 3/4

Once you have the slope, use that and one of the coordinates to put the equation in point-slope form.

Point-slope form is: y-y1=m(x-x1), with m representing slope and y1 and x1 representing the x and y coordinates.

y-3=1(x-2)

With the equation in point-slope form, you can now distribute 1 to x-2 and simplify to put it in slope-intercept form.

y-3=1x-2

y=x+1

3 0

3 years ago

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saul85 [17]

$\large\underline{\sf{Solution-}}$

Let us assume that:

$\sf \longmapsto x = \sqrt{6 + \sqrt{6 + \sqrt{6 + ... \infty } } }$

We can also write it as:

$\sf \longmapsto x = \sqrt{6 + x }$

Squaring both sides, we get:

$\sf \longmapsto {x}^{2} =6 + x$

$\sf \longmapsto {x}^{2} - x - 6 =0$

By splitting the middle term:

$\sf \longmapsto {x}^{2} - 3x + 2x - 6 =0$

$\sf \longmapsto x(x - 3) + 2(x - 3 ) =0$

$\sf \longmapsto (x+ 2)(x - 3 ) =0$

Therefore:

$\longmapsto\begin{cases} \sf (x+ 2) =0 \\ \sf (x - 3) = 0 \end{cases}$

$\sf \longmapsto x = - 2,3$

But x cannot be negative. 

$\sf \longmapsto x = 3$

Therefore, the value of the expression is 3.

$\large\underline{\sf{Verification-}}$

Given:

$\sf\longmapsto x=3$

We can also write it as:

$\sf\longmapsto x = \sqrt{9}$

$\sf\longmapsto x = \sqrt{6+3}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{9}}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+3}}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{9}}}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+3}}}$

This pattern will continue.

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+...\infty}}}$

7 0

3 years ago

Given: quadrilateral MNOL with MN ≅ LO and ML ≅ NO

mr_godi [17]

The answers are ONM, NOM, and alternate interior angles.
Based on the SSS postulate, the two triangles △MLO and △ONM are congruent since three sides of △MLO are respectively equal to the three sides of △ONM.
Based on CPCTC, all of the corresponding angles of △MLO and △ONM are congruent as well since the two triangles are congruent, that is,
∠LMO≅∠NOM and ∠NMO≅∠LOM.
Since the pair ∠LMO and ∠NOM as well as ∠NMO and ∠LOM are angles on the inner side of two lines but on opposite sides of the transversal MO, these pairs of angles are also alternate interior angles.

5 0

3 years ago

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