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3 years ago

(3^2+4−2)(2^2−3−4) can you guys please solve this for me please.

Mathematics

1 answer:

Degger [83]3 years ago

6 0

Answer:

-33

Step-by-step explanation:

$(3^2+4-2)(2^2-3-4)$

$(9+4-2)(4-3-4)$

$(13-2)(1-4)$

$(11)$ · $(-3)$

(-33)

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Can someone answer this question please answer it correctly if it’s corect I will mark you brainliest

Karolina [17]

Answer:

1/2 or 50%

Step-by-step explanation:

There are 4 total cards

The number of odd or 5 is 3 or 5 so a total of 2 choices

P(5 or odd) = { 5 or odd}/ total

= 2/4

=1/2

8 0

3 years ago

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Please help, I will give brainliest to best answer

tester [92]

Answer:

060.28

Step-by-step explanation:

3 can't go into 1, but it goes into 18 six times with 0 left over. 3 goes into 0 zero times. 3 goes into 5 one time with 2 left over. 3 goes into 26 (because you pull the 6 down) 8 times with 2 left over.

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2 years ago

Determine the standard form of the equation of the line that passes through (-5,0) and (0,-9)

bonufazy [111]

Answer:

9x+5y=-45

Step-by-step explanation:

To write the equation of a line we must have a slope and a point. To find the slope we use the slope formula and substitute (x,y) points in it as shown below:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{-9-0}{0--5}=\frac{-9}{0+5}=\frac{-9}{5}=\frac{-9}{5}$

Now that we have the slope, plug in the slope and choose one point to plug into the point slope formula. Use the point-slope form to write the equation, then simplify and convert into the slope intercept form.

(y-0)=-9/5(x--5)

y=-9/5(x+5)

y=-9/5x-9

Now convert to standard form, Ax+By=C.

y=-9/5x-9

9/5x+y=-9

9x+5y=-45

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3 years ago

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aleksandrvk [35]

Answer: Irregular Hexagon

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3 years ago

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Find the vertices and foci of the hyperbola with equation quantity x minus 5 squared divided by 144 minus the quantity of y minu

Serjik [45]

$\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases}\\\\ -------------------------------$

$\bf \cfrac{(x-5)^2}{144}-\cfrac{(y-4)^2}{81}=1\implies \cfrac{(x-5)^2}{12^2}-\cfrac{(y-4)^2}{9^2}=1 \\\\\\ \begin{cases} h=5\\ k=4\\ a=12\\ b=9 \end{cases}\implies c=\sqrt{144+81}\implies c=\sqrt{225}\implies c=15 \\\\\\ \stackrel{center}{(5,4)}\qquad \qquad \stackrel{\textit{because is a horizontal traverse hyperbola}}{\stackrel{foci}{\stackrel{(5\pm 15,4)}{(20,4),(-10,4)}}\qquad \qquad \stackrel{vertices}{\stackrel{(5\pm 12,4)}{(17,4),(-7,4)}}}$

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3 years ago