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3 years ago

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Mathematics

1 answer:

Alika [10]3 years ago

4 0

When Stacy draws the first marble, her probability of drawing one of the 7 marbles from the 14 in the bag is 7/14. When she draws the second marble, her probability of drawing one of the 6 remaining white marbles from the remaining 13 in the bag is 6/13.

Her probability of drawing two white marbles in succession from the bag is ...
$\dfrac{7}{14}\cdot\dfrac{6}{13}$ matches the 2nd selection

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4x + 5 + 5x + 6 = 29. Whats X?

BigorU [14]

Answer:

Step-by-step explanation:

the full answer is 9x+11=29

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3 years ago

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Use the figure to find the measure of angle 2.

ale4655 [162]

Answer:

∠ 2 = 70°

Step-by-step explanation:

110° and ∠ 1 are corresponding angles and congruent, thus

∠ 1 = 110°

∠ 1 and ∠ 2 are adjacent angles and are supplementary, thus

∠ 2 = 180° - ∠ 1 = 180° - 110° = 70°

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3 years ago

Please i need your help...Thanks

pishuonlain [190]

Lol I’m not sure but I’m logging in so it’s making me “answer”

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3 years ago

Plot (−2 3/4, −4 1/2) on the coordinate plane.

Makovka662 [10]

Because -2 3/4 is on the x axis, it is the diagonal one, -4 1/2 is on the y axis it’s vertical.

The first number is x, the second is y.

Brainliest answer please?

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3 years ago

The area of the triangle formed by x− and y− intercepts of the parabola y=0.5(x−3)(x+k) is equal to 1.5 square units. Find all p

Juliette [100K]

Check the picture below.

based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).

and that will give us two x-intercepts, at x = 3 and x = k.

since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.

now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.

let's find the y-intercept by setting x = 0 now,

$\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)$

$\bf \cfrac{3}{2}=\cfrac{3+k}{2}\left( -\cfrac{3k}{2} \right)\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{2}}{3=\cfrac{(3+k)(-3k)}{2}}\implies 6=-9k-3k^2 \\\\\\ 6=-3(3k+k^2)\implies \cfrac{6}{-3}=3k+k^2\implies -2=3k+k^2 \\\\\\ 0=k^2+3k+2\implies 0=(k+2)(k+1)\implies k= \begin{cases} -2\\ -1 \end{cases}$

now, we can plug those values on A = (1/2)bh,

$\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5$

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3 years ago