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nataly862011 [7]
3 years ago
10

HELPPPPPPP asap Condense each expression. show work

Mathematics
1 answer:
Karolina [17]3 years ago
6 0

Answer:

No question?

Step-by-step explanation:

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cluponka [151]

the annnsweeeer is 14.24 degrees

7 0
4 years ago
1. Simplify. (343)1/3
elena-s [515]

1) 343/32 or 10.71875 or 10 23/32
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3)7
8 0
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

7 0
3 years ago
A particle moves on a line away from its initial position so that after t hours it is s = 2t^2 +3t miles from its initial positi
Vaselesa [24]

Answer:

Average speed of the particle is 13 miles per hour.

Step-by-step explanation:

From the given question expression for distance traveled by the particle is

s=2t^{2}+3t

Then fro the time interval (1, 4) we have to calculate the average speed.

At t=1, s=2(1)²+3(1) =2+3 =5 miles

And at t = 4 s=2(4)²+3(4) = 2(16)+12 = 32+12 =44 miles

So displacement of the particle in the time span from 1 to hours.

s = (44-5) = 39 miles

And the times taken = (4-1) =3 hours

From the formula speed = displacement/(time taken for displacement)

V = 39/3 = 13 miles/hour



4 0
3 years ago
In the figure, AngleRQS Is-congruent-to AngleQLK. 3 lines are shown. Lines S P and K N are parallel. Line R M intersects line S
stepan [7]

Answer:

x=108

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

m∠RQS≅m∠QLK -----> by corresponding angles

m∠KLM+m∠QLK=180° -----> by supplementary angles (consecutive interior angles)

we have that

m∠RQS=x° ----> given problem

so

m∠QLX=x°

m∠KLM=(x-36)° ----> given problem

substitute

(x-36)\°+x\°=180\°\\2x=180+36\\2x=216\\x=108

3 0
3 years ago
Read 2 more answers
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