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sveticcg [70]
3 years ago
13

HELP ME ON THIS PLEASE!!!! SHOW WORK!!!!

Mathematics
1 answer:
Mumz [18]3 years ago
3 0

Answer:

<h2>1. x = 4</h2><h2>2. x = 20</h2>

Step-by-step explanation:

1.

ΔABC and ΔAJK are similar (AA). Therefore the sides are in proportion:

\dfrac{AC}{AJ}=\dfrac{AB}{AK}

We have:

AC = 1 + 4 = 5

AJ = 1

AB = 1 + x

AK = 1

Substitute:

\dfrac{5}{1}=\dfrac{1+x}{1}

5=1+x           <em>subtract 1 from both sides</em>

4=x\to x=4

2.

ΔVUT and ΔVMN are similar (AA). Therefore the sides are in proportion:

\dfrac{VU}{VM}=\dfrac{VT}{VN}

We hve:

VU = x + 8

VM = x

VT = 49

VN = 49 - 14 = 35

Substitute:

\dfrac{x+8}{x}=\dfrac{49}{35}              <em>cross multiply</em>

35(x+8)=49x          <em>use the distributive property a(c + b) = ab + ac</em>

35x+280=49x              <em>subtract 35x from both sides</em>

280=14x         <em>divide both sides by 14</em>

20=x\to x=20

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7 0
3 years ago
Read 2 more answers
1. Given points ????(3, −5) and ????(19, −1), find the coordinates of point ???? that sit 3/8 of the way along AB ,close to A th
Nastasia [14]

Answer:  The required co-ordinates of point C are (9, -3.5).

Step-by-step explanation:   We are given the points A(3, -5) and B(19, -1).

We are to find the co-ordinates of point C that sit 3/8 of the way along AB, where the point P is close to A than to B.

According to the given information, we have

\dfrac{AC}{AB}=\dfrac{3}{8}\\\\\Rightarrow \dfrac{AC}{AC+BC}=\dfrac{3}{8}\\\\\Rightarrow 8AC=3AC+3BC\\\\\Rightarrow 5AC=3BC\\\\\Rightarrow AC:BC=3:5.

So, point C divides the line segment AB in the ratio 3 : 5.

We know that

if a point Q divides a line segment joining the points S(a,b) and T(c,d), in the ratio m : n, then the co-ordinates of Q are

\left(\dfrac{mc+na}{m+n},\dfrac{md+nb}{m+n}\right).

Therefore, the co-ordinates of point C are

\left(\dfrac{3\times19+5\times3}{3+5},\dfrac{3\times(-1)+5\times(-5)}{3+5}\right)\\\\\\=\left(\dfrac{57+15}{8},\dfrac{-3-25}{8}\right)\\\\=(9,-3.5).

Thus, the required co-ordinates of point C are (9, -3.5).

3 0
4 years ago
A cup of coffee contains 130 mg of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how much caffein
german
---<span>
y(t) = a*e^(kt)</span><span>
---
now:</span><span>
130 mg</span><span>
---
one hour from intake:</span><span>
130 * (1 - 0.11) = 115.7 mg</span><span>
---
y(1) = 115.7 = 130*e^(k(1))</span><span>
115.7 = 130*e^k</span><span>
115.7/130 = e^k</span><span>
ln( 115.7/130 ) = ln( e^k )</span><span>
ln( 115.7/130 ) = k</span><span>
k = -0.1165338

</span><span>---
three hours from intake:</span><span>
y(3) = 130*e^(3k)</span><span>
y(3) = 91.64597</span><span>
---
answer:</span><span>
three hours after drinking a cup of coffee, 91.6 mg of caffeine will remain in the body

</span>
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4 years ago
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The first on is this answer here

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A student pulls a rope with a force of 20 N to the left. Another student pulls on the
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Answer:

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