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mrs_skeptik [129]
3 years ago
11

What is the solution to the equation 41/5 =15-(2x+3)

Mathematics
2 answers:
likoan [24]3 years ago
4 0
\frac{41}{5}=15-(2x+3) \\\\ \frac{41}{5}=15-2x-3 \\\\ \frac{41}{5}=12^{(5}-2x^{(5} \\\\ 41=60-10x \\\\ 41-60=-10x \\\\ -19=-10x \\\\ 19=10x \\\\ \boxed{x=\frac{19}{10}}
Sunny_sXe [5.5K]3 years ago
3 0
\dfrac{41}{5}=15-(2x+3)\\
\dfrac{41}{5}=15-2x-3\\
\dfrac{41}{5}=12-2x|\cdot5\\
41=60-10x\\
10x=19\\
x=\dfrac{19}{10}

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Which number line model represents the expression-1 1/2 + 1 1/2
Luba_88 [7]

Answer:

I can't see the other answers, but find the one that goes to 3. It will look like C, but only arrows going up.

Step-by-step explanation:

1 + 1 = 2

1/2 + 1/2 = 1

2 + 1 = 3

7 0
3 years ago
Please help me <br> Show your work <br> 10 points
Svet_ta [14]
<h2>Answer</h2>

After the dilation \frac{5}{3} around the center of dilation (2, -2), our triangle will have coordinates:

R'=(2,3)

S'=(2,-2)

T'=(-3,-2)

<h2>Explanation</h2>

First, we are going to translate the center of dilation to the origin. Since the center of dilation is (2, -2) we need to move two units to the left (-2) and two units up (2) to get to the origin. Therefore, our first partial rule will be:

(x,y)→(x-2, y+2)

Next, we are going to perform our dilation, so we are going to multiply our resulting point by the dilation factor \frac{5}{3}. Therefore our second partial rule will be:

(x,y)→\frac{5}{3} (x-2,y+2)

(x,y)→(\frac{5}{3} x-\frac{10}{3} ,\frac{5}{3} y+\frac{10}{3} )

Now, the only thing left to create our actual rule is going back from the origin to the original center of dilation, so we need to move two units to the right (2) and two units down (-2)

(x,y)→(\frac{5}{3} x-\frac{10}{3}+2,\frac{5}{3} y+\frac{10}{3}-2)

(x,y)→(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})

Now that we have our rule, we just need to apply it to each point of our triangle to perform the required dilation:

R=(2,1)

R'=(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})

R'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(1)+ \frac{4}{3})

R'=(\frac{10}{3} -\frac{4}{3} ,\frac{5}{3}+ \frac{4}{3})

R'=(2,3)

S=(2,-2)

S'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})

S'=(\frac{10}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})

S'=(2,-2)

T=(-1,-2)

T'=(\frac{5}{3} (-1)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})

T'=(-\frac{5}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})

T'=(-3,-2)

Now we can finally draw our triangle:

8 0
3 years ago
How many more poodles did ava observe than bulldogs?
Andreas93 [3]
You need to put the Answers and picture/problem before asking someone to answer it.
8 0
3 years ago
**A rock dropped from a 100 foot tower.
mars1129 [50]

Answer:

2.83 s

Step-by-step explanation:

We are given that

h(t)=-16t^2+10t+100

When a rock reached the ground then h(t)=0

-16t^2+10t+100=0

16t^2-10t-100=0

Using quadratic formula

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

t=\frac{10\pm\sqrt{(-10)^2-4\times 16\times (-100)}}{2(16)}

t=\frac{10\pm\sqrt{100+6400}}{32}

t=\frac{10+\sqrt{6500}}{32}=2.83

t=\frac{10-\sqrt{6500}}{32}=-2.2

Time cannot be negative .Therefore, negative value of t can not be possible.

Hence, the rock takes 2.83 sec to reach the ground.

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Nezavi [6.7K]

Answer:

thanks for the points !!! ;)

Step-by-step explanation:

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3 years ago
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