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Komok [63]
3 years ago
11

PLEASE HELP DUE TOMORROW

Mathematics
2 answers:
maxonik [38]3 years ago
7 0
The answers to these questions are C and B
Whitepunk [10]3 years ago
3 0
C and b it would be
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A camper attaches a rope from the top of her tent, 4 feet above the ground, to give it more support. If the rope is 8 feet long,
artcher [175]
8ft is the length of the rope on the outside of the tent. I hope its right!
7 0
3 years ago
I need this answer plzzz!!
Dmitriy789 [7]

Answer: True, False, False, False, False

<u>Step-by-step explanation:</u>

a) 5x - 7(x - 1)

   5x - 7x + 7

         -2x + 7   ⇒  a = -2, b = 7   <em>One Solution</em>

b) 3(x - 5) - 7

   3x - 15 - 7

       3x - 22   ⇒  a = 3, b = -22   <em>One Solution</em>

c) 2 - 7x + 3 + 4x

  4x - 7x + 3 + 2  

              -3x + 5  ⇒   a = -3, b = 5   <em>One Solution</em>    

d) -3(x - 3) - 1

   -3x + 9 - 1

       -3x + 8   ⇒   a = -3, b = 8   <em>One Solution</em>

e) -5x + 2 + 2x + 4

   -5x + 2x + 2 + 4

                  -3x + 6  ⇒   a = -3, b = 6   <em>One Solution</em>  

4 0
3 years ago
1. Gavin and Seiji both worked hard over the summer. Together, they earned a total of $425. Gavin earned $25 more than Seiji. Ho
Gre4nikov [31]
Let x = the amount that Seini earned
Then x + 25 = the amount that Gavin
The amount Gavin earned would also = 425 - x (your directions asked you to have 2 equations.)
Together they earned 425:
x + (x + 25) = 425
2x + 25 = 425
2x = 400
Now divide both sides by 2 to find x.

Use all of this data to satisfy the directions.
8 0
3 years ago
I need to find sec(theta)= 2(sqrt3)/3
Sloan [31]

Answer: 30° and 330°

<u>Step-by-step explanation:</u>

sec θ = \frac{2\sqrt{3}}{3}

sec θ = \frac{1}{cos(\theta)}

\frac{1}{cos(\theta)} = \frac{2\sqrt{3}}{3}

cos θ = \frac{3}{2\sqrt{3}}

cos θ = \frac{3}{2\sqrt{3}}(\frac{\sqrt{3}}{\sqrt{3}})

cos θ = \frac{3\sqrt{3}}{3*2}

cos θ = \frac{\sqrt{3}}{2}

Look at the Unit Circle to see that cos = \frac{\sqrt{3}}{2} at 30° and 330°  ( which is equivalent to π/6 and 11π/6)


4 0
3 years ago
Read 2 more answers
What is the similarity ratio of the smaller to the larger similar cylinders?
lbvjy [14]
\bf \qquad \qquad \textit{ratio relations}&#10;\\\\&#10;\begin{array}{ccccllll}&#10;&Sides&Area&Volume\\&#10;&-----&-----&-----\\&#10;\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}&#10;\end{array} \\\\&#10;-----------------------------\\\\&#10;\cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\&#10;-------------------------------\\\\

\bf \cfrac{smaller}{larger}\qquad \cfrac{s}{s}=\cfrac{\sqrt{48\pi }}{\sqrt{75\pi }}\implies \cfrac{s}{s}=\cfrac{\sqrt{(2^2)^2\cdot 3}}{\sqrt{5^2\cdot 3}}\implies \cfrac{s}{s}=\cfrac{4\sqrt{3}}{5\sqrt{3}}&#10;\\\\\\&#10;\cfrac{s}{s}=\cfrac{4}{5}
8 0
3 years ago
Read 2 more answers
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