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seropon [69]
1 year ago
8

Diego is monitoring his weight. On week 1 he weighed 160 pounds, the next week he weighed 158 pounds, and the third week he weig

hed 156 pounds. If his weight
continues to decrease at the same rate, what will his weight be on week 6?
Mathematics
1 answer:
castortr0y [4]1 year ago
4 0

Answer:

150 pounds

Step-by-step explanation:

Subtract two pounds each week.

Week 1- 160

Week 2-158

Week 3 -156

Week 4 -154

Week 5 -152

Week 6 -150

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Answer:

I believe your answer would be A.

Step-by-step explanation:

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3 years ago
Y'ALL, WOULD A KIND SOUL PLEASE PLEASE HELP ME???!!!!!!!!!!! im running out of points
Tasya [4]

1. ∠A and ∠B are complementary

2. ∠A + ∠B = 90° and ∠B + ∠C = 90°

3. ∠A = 90° - ∠B

4. ∠C = 90° - ∠B

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2 years ago
A rectangular piece of metal is 30 in longer than it is wide. Squares with sides 6 in long are cut from the four corners and the
Alexandra [31]

Answer:

length=53 in

width= 23 in

Step-by-step explanation:

First, we know that each corner has now a square of 6 in long, so the heigth of the box will be of this size (as you can see in the picture attached). Then, the length and the width of the box would be the same as the piece, but with 12 in less. Therefore the equations to solve the problem are:

h=6 in\\w=x-12\\l=x+30-12\\l=x+18\\\\V: volume of the box\\V=6*(x-12)*(x+18)\\2706=6x^{2}+36x-1296\\0=6x^{2}+36x-4002\\\\

We factorize the 6 and we get:

0=x^{2}+6x-667\\

Finally we solve for x, and we get the initial dimensions of the piece of metal:

x=23l=30+x=30+23=53 in\\w=x= 23 in

7 0
3 years ago
PLEASE HELP ME:<br><br> Find the perimeter of the figure. Round to the nearest tenth.
Ksenya-84 [330]

Answer:

perimeter = 20.9 units

Step-by-step explanation:

perimeter

perimeter = distance around two dimensional shape

= addition of all sides lengths

<h2>perimeter of the figure</h2><h2>= AB+BC+CD+AD</h2>

distance formula:

d =  \sqrt{(  x_{2} -  x_{1})  {}^{2}   + ( y_{2} -  y_{1}) {}^{2}  }

<h3>1) distance of AB</h3>

A(-3,0) B(2,4)

x1 = -3 x2 = 2

y1 = 0 y2 = 4

(substitute the values into the distance formula)

ab =  \sqrt{(2 - ( - 3)) {}^{2}  + (4 - 0) {}^{2} }

ab =  \sqrt{5 {}^{2} + 4 {}^{2}  }

ab =  \sqrt{41}AB = 6.4 units

<h3>2) distance of BC</h3>

B(2,4) C(3,1)

x1 = 2 x2 = 3

y1 = 4 y2 = 1

bc =  \sqrt{(3 - 2) {}^{2}  + (1 - 4) {}^{2} }

bc = \sqrt{1 {}^{2} + ( - 3) {}^{2}  }

bc =  \sqrt{10}

BC = 3.2 units

<h3>3) distance of CD</h3>

C(3,1) D(-4,-3)

x1 = 3 x2 = -4

y1 = 1 y2 = -3

cd =  \sqrt{( - 4 - 3) {}^{2}  + ( - 3 - 1)) {}^{2} }

cd =  \sqrt{( -   7) {}^{2}  + ( - 4 ){}^{2} }

cd =   \sqrt{65}

CD = 8.1 units

<h3>4) distance of AD</h3>

A(-3,0) D(-4,-3)

x1 = -3 x2 = -4

y1 = 0 y2 = -3

ad =  \sqrt{(  - 4 - ( - 3)) {}^{2}  +  ( - 3 - 0) {}^{2} }

ad =  \sqrt{( - 1) {}^{2} + ( - 3) {}^{2}  }

ad =  \sqrt{10}

AD = 3.2 units

<h2>perimeter of figure</h2>

= AB+BC+CD+AD

= 6.4 + 3.2 + 8.1 + 3.2

= 20.9 units

8 0
3 years ago
Solve for the variable <br>3(h - 9) =36
Diano4ka-milaya [45]
3h - 27 = 36
3h = 63
h = 21
6 0
3 years ago
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