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Anni [7]
3 years ago
8

8x-28=44 How do you solve and graph the solution? HELP ASAP!!

Mathematics
1 answer:
viva [34]3 years ago
6 0
So, idrk how to graph it, but to solve it by multiplying 8 by 8, then carrying 6. Now, you multiply 2 by eight and add 6 which will give you 224. That means somehow the answer u typed in i think was wrong. 
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What value should go in the empty boxes to complete the calculation for finding the product of 0.61 × 0.45? Both numbers are the
nalin [4]

its 0.106 because I said so

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3 years ago
A furniture store had a chair that cost $33. After a few months the owner owner took 9% off the price. How much is the chair now
yawa3891 [41]

Answer:

$30.03

Step-by-step explanation:

To find what 9% of 33 is we need to divide 9% by 1

9% ÷ 1 = 2.97

33 - 2.97 = 30.03

4 0
3 years ago
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Please help me pretty please
dlinn [17]

Answer:

the correct anwser is B................

7 0
3 years ago
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What is the area of a rectangle with vertices at ​ (0, −4) ​, ​ (−1, −3) ​ , (2, 0) , and (3, −1) ? Enter your answer in the box
Bad White [126]

We are given vertices of a rectangle (0, −4) ​, ​ (−1, −3) ​ , (2, 0) , and (3, −1).

Length is the distance between (0, −4) ​and ​ (−1, −3) points.

Width is distance between  (−1, −3) ​ and (2, 0) points.

<u>Computing length:</u>

\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}

=\sqrt{\left(-1-0\right)^2+\left(-3-\left(-4\right)\right)^2}

=\sqrt{2}

<u>Computing Width :</u>

\mathrm{The\:distance\:between\:}\left(-1,\:-3\right)\mathrm{\:and\:}\left(2,\:0\right)\mathrm{\:is\:}

=\sqrt{\left(2-\left(-1\right)\right)^2+\left(0-\left(-3\right)\right)^2}

=3\sqrt{2}

<h3>Area of the rectangle = Length × Width  </h3>

= \sqrt{2} \times 3\sqrt{2} = 3 \times 2 = 6 \ squares \ units..


7 0
3 years ago
PLEASE HELP, WILL MARK BRAINLIEST!
Leokris [45]

Answer:

8,5 feet

Step-by-step explanation:

The equation of the form

x² + y² = (r)²

Is the equation of the circumference

We have for the pool:

x²  +  y²  = 2500      ⇒  x²  +  y²  = (50)²

And for the outside edge of the footpath

x²  +  y²  = 3422,25  ⇒  x² + y²  = (58,5)²

So we obtain the radius of each circumference

For the outside edge of the pool  r₁  = 50 feet

For the outside edge of the footpath r₂ = 58,5  feet

Then the width of the footpath is  58,5  - 50  =  8,5 feet

8 0
3 years ago
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