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padilas [110]
2 years ago
14

Write 4% as a fraction in lowest terms.

Mathematics
1 answer:
AURORKA [14]2 years ago
5 0

Answer:

1/25

Step-by-step explanation:

4 over 100 = 1/25

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Expand this expression (x-y)²<br>​
pochemuha

Answer:

x^2 - 2xy + y^2.

Step-by-step explanation:

(x - y)^2

= (x - y)(x - y)

= x(x -y) - y(x - y)

= x^2 - xy - xy + y^2

= x^2 - 2xy + y^2.

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2 years ago
The Bengals football team scored 2 touchdowns for 6 points each, one extra point, and 3 field goals for 3 points each. The Raven
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An ice machine uses 3 gallons of water every 9 hours. How many gallons of water does it use each hour? How many hours does it ta
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3 years ago
Solve each triangle using the information provided. If necessary, round to the<br> nearest tenth.
Goryan [66]

Answer:

1) B = 66.5° c = 10.9

Step-by-step explanation:

I will do question one as an example. In general, for these questions you want to use the appropriate trigonometric ratios to solve for the variables and/or apply logic using rules regarding triangles. See attached image for all solving steps.

For side c, we can use Cosine of angle A for a ratio between 10 and c. When we write out the equation, we can solve for side c. So when we write it out, we get the equation:

cos23.5 = ¹⁰⁄c

c = ¹⁰⁄cos₂₃.₅

c = 10.9044 (make sure to round to the nearest tenth, which is one decimal place)

For angle B, since they have given two angles, you can solve for B since all angles of a triangle add up to 180 degrees.

So b = 180 - (90 + 23.5) = 180 - 113.5

b = 66.5

  • It is also possible to solve this using sine of angle B and solve it from there, but applying the theory this way is much simpler. (this is on the image if you're curious about it)

I hope this helps you with the other 3 questions.

4 0
3 years ago
PLS HELP TIMED QUESTION (I'll mark brainiest)<br> Please show work if possible.
elena-s [515]

Hello,

sin(45^o)=\dfrac{\sqrt{2} }{2} \\\\sin(45^o)=\dfrac{a}{c} \\\\\dfrac{\sqrt{2} }{2}=\dfrac{a}{6} \\\\a=6*\dfrac{\sqrt{2} }{2}=3\sqrt{2}\\

Answer A

5 0
2 years ago
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