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gtnhenbr [62]
2 years ago
6

Marco is measuring two pieces of strings. One piece of string is 15 cm long. The other string is 250 mm long. In centimeters, ho

w long are the pieces of strings together?​
Mathematics
1 answer:
faltersainse [42]2 years ago
5 0

Answer:

400 mm

Step-by-step explanation:

to convert cm to mm, you add a 0, or more specifically, multiply by 10. so, 15 × 10= 150. 250+150=400.

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Simplify.<br><br> 16 - 8 + 5x - 3x<br><br> 16 + 2x<br> 8 + 8x<br> 8 + 2x<br> 10x
klio [65]

Answer:

8+2x

Step-by-step explanation:

Combine Like Terms:

<u>1</u><u>6</u><u>-</u><u>8</u>+5x-3x

8+<u>5</u><u>x</u><u>-</u><u>3</u><u>x</u>

8+2x

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3 years ago
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8 0
3 years ago
An empty 5-gal water jug weighs 0.75 lb. With 3 c of water inside, the jug weighs 2.25 lb. Which equation models the jug's weigh
Deffense [45]
Weight of 1 c
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8 0
3 years ago
A certain bridge arch is in the shape of half an ellipse 106 feet wide and 33.9 feet high. At what horizontal distance from the
nata0808 [166]

Answer:

The horizontal distance from the center is 49.3883 feet

Step-by-step explanation:

The equation of an ellipse is equal to:

\frac{x^2}{a^{2} } +\frac{y^2}{b^{2} } =1

Where a is the half of the wide, b is the high of the ellipse, x is the horizontal distance from the center and y is the height of the ellipse at that distance.

Then, replacing a by 106/2 and b by 33.9, we get:

\frac{x^2}{53^{2} } +\frac{y^2}{33.9^{2} } =1\\\frac{x^2}{2809} +\frac{y^2}{1149.21} =1

Therefore, the horizontal distances from the center of the arch where the height is equal to 12.3 feet is calculated replacing y by 12.3 and solving for x as:

\frac{x^2}{2809} +\frac{y^2}{1149.21} =1\\\frac{x^2}{2809} +\frac{12.3^2}{1149.21} =1\\\\\frac{x^2}{2809}=1-\frac{12.3^2}{1149.21}\\\\x^{2} =2809(0.8684)\\x=\sqrt{2809(0.8684)}\\x=49.3883

So, the horizontal distance from the center is 49.3883 feet

8 0
3 years ago
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