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Ksivusya [100]
3 years ago
9

30 tens is the same value as what number

Mathematics
2 answers:
Fudgin [204]3 years ago
7 0
30 tens is the same value as 300.  30 x 10 = 300
RSB [31]3 years ago
7 0
The answer to 30 tens is 300 <span>
</span>

The answer to that problem would be 30 tens times 10 = 300 having the same value as 30 tens.

Hope that helps!

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What's the answer to 1&amp;2 plzz
Dmitrij [34]
1. 2880 
2. 67%
 Its simple math just put in effort

4 0
4 years ago
Please help if possible its important!!​
Otrada [13]

Answer:

600 ft³

Step-by-step explanation:

Base area: B

Base is triangle

base = 12 ft and height = 5 ft

Area of triangle = B = \frac{1}{2}* \ base * \  height

                                 = \frac{1}{2}*12 * 5\\\\= 6 * 5\\\\= 30 \ ft^{2}

H = 20 ft

Volume = B * H

             = 30 * 20

             = 600 ft³

7 0
3 years ago
9) Nancy starts a race at the start line and she is running 3 meters per second. Juan starts the same race 3 meters ahead of Nan
Travka [436]

Given:

Nancy is running 3 meters per second.

Juan starts the same race 3 meters ahead of Nancy but he is going at 2 meters per second.

To find:

The equations for Nancy and Juan.

Solution:

Let x be the number of seconds.

Nancy is running 3 meters per second. So, the total distance covered by Nancy in the race is

y=3x

Juan starts the same race 3 meters ahead of Nancy but he is going at 2 meters per second. So, the total distance covered by Juan in the race is

y=3+2x

Therefore, the equations of Nancy and Juan are y=3x and y=3+2x respectively.

5 0
3 years ago
Find the volume of the solid under the plane 5x + 9y − z = 0 and above the region bounded by y = x and y = x4.
svp [43]
<span>For the plane, we have z = 5x + 9y

For the region, we first find its boundary curves' points of intersection.
x = x^4 ==> x = 0, 1.

Since x > x^4 for y in [0, 1],

The volume of the solid equals

\int\limits^1_0 { \int\limits_{x^4}^x {(5x+9y)} \, dy } \, dx = \int\limits^1_0 {\left[5xy+ \frac{9}{2} y^2\right]_{x^4}^{x}} \, dx  \\  \\ =\int\limits^1_0 {\left[\left(5x(x)+ \frac{9}{2} (x)^2\right)-\left(5x(x^4)+ \frac{9}{2} (x^4)^2\right)\right]} \, dx  \\  \\ =\int\limits^1_0 {\left(5x^2+ \frac{9}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx =\int\limits^1_0 {\left( \frac{19}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx \\  \\ =\left[ \frac{19}{6} x^3- \frac{5}{6} x^6- \frac{1}{2} x^9\right]^1_0

=\frac{19}{6} - \frac{5}{6} - \frac{1}{2} =\bold{ \frac{11}{6} \ cubic \ units}</span>
8 0
3 years ago
Which one is it ? I need help<br> 1. 60<br> 2. 55<br> 3. 65<br> 4. 180
Alexeev081 [22]
The answer should be 55
6 0
3 years ago
Read 2 more answers
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