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d1i1m1o1n [39]
3 years ago
8

What is the LCM of 120 and 150​

Mathematics
1 answer:
Mrac [35]3 years ago
4 0
The LCM of 120 and 150 is 600
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Which is a solution to the equation?<br> (x-2)(x=5)=18
kobusy [5.1K]
Expand
x^2+3x-10=18
minus 18 both sides
x^2+3x-28=0
find what 2 numbers mutiply to -28 and add to 3
7 and -3 
(x+7)(x-3)=0
set each to zero

x+7=0
x=-7
x-3=0
x=3

x=-7 and 3
6 0
4 years ago
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Can someone solve this for me please?
Ksivusya [100]
The y-in is -6
The x- in is -1 and -6
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5 0
4 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
Macy is tutoring students in mathematics during her summer break. She charges $20 for a 30 minute session and $35 for an hour. T
Zanzabum
If she only had 1 hour sessions she had 8 sessions in the first week because if you do 35 times 8 you get 280. 
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4 years ago
What is an equation in point-slope form the line that passes through the point (-3,5) and (2,-3)?
lyudmila [28]

Answer:

Y-5=-8/5 (x+3)

Step-by-step explanation:

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