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Julli [10]
3 years ago
8

46 decreased by twice Chau's savings. use the variable c to represent Chau's savings.

Mathematics
1 answer:
dedylja [7]3 years ago
4 0
46 -2c

46 decreased by means 46 minus
Twice c would be c times 2
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In a classroom of 33 students, the ratio of boys to girls is 3:8. How many boys are in the class?
Hunter-Best [27]
X=number of boys
y=number of girls.

We can suggest the following system of equations:
x+y=33
x/y=3/8

We solve this system of equations by substitution method.
x/y=3/8       ⇒x=3y/8

(3y/8)+y=33
least common multiple=8
3y+8y=33*8
11y=33*8
y=(33*8)/11
y=3*8
y=24

x+y=33
x+24=33
x=33-24
x=9

number of boys /total number of students=9/33=3/11

There are 9 boys and 24 girls in the class. 

Answer: C: 9. since 3/11 of the students are boys. 

3 0
3 years ago
Read 2 more answers
A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, th
Yuri [45]

Answer:

15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = ± 170 cos(5t)

y(0)=0, y'(0)=0

Step-by-step explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: F_{r} (t) that correspond to the force of resistance on the mass by the action of the spring and F(t) that is an external force with unknown direction (that does not specify in the enounce).

For determinate F_{r} (t) we can use Hooke's Law given by the formula F_{r} (t) = k y(t) where k correspond to the elastic constant of the spring and y(t) correspond to  the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m}  = 441.45 \frac{N}{m}

Now we apply Newton's 2nd Law and obtaint that...

F_{r} (t) ± F(t) = ma(t)

F_{r} (t) = ky(t) = 441.45y(t)

F(t) = 170 cos(5t)

m = 15 kg

a(t) = \frac{d^{2}y(t)}{dt^{2} }

Finally... 15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = ± 170 cos(5t)

We know from the problem that there's not initial displacement and initial velocity, so... y(0)=0 and y'(0)=0

Finally the Initial Value Problem that models the situation describe by the problem is

\left \{ 15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.

6 0
3 years ago
What is the scale factor from the original triangle to copy?
topjm [15]

Answer:

2

Step-by-step explanation:

Copy 1 is a scaled copy of the original triangle. The scale factor is 2, because each side in Copy 1 is twice as long as the corresponding side in the original triangle. , , Copy 2 is a scaled copy of the original triangle.

5 0
3 years ago
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The rationalisation factor of 2 + √3 is<br><br>step by step for BRAINLIST ​
dangina [55]

Answer:

rationalising factor wud be

2 - root3

as on multiying both and applying identity we end up

2^2 - (root3)^2

4 - 3 = 1

we got a rational number so rationalisng factor is

2 - root3

5 0
3 years ago
How to do this? Help please
Svetradugi [14.3K]
Write it out and use photo math
7 0
3 years ago
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