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4 years ago

4(6x^3-4x^2+7x+1)-9(4x^3-2x^2-6x+1

Mathematics

1 answer:

eimsori [14]4 years ago

7 0

Answer:not the answer

Step-by-step explanation:

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Use the Distributive Property to write and simplify an expression for the cost.

NNADVOKAT [17]

Answer:

240+30x

Step-by-step explanation:

240+30x because you multiply 30 x 8 and get 240 plus 30 x X and get 30x

5 0

3 years ago

Read 2 more answers

The band has 30 more members than the school chorale. If each group had 10 more members, the ratio of their membership would be

vladimir1956 [14]

Answer:

80 members

Step-by-step explanation:

Let the band members be= b and the chorale members= c

Then, according to question, b=c+30 and $\frac{b+10}{c+10}=\frac{3}{2}$

⇒ $2(b+10)=3(c+10)$

⇒ $2b+20=3c+30$

⇒ $2b-3c=30-20$

⇒ $2b-3c=10$

Now, substituting the value of b=c+30, we get

⇒ $2(c+30)-3c=10$

⇒ $2c+60-3c=10$

⇒ $-c=-50$

⇒ $c=50$

Therefore, $b=c+30$

$b=50+30=80$

Therefore, there are 80 members in the band.

8 0

3 years ago

The junior high band sold candy for a fundraiser the profit from the candy sale was 180.00 the girls sold 3 times as much as the

kaheart [24]

Let the amount that the boys sold be x, the amount that girls sold will be 3x. The total will be given by:
x+3x=180
4x=180
solving for x we get:
x=180/4
x=45
Thus, boys sold $45, the girls provided 45*3=$135

3 0

3 years ago

A spherical balloon is inflated with a gas at the rate of 500 cm3/min. How fast is the radius of the balloon increasing at the i

Burka [1]

The volume of the sphere is expressed in the formula V = 4/3 pi r^3. The rate of change of volume is determined by differentiating the formula: dV/dt = 4pi r^2 dr/dt. When we substitute 500 cm3/min as dV/dt and 30 cm as r. Then dr/dt is equal to 0.0442 cm/min

7 0

4 years ago

A random sample of 401 student were recent survey regarding their class standing freshmen software junior senior and their major

SOVA2 [1]

the Solution

$Pr(\text{STEM)}=\frac{158}{401}=0.394$ $Pr(\text{Sophomore)}=\frac{87}{401}=0.217$ $Pr(sophomore|Non-stem)=\frac{68}{401}=0.170$ $Pr(\text{sophomore and Non-stem)=}\frac{87}{401}\times\frac{243}{401}=\frac{21141}{160801}=0.1315$

To ascertain whether Sophomore and Non-stem are dependent, we have to test the following:

$\begin{gathered} If\text{ Pr(sophomore or Non-stem) = Pr(sophomore and Non-stem),} \\ \text{then we conclude that both events are Independent,} \\ \text{otherwise, they are dependent.} \end{gathered}$ $\begin{gathered} Pr(\text{sophomore or Non-stem)=Pr(sophomore)+Pr(Non-stem)} \\ -Pr(\text{sophomore and Non-stem)} \end{gathered}$ $Pr(\text{Sophomore or Non-stem)=}\frac{87}{401}+\frac{243}{401}-(\frac{87}{401}\times\frac{234}{401})$ $=\frac{330}{401}-\frac{21141}{160801}=0.82294-0.13147=0.69147$

Cleary, we have that sophomore and non-stem events are dependent events since 0.69147 is not the same as 0.13147.

4 0

1 year ago