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3 years ago

Find the quotient of x^3 + 4x^2 - x - 4 divided by x + 1

Mathematics

1 answer:

svlad2 [7]3 years ago

4 0

Answer:

x^2 + 3x - 4

Step-by-step explanation:

solved using long division

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Which of the following high speed trains are traveling at a rate of 150 miles per hour? A: a train that travels 100 miles in 2/3

kondor19780726 [428]

Answer:

Step-by-step explanation:

youjust have to do this:

100 miles-------------------2/3 hour

x-------------------------------3/3 hour or 1 hour

x=100x1/0.6666666667 2/3=0.6666666667

x= 150 miles

8 0

3 years ago

Maths people please write the answers !<br><br><br> Equations- Maths

Korvikt [17]

Answer:

length = 20cm

breadth = 8cm

Step-by-step explanation:

solution;

length(l) = x + 12

breadth(b) = x

perimeter (p) = 56 cm

we know that,

perimeter(p) = 2(l + b)

or, 56 = 2(x + 12 + x)

or, 56 = 2(2x + 12)

or, 56 = 4x +24

or, 56 - 24 = 4x

or, 32 = 4x

or, 32 / 4 = x

x = 8

now

putting the value of x in length and breadth

length = x+ 12

=8 + 12

=20cm

breadth = 8cm

7 0

3 years ago

Read 2 more answers

Convenient store sells prepaid mobile phones a purchase of them for $12 each and uses a markup rate of 250% what is the markup r

kozerog [31]

250%=2.5

Explanation: it just does

4 0

3 years ago

Tesla Battery Recharge Time.The electric vehicle manufacturing company Tesla estimates that a driver who commutes 50 miles per

madam [21]

Answer:

a.) f(x) = $\frac{1}{30}$ where 90 < x < 120

b.) $\frac{2}{3}$

c.) $\frac{2}{3}$

d.) $\frac{1}{2}$

Step-by-step explanation:

Let

X be a uniform random variable that denotes the actual charging time of battery.

Given that, the actual recharging time required is uniformly distributed between 90 and 120 minutes.

⇒X ≈ ∪ ( 90, 120 )

a.)

Probability density function , f (x) = $\frac{1}{120 - 90} = \frac{1}{30}$ where 90 < x < 120

b.)

P(x < 110) = $\int\limits^{110}_{90} {\frac{1}{30} } \, dx$

= $\frac{1}{30}[x]\limits^{110}_{90} = \frac{1}{30} [ 110 - 90 ] = \frac{1}{30} [ 20] = \frac{2}{3}$

c.)

P(x > 100 ) = $\int\limits^{120}_{100} {\frac{1}{30} } \, dx$

= $\frac{1}{30}[x]\limits^{120}_{100} = \frac{1}{30} [ 120 - 100 ] = \frac{1}{30} [ 20] = \frac{2}{3}$

d.)

P(95 < x< 110) = $\int\limits^{110}_{95} {\frac{1}{30} } \, dx$

= $\frac{1}{30}[x]\limits^{110}_{95} = \frac{1}{30} [ 110 - 95 ] = \frac{1}{30} [ 15] = \frac{1}{2}$

7 0

3 years ago

A cargo truck weighs 8750 lb the weight limit for a certain bridge is 5 tons how many pounds of cargo can be added to the truck

Vaselesa [24]

Answer:

1,250 lbs

Step-by-step explanation:

There are 2,000 in 1 ton, so 5 * 2,000 = 10,000 lbs weight limit for the bridge.

10,000-8750 = 1,250 lbs can be added to the truck before they reach the bridge weight limit.

7 0

3 years ago

Read 2 more answers