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liberstina [14]
3 years ago
8

Factor completely: 7a^2+49a+14

Mathematics
1 answer:
viktelen [127]3 years ago
4 0

Answer:

7(a^2+7a+2)

Step-by-step explanation:

Divide out the 7. This becomes 7(a^2+7a+2). You can't factor any further.

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A rectangular window is topped with a semicircle on top. the height of the rectangular part is 1 more than 3 times its width , w
tensa zangetsu [6.8K]

In general for a rectangle of dimensions w by h and a semicircle of diameter w we get an area:


A = wh + \frac 1 2 \pi (\frac w 2)^2


The halfs are because a semicircle is half a circle and a radius is half a diameter.


We're told


h = 1+ 3w


Substituting,


A(w) = w(1+3w) + \frac 1 2 \pi (\frac w 2)^2


Answer: B



5 0
3 years ago
This one is very hard thank you
makvit [3.9K]

check the picture below.


if we run a vertical line test on that graph, like you see there, namely run vertical lines through it, we can see that one vertical line only hits the graph once, not twice or more.


on the vertical line test, if the graph is hit only once by each of the vertical lines, then the graph is a function.

3 0
4 years ago
Can I get some help ​
dexar [7]

Answer:

c.13.7mi, A=33.3°, B=56.7°

6 0
3 years ago
Read 2 more answers
Seattle-Pipes Co. produces pipes to be supplied to a Seattle utility company. The requirement of the utility company is that the
g100num [7]

Answer:

Step-by-step explanation:

Hello!

The study variable is

X: Pipe length.

It is known that this variable has a normal distribution and that the distribution parameter varies depending on the process used to manufacture the pipes.

Process A: μ= 200cm δ= 0.5cm

Process B: μ=201cm δ= 1cm

Process C: μ=202cm δ= 1.5cm

Pipes with a length of 200cm or more will be accepted by the utility company (X≥200), but pipes with less than 200cm length will be rejected (X<200)

a. Using Process C, you need to calculate the probability that the pipe will be rejected, symbolically:

P(X<200)

Using the distribution data of process C you have to standardize the value:

P(Z<(200-202)/1.5)= P(Z<-1.33)= 0.09176

b. The requirements change, accepting any pipe between 199 and 202, you have to calculate the probabilities of the pipes being between those lengths using the three process:

Process A:

P(199≤X≤202) = P(X≤202) - P(X≤199)

P(Z≤(202-200)/0.5)) - P(Z≤(199-200)/0.5))

P(Z≤4) - P(Z≤-2) = 1 - 0.02275 = 0.97725

The probability of the pipe being rejected is 0.02275

Process B:

P(199≤X≤202) = P(X≤202) - P(X≤199)

P(Z≤(202-201)/1)) - P(Z≤(199-201)/1))

P(Z≤1) - P(Z≤-2) = 0.84134 - 0.02275 = 0.81859

The probability of the pipe being rejected is 1-0.81859= 0.18141

Process C:

P(199≤X≤202) = P(X≤202) - P(X≤199)

P(Z≤(202-202)/1.5)) - P(Z≤(199-202)/1.5))

P(Z≤0) - P(Z≤-2) = 0.5 - 0.02275 = 0.47725

The probability of the pipe being rejected is 1-0.47725= 0.52275

The pipes manufactured using process A has fewer chances of being rejected.

c.

Process A costs $140

P(X≥200)= 1 - P(X<200)= 1 - P(Z<0)= 1 - 0.5= 0.5

Process B costs $160

P(X≥200)= 1 - P(X<200)= 1 - P(Z<-1)= 1 - 0.15866= 0.84134

Process C costs $177

P(X≥200)= 1 - P(X<200)= 1 - P(Z<-1.33)= 1 - 0.09176= 0.90824

If they where to make 100 pipes:

Using process A: 100*0.5= 50 pipes will be accepted, so they'll win 50*($200-$140)= $3000

Using process B: 100*0.84134= 84.134≅ 84 pipes will beaccepted, so they'll win 84*($200-$160)= $3360

Using the process C: 100*0.90824= 80.824≅ 90 pipes will be accepted, so they'll win 90*($200-$177)= $2070

As you can see, using process B will maximize the profits.

I hope it helps!

4 0
3 years ago
If f(x) = x2 – 3 and g(x) = 4x +3, then<br> f(g(2)) =
WARRIOR [948]

Answer:

4*+1

Step-by-step explanation:

To find

g

(

f

(

x

)

)

you must substitute

f

(

x

)

or

4

x

+

3

for each occurrence of

x

in

g

(

x

)

.

g

(

x

)

=

x

−

2

becomes:

g

(

f

(

x

)

)

=

(

4

x

+

3

)

−

2

g

(

f

(

x

)

)

=

4

x

+

3

−

2

g

(

f

(

x

)

)

=

4

x

+

1

8 0
3 years ago
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