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3 years ago

What are the values of k that make 2x^2 + kx + 11 factorable?

1 answer:

8 0

Well K is the sum of two integers so the product equals 2*11, as a Hero pointed out:
ac=2*11=22 k=sum of two integers so now it = to 22

I hope this helps :D

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FInd the value of x. Y D W X (15x-8) X = N 226

Oksi-84 [34.3K]

Let's start with the given arc and its angle.

The angle YWX is going to be half the arc length.

YWX = 1/2 (226) = 113

Angles VWX and YWX form a linear pair (or are supplementary angles), which means that their sum is 180 degrees.

(15x - 8) + 113 = 180

15x + 105 = 180

15x = 75

x = 5

Hope this helps!

8 0

2 years ago

Hey can you help me out with this problem?

Reika [66]

Answer:

Step-by-step explanation:

20/2=10 smores

5 0

2 years ago

6w-y=2z, Solve for the value of w

Lerok [7]

6w-y=2z 6w=2z+y w=(2z+y):6 w=(z:3)+(y:6)

7 0

2 years ago

Read 2 more answers

If the rate of inflation is 3.4%per year, the future price pt(in dollars) of a certain item can be modeled by the following expo

Doss [256]

Answer:

The current price is $3000 and price after 9 years from today is $4054.

Step-by-step explanation:

The future price pt(in dollars) of a certain item can be modeled by the following exponential function

$p(t)=3000(1.034)^t$

where, t is the number of years from today.

Substitute t=0 to find the current price.

$p(0)=3000(1.034)^0=3000$

Therefore the current price is $3000.

Substitute t=9 to find the price after 9 years from today.

$p(9)=3000(1.034)^9$

$p(9)=3000(1.35109177087)$

$p(9)=4053.27531261$

$p(9)\approx 4054$

Therefore the price after 9 years from today is $4054.

4 0

2 years ago

From the lighthouse 1000 ft above sea level the angle of depression to boat a is 29 degress.a little bit later the boat moved cl

AnnyKZ [126]

Answer: The boat moved 768.51 feet in that time .

Step-by-step explanation:

Since we have given that

Height of the lighthouse = 1000 feet

Angle depression to boat 'a' = 29°

Angle of depression to shore 'b' = 44°

Consider ΔABC,

$\tan 44\textdegree=\frac{AB}{BC}\\\\\tan 44\textdegree=\frac{1000}{BC}\\\\BC=\frac{1000}{\tan 44\textdegree}\\\\BC=1035.53\ feet$

Now, Consider, ΔABD,

$\tan 29\textdegree=\frac{AB}{BD}\\\\\tan 29\textdegree=\frac{1000}{BD}\\\\BD=\frac{1000}{\tan 29\textdegree}\\\\BD=1804.04\ feet$

We need to find the distance that the boat moved in that time i.e. BC

so,

$DC=BD-BC\\\\CD=1804.04-1035.53\\\\CD=768.51\ feet$

Hence, the boat moved 768.51 feet in that time .

6 0

3 years ago