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sp2606 [1]
3 years ago
13

For which x-values is the graph below discontinuous?

Mathematics
2 answers:
aleksandr82 [10.1K]3 years ago
3 0

Answer:

A

Step-by-step explanation:

stealth61 [152]3 years ago
3 0

Answer:

b)

Step-by-step explanation:

The LHL and RHL at these places are different.

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Find the interest due on $600 at 9.5% for 120 days.
KATRIN_1 [288]

Answer:

Interest= $ 18.73

Step-by-step explanation:

Given : $600 at 9.5% for 120 days

To find : Find the interest due

Solution :

Simple interest formula  I=P\times r\times t

Principle(P)=$600 , rate(r)=9.5%=0.095 , time (t)= 120 days

In years, 1 year = 365 days

1 day = \frac{1}{365} year

120 days = \frac{120}{365} year

Put values in the formula

I=P\times r\times t

I=600\times 0.095\times\frac{120}{365}

 I=\frac{6840}{365}=18.73

Therefore, Interest= $ 18.73




5 0
3 years ago
Read 2 more answers
Solve |2x - 1| + 3 = 6.
umka21 [38]

Answer:

x =2                         x = -1

Step-by-step explanation:

|2x - 1| + 3 = 6.

Subtract 3 from each side

|2x - 1| + 3-3 = 6-3

|2x - 1|= 3

An absolute value has two solutions, one positive and one negative

2x-1 =3                   2x -1 = -3

Add 1 to all sides

2x-1+1 =3+1             2x-1+1 = -3+1

2x = 4                      2x =-2

Divide by 2

2x/2 = 4/2                 2x/2 = -2/2

x =2                         x = -1

5 0
4 years ago
Read 2 more answers
What is the vertex of the absolute value function below?
laiz [17]

Answer:

(-3, 2)

Step-by-step explanation:

I think because of the vertex

3 0
3 years ago
susan englared a rectangle with a height of 3cm and length of 10cm on her computer. The length of the new rectangle is 15cm.Find
ExtremeBDS [4]

\frac{3}{x}  \times  \frac{10}{15}
when you cross multiply the answer is 4.5
8 0
4 years ago
(cot^2x - 1)/(csc^2x) = cos2x​
Alex787 [66]

Answer:

Step-by-step explanation:

The idea here is to get the left side simplified down so it is the same as the right side. Consequently, there are 3 identities for cos(2x):

cos(2x)=cos^2x-sin^2x,

cos(2x)=1-2sin^2x, and

cos(2x)=2cos^2x-1

We begin by rewriting the left side in terms of sin and cos, since all the identities deal with sines and cosines and no cotangents or cosecants.  Rewriting gives you:

\frac{\frac{cos^2x}{sin^2x} -\frac{sin^2x}{sin^2x} }{\frac{1}{sin^2x} }

Notice I also wrote the 1 in terms of sin^2(x).

Now we will put the numerator of the bigger fraction over the common denominator:

\frac{\frac{cos^2x-sin^2x}{sin^2x} }{\frac{1}{sin^2x} }

The rule is bring up the lower fraction and flip it to multiply, so that will give us:

\frac{cos^2x-sin^2x}{sin^2x} *\frac{sin^2x}{1}

And canceling out the sin^2 x leaves us with just

cos^2x-sin^2x which is one of our identities.

5 0
3 years ago
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