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Sveta_85 [38]
2 years ago
6

Analyze the graphed function to find the local minimum and the local maximum for the given function. On a coordinate plane, a cu

rved line with minimum values of (0.6, negative 8) and (3.4, negative 8), and a maximum value of (2, 0), crosses the x-axis at (0, 0), (2, 0), and (4, 0), and crosses the y-axis at (0, 0). Which statements about the local maximums and minimums for the given function are true? Choose three options. Over the interval [1, 3], the local minimum is 0 Over the interval [2, 4], the local minimum is –8. Over the interval [3, 5], the local minimum is –8. Over the interval [1, 4], the local maximum is 0. Over the interval [3, 5], the local maximum is 0.
Mathematics
2 answers:
kozerog [31]2 years ago
8 0

Answer:

b

Step-by-step explanation:

Kisachek [45]2 years ago
7 0

Answer:

B C D

Step-by-step explanation:

I hope this is correct

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What are the x-intercepts of the function f(x) = –2x2 – 3x + 20?
Hatshy [7]

The root sof this quadratic equation are -4 and 2.5.


The roots of any quadratic can be found by using the quadratic equation. The equation is below for you.


\frac{-b +/- \sqrt{b^{2}- 4ac}}{2a}


In this equation you use the number attached to x^2 as the a, which in this case is -2. The number attached to x as b, which is in this case -3. And the number at the end as c, which is 20. From there you solve for the answers.


\frac{3 +/- \sqrt{-3^{2}- 4(-2)(20)}}{2(-2)}

\frac{3 +/- \sqrt{9 + 160}}{-4}

\frac{3 +/- 13}}{4}


Now you separate and get the two separate answers. First the positive.


\frac{3 + 13}}{4}

\frac{16}}{4}

4


Now the negative


\frac{3 - 13}}{4}

\frac{-10}}{4}

-2.5

5 0
2 years ago
Read 2 more answers
Which terms could be used as the first term of the expression below to create a polynomial written in standard form? Select five
kondaur [170]

Answer:

The last two terms of the expression are

  ----+8r^2s^4-3r^3s^3

Both the last terms has variable of degree equal to (2+4=6) and (3+3=6).So, the first term must have degree greater than 6.

Correct Options are

 1.\rightarrow 3r^4s^5\\\\2.\rightarrow -r^4s^6

3 0
2 years ago
Compare the values of the underlined digits. 2,783 and 7,283.
netineya [11]

The value of the 2 in the thousands place is 10 times the value of 2 in the hundreds place.

200 x 10 =  2000

hope this helps

4 0
2 years ago
Eli, Freda and Geoff were given £800 to share in the ratio of their ages.
algol13
First, we are going to find the sum of their age. To do that we are going to add the age of Eli, the age Freda, and the age of <span>Geoff:
</span>9+13+18=40
The combined age of Eli, Freda, and Geoff is 40, so the denominator of each ratio will be 40.

Next, we are going to multiply the ratio between the age of the person and their combined age by <span>£800:
For Eli: </span>\frac{9}{40} *800=180
For Freda: \frac{13}{40} *800=260
For Geoff: \frac{18}{40} *800=360
<span>
We can conclude that Eli will get </span>£180, Freda will get £260, and Geoff will get <span>£360.</span>
4 0
2 years ago
At an intersection, the red-light times are normally distributed with a mean time of 3 minutes and a standard deviation of 0.25
denpristay [2]

Percent of red lights last between 2.5 and 3.5 minutes is 95.44% .

<u>Step-by-step explanation:</u>

Step 1: Sketch the curve.

The probability that 2.5<X<3.5 is equal to the blue area under the curve.

Step 2:

Since μ=3 and σ=0.25 we have:

P ( 2.5 < X < 3.5 ) =P ( 2.5−3 <  X−μ < 3.5−3 )

⇒ P ( (2.5−3)/0.25 < (X−μ)/σ < (3.5−3)/0.25)

Since, Z = (x−μ)/σ , (2.5−3)/0.25 = −2 and (3.5−3)/0.25 = 2 we have:

P ( 2.5<X<3.5 )=P ( −2<Z<2 )

Step 3: Use the standard normal table to conclude that:

P ( −2<Z<2 )=0.9544

Percent of red lights last between 2.5 and 3.5 minutes is 0.9544(100) = 95.44% .

3 0
2 years ago
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