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yan [13]
2 years ago
13

NO BULL SH*T REAAAAAAAL A F

Mathematics
1 answer:
Mariulka [41]2 years ago
4 0
The answer is A! 4%
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Answer asap<br> Thanks<br> ;)
Fofino [41]

Answer:

6, 4, 5, 1, 3, 2, 7

Step-by-step explanation:

Sorry if I’m wrong

5 0
2 years ago
80=-8a please I need the answer
Free_Kalibri [48]
80=-8a 
We are trying to solve for a. So we need to get a alone. To do that we have to divide each side by -8. 
80/-8= -8a/-8
Once you have solved each side you will get the following. 
-10=a
6 0
3 years ago
Use the remainder theorem to determine the remainder when 3t2 + 5t − 7 is divided by t − 5. A. 93 B. 43 C. 107 D. 57
Usimov [2.4K]
Using remainder theorem, we get:

P(t) = 3t^{2} + 5t - 7

Substitute t = 5 into the equation:
P(5) = 3(5)^{2} + 5^{2} - 7
P(5) = 75 + 25 - 7 = 93

Thus, we get a remainder of 93 or (A)
7 0
3 years ago
Read 2 more answers
Find a polynomial function of degree 3 with 2, i, -i as zeros.
sergiy2304 [10]

Answer:

p(x)= x^3-2x^2+x-2

Step-by-step explanation:

Here we are given that a polynomial has zeros as 2 , i and -i . We need to find out the cubic polynomial . In general we know that if \alpha , \ \beta \ \& \ \gamma are the zeros of the cubic polynomial , then ,

\sf \longrightarrow p(x)= (x -\alpha )(x-\beta)(x-\gamma)

Here in place of the Greek letters , substitute 2,i and -i , we get ,

\sf\longrightarrow p(x)= (x -2 )(x-i)(x+i)

Now multiply (x-i) and (x+i ) using the identity (a+b)(a-b)=a² - b² , we have ,

\sf  \longrightarrow p(x)= (x-2)\{ x^2 - (i)^2\}

Simplify using i = √-1 ,

\sf \longrightarrow p(x)= (x-2)( x^2 + 1 )

Multiply by distribution ,

\sf \longrightarrow p(x)= x(x^2+1) -2(x^2+1)

Simplify by opening the brackets ,

\sf\longrightarrow p(x)= x^3+x-2x^2-2

Rearrange ,

\sf\longrightarrow \underline{\boxed{\blue{\sf p(x)= x^3-2x^2+x-2}}}

4 0
2 years ago
The graph of f ′ (x), the derivative of f(x), is continuous for all x and consists of five line segments as shown below. Given f
Nataliya [291]

The maxima of f(x) occur at its critical points, where f '(x) is zero or undefined. We're given f '(x) is continuous, so we only care about the first case. Looking at the plot, we see that f '(x) = 0 when x = -4, x = 0, and x = 5.

Notice that f '(x) ≥ 0 for all x in the interval [0, 5]. This means f(x) is strictly increasing, and so the absolute maximum of f(x) over [0, 5] occurs at x = 5.

By the fundamental theorem of calculus,

\displaystyle f(5) = f(0) + \int_0^5 f'(x) \, dx

The definite integral corresponds to the area of a trapezoid with height 2 and "bases" of length 5 and 2, so

\displaystyle \int_0^5 f'(x) \, dx = \frac{5+2}2 \times 2 = 7

\implies \max\{f(x) \mid 0\le x \le5\} = f(5) = f(0) + 7 = \boxed{13}

8 0
2 years ago
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