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lara31 [8.8K]
3 years ago
14

Evaluate the expression 5x-y when x=2 and y=4

Mathematics
1 answer:
mariarad [96]3 years ago
5 0

Answer:

6

Step-by-step explanation:

5 times 2 (×) 10- 4(y)

when a varbile (letter) is besides then you multiply

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Zigmanuir [339]
Point A means the smallest value (and = 4)

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<span>C. the smallest value</span>
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I am a number greater than 20 but smaller than 30. I have two decimal places. The digit in the ones place is 1 less than the dig
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Answer:

The answer for this problem is 22.36

4 0
3 years ago
F(x) = x^2. What is g(x)
Alexeev081 [22]

Answer:

g(x) = - x² - 4 ⇒ A

Step-by-step explanation:

Let us revise the reflection and translation of a function

  • If the function f(x) reflected across the x-axis, then its image is g(x) = - f(x)
  • If the function f(x) reflected across the y-axis, then its image is g(x) = f(-x)
  • If the function f(x) translated horizontally to the right by h units, then its image is g(x) = f(x - h)  
  • If the function f(x) translated horizontally to the left by h units, then its image is g(x) = f(x + h)  
  • If the function f(x) translated vertically up by k units, then its image is g(x) = f(x) + k  
  • If the function f(x) translated vertically down by k units, then its image is g(x) = f(x) – k  

f(x) = x² is the blue curve

g(x) is its image is the red curve

∵ g(x) is the image of f(x)

∵ f(x) is opened upward

∵ g(x) is opened downward

→ That means the sign of y-coordinates of all points on the blue

   graph are opposite

∴ f(x) is reflected about the x-axis

∴ Its image is - f(x)

∵ The vertex of f(x) is (0, 0)

∵ The vertex of g(x) = (0, -4)

→ That means the function translated 4 units down

∴ - f(x) is translated 4 units down

∴ Its image is - f(x) - 4

∴ g(x) = - f(x) - 4

∵ f(x) = x²

∴ g(x) = - x² - 4

4 0
3 years ago
(a) Use the definition to find an expression for the area under the curve y = x3 from 0 to 1 as a limit. lim n→∞ n i = 1 Correct
Luba_88 [7]

Splitting up the interval of integration into n subintervals gives the partition

\left[0,\dfrac1n\right],\left[\dfrac1n,\dfrac2n\right],\ldots,\left[\dfrac{n-1}n,1\right]

Each subinterval has length \dfrac{1-0}n=\dfrac1n. The right endpoints of each subinterval follow the sequence

r_i=\dfrac in

with i=1,2,3,\ldots,n. Then the left-endpoint Riemann sum that approximates the definite integral is

\displaystyle\sum_{i=1}^n\frac{{r_i}^3}n

and taking the limit as n\to\infty gives the area exactly. We have

\displaystyle\lim_{n\to\infty}\frac1n\sum_{i=1}^n\left(\frac in\right)^3=\lim_{n\to\infty}\frac{n^2(n+1)^2}{4n^3}=\boxed{\frac14}

6 0
3 years ago
Determine the maximum number of zeros of the polynomial function -8x^4-x^2+2 a. 1 c. 3 b. 4 d. 2 Please select the best answer f
alexgriva [62]
Generally speaking, the number of solutions to an algebraic equation is equal to the exponent of the highest power.  so the answer is b.4


6 0
3 years ago
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