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Anvisha [2.4K]
3 years ago
14

Find the volume of the given prism. Round to the nearest tenth if necessary.

Mathematics
1 answer:
alex41 [277]3 years ago
3 0

Answer:

17m³

Step-by-step explanation:

==>Given:

Prism with right triangular base with the following dimensions:

Height = 17m

Base area = area of the right triangle = ½×2×1 = 1m²

==>Required:

Volume of prism

==>Solution:

Volume of prism = Base area × height of prism

Volume of prism = 1m² × 17m

Volume = 17m³

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PLS HELP PLS
77julia77 [94]

Red Perimeter= 4u X π= 4uπ

Blue Perimeter= u X π= uπ

Red Area= 2u X 2u X π= 4u2π

Blue Area= u X u X π= u2π

4 0
3 years ago
Apple juice has 216 grams of sugar for every 8 servings. The soda has 234 grams of sugar for every 9 servings
pav-90 [236]

Answer:

Apple juice contains 1728 g sugar whereas soda contains 1404 g sugar

Step-by-step explanation:

3 0
3 years ago
At an art exhibition there are 12 paintings of which 10 are original. A visitor selects a painting at random and before decides
Reika [66]

Answer:

a) 0.8333

b) 0.75

c) 0.8181 or 0.9090

Step-by-step explanation:

a)

The probability the visitor selects an authentic painting is  

10/12 = 0.8333

b)

Since the opinion of the expert does not depend on your choice, the events are <em>independent</em>, so the probability that the expert says is authentic and it really is, is

0.8333*0.9 = 0.75

c)

If the expert decides the painting is a copy and it is not, then there are 11 paintings of which 9 are authentic, so the probability the visitor selects a new original painting is

9/11= 0.8181

If the expert decides the painting is a copy and it is, then there are 11 paintings of which 10 are authentic, so the probability the visitor selects a new original painting is

10/11= 0.9090

8 0
3 years ago
What is the factored form of x^3 -1
lianna [129]

Answer:

The factored form of x^3 -1 will be:

x^3-1^3=\left(x-1\right)\left(x^2+x+1\right)

Step-by-step explanation:

Given the expression

x^3-1

Rewrite 1 as 1³

=x^3-1^3

\mathrm{Apply\:Difference\:of\:Cubes\:Formula:\:}x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)

x^3-1^3=\left(x-1\right)\left(x^2+x+1\right)

            =\left(x-1\right)\left(x^2+x+1\right)

Thus, the factored form of x^3 -1 will be:

x^3-1^3=\left(x-1\right)\left(x^2+x+1\right)

8 0
3 years ago
Apply The Remainder Theorem, Fundamental Theorem, Rational Root Theorem, Descartes Rule, and Factor Theorem to find the remainde
Over [174]

9514 1404 393

Answer:

  possible rational roots: ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12}

  actual roots: -1, (2 ±4i√2)/3

  no turning points; no local extrema

  end behavior is same-sign as x-value end-behavior

Step-by-step explanation:

The Fundamental Theorem tells us this 3rd-degree polynomial will have 3 roots.

The Rational Root Theorem tells us any rational roots will be of the form ...

  ±{factor of 12}/{factor of 3} = ±{1, 2, 3, 4, 6, 12}/{1, 3}

  = ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12} . . . possible rational roots

Descartes' Rule of Signs tells us the two sign changes mean there will be 0 or 2 positive real roots. Changing signs on the odd-degree terms makes the sign-change count go to 1, so we know there is one negative real root.

The y-intercept is 12. The sum of all coefficients is 22, so f(1) > f(0) and there are no positive real roots in the interval [0, 1]. Synthetic division by x-1 shows the remainder is 22 (which we knew) and all the quotient coefficients are all positive. This means x=0 is an upper bound on the real roots.

The sum of odd-degree coefficients is 3+8=11, equal to the sum of even-degree coefficients, -1+12=11. This means that -1 is a real root. Synthetic division by x+1 shows the remainder is zero (which we knew) and the quotient coefficients alternate signs. This means x=-1 is a lower bound on real roots. The quotient of 3x^2 -4x +12 is a quadratic factor of f(x):

  f(x) = (x +1)(3x^2 -4x +12)

The complex roots of the quadratic can be found using the quadratic formula:

  x = (-(-4) ±√((-4)^2 -4(3)(12)))/(2(3)) = (4 ± √-128)/6

  x = (2 ± 4i√2)/3 . . . . complex roots

__

The graph in the third attachment (red) shows there are no turning points, hence no relative extrema. The end behavior, as for any odd-degree polynomial with a positive leading coefficient, is down to the left and up to the right.

4 0
3 years ago
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