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borishaifa [10]
3 years ago
5

Given that a = 10,

Mathematics
1 answer:
tankabanditka [31]3 years ago
3 0

Since a = 10 and b = 2x + 5, we have

ab=10(2x+5)=20x+50

So, the equation becomes

ab = 6x^2 +11x-10 \iff 20x+50= 6x^2 +11x-10 \iff 6x^2-9x-60=0

The solutions to this equation are 4 and -5/2, so I'm afraid there's a typo in your question.

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Simplify:<br> -z3+5k6-(-z3+10k6)
viva [34]

Answer:-5k6

Step-by-step explanation:

8 0
3 years ago
HELP PLS ASAP!
Leokris [45]

Step-by-step explanation:

The horizontal stretch or compression for a function f(x) is given by  g = f(bx) where b is a constant. If b> 0 then the graph of a function is compressed.

As it is given in the question that the function is transformed by a compression factor of 3.

Given function

The value of k will be 3 if the function is transformed by a compression factor of 3

3 0
3 years ago
Please help me with this question. Thanks!
olasank [31]

Assuming that all the grapes have the same probability of being randomly picked:

  • a) 0.11
  • b)  0.24

<h3>How to find the probabilities?</h3>

We know that there are 8 green grapes and 15 red grapes on the bowl, so there is a total of 23 grapes.

a) Here we need to find the probability that both grapes are green. Remember that the probability of getting a green grape is equal to the quotient between the number of green grapes and the total number of grapes, this is:

P = 8/23

But then he must take another, because he already took one, the number of green grapes is 7, and the total number of grapes is 22, so for the second grape the probability is:

P' = 7/22.

The joint probability is the product of the two individual probabilities:

prob = (8/23)*(7/22) = 0.11

b) For the first green grape we know that the probability is:

P = 8/23

Then he must get a red grape. There are 15 red grapes and 22 grapes in total, so the probability is:

P' = 15/22

Then the joint probability is:

prob = (8/23)*(15/22) = 0.24

If you want to learn more about probability:

brainly.com/question/25870256

#SPJ1

8 0
3 years ago
Find the volume of the largest circular cone that can<br> beinscribed in a shpere of radius 3.
Aleksandr [31]

Answer:

V =\dfrac{32}{3}\pi

Step-by-step explanation:

given,

radius of sphere = 3

volume of cone:

V = \dfrac{1}{3}\pi r^2h

r is the radius of circular base

h is the height of the cone

here r = x and h = 3 + y

now, volume in term of x and y

V = \dfrac{1}{3}\pi x^2(3+y)

Applying Pythagoras theorem

x² + y² = 3²

x = \sqrt{9-y^2}

V = \dfrac{1}{3}\pi ( \sqrt{9-y^2})^2(3+y)

V = \dfrac{1}{3}\pi ( 9-y^2)(3+y)

V = \dfrac{1}{3}\pi (27 + 9 y - 3 y^2-y^3)

differentiating both side

\dfrac{dV}{dy} =\dfrac{1}{3}\pi ( 9-6y- 3y^2)

for maxima  \dfrac{dV}{dy} = 0

\pi ( 3-2 y - y^2)=0

 y² + 2 y - 3 = 0

(y+3)(y-1)=0

 y = 1,-3

y cannot be negative so, volume at y = 1

V = \dfrac{1}{3}\pi (27 + 9 (1)- 3(1)^2-(1)^3)

V =\dfrac{32}{3}\pi

Hence, the largest cone which can be inscribed in the spheres of the radius 3 has volume  V =\dfrac{32}{3}\pi

5 0
3 years ago
What is the reciprocal of 5/6​
mestny [16]

Answer:

-6/5

Hope this helps!

5 0
3 years ago
Read 2 more answers
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