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andre [41]
3 years ago
7

Evaluate v+t when t=4x-1 and v=2x+1

Mathematics
1 answer:
Rainbow [258]3 years ago
6 0

You would simply add the last two equations together. If v=2x+1 and t=4x-1, you can add them up. This would make 2x+1+4x-1. If you combine like terms, 1 and -1 cancel out and you are left with 6x. This means x can be anything.
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Step-by-step explanation:

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(05.01) Which statement best describes the area of the triangle shown below?​
alukav5142 [94]

Answer:

A

Step-by-step explanation:

Area of a triangle:

A=\frac{b*h}{2}

In our case:

b=4

h=2

Plug in what we know:

A=\frac{(4)(2)}{2} \\A=4units

Find the matching solution:

A.) it is 1/2 the area of a rectangle of length 4 units and width 2 units

X B.) it is twice the area of a rectangle of length 4 units and width two units

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X D.) it is twice the area of a square of side length 4 units

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2 years ago
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nata0808 [166]
  • \sf \vec{w}=16km
  • \sf \vec{s}=9km

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\\ \sf\longmapsto PQ=16km-9km

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4 0
2 years ago
Suppose that a box contains one fair coin (Heads and Tails are equally likely) and one coin with Heads on each side. Suppose fur
stealth61 [152]

Using conditional probability, it is found that there is a 0.1 = 10% probability that the chosen coin was the fair coin.

Conditional Probability

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • P(A \cap B) is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem:

  • Event A: Three heads.
  • Event B: Fair coin.

The probability associated with 3 heads are:

  • 0.5^3 = 0.125 out of 0.5(fair coin).
  • 1 out of 0.5(biased).

Hence:

P(A) = 0.125 + 0.5 = 0.625

The probability of 3 heads and the fair coin is:

P(A \cap B) = 0.5(0.125) = 0.0625

Then, the conditional probability is:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0625}{0.625} = 0.1

0.1 = 10% probability that the chosen coin was the fair coin.

A similar problem is given at brainly.com/question/14398287

4 0
3 years ago
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