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2 years ago

I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

Mathematics

1 answer:

Lubov Fominskaja [6]2 years ago

3 0

Answer:

okay let's start

Step-by-step explanation:

$use \: sine \: rule \\ \frac{x}{ \sin(31) } = \frac{42}{ \sin(59) } \: \: taking \: angle \: at \: c \: as \: 90 \\ x = \frac{ 42 \sin(31)}{ \sin(59)} \\ x = 25.236146$

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<img src="https://tex.z-dn.net/?f=494%20divided%20by%2019%2C594" id="TexFormula1" title="494 divided by 19,594" alt="494 divided

Vaselesa [24]

0.02521179 jdjdjdduhdbdjdhdhj

3 0

3 years ago

9/20

Stella [2.4K]

<u>Question:</u>

Find the number of real number solutions for the equation. x^2 + 5x + 7 = 0

<u>Answer:</u>

The number of real solutions for the equation $x^{2}+5 x+7=0$ is zero

<u>Solution:</u>

For a Quadratic Equation of form : $a x^{2}+b x+c=0$ ---- eqn 1

The solution is $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Now , the given Quadratic Equation is $x^{2}+5 x+7=0$ ---- eqn 2

On comparing Equation (1) and Equation(2), we get

a = 1 , b = 5 and c = 7

In $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ , $b^2 - 4ac$ is called the discriminant of the quadratic equation

Its value determines the nature of roots

Now, here are the rules with discriminants:

1) D > 0; there are 2 real solutions in the equation

2) D = 0; there is 1 real solution in the equation

3) D < 0; there are no real solutions in the equation

Now let solve for given equation

$D= b^2 - 4ac\\\\D = 5^2 - 4(1)(7)\\\\D = 25 - 28 \\\\D = -3$

Since -3 is less than 0, this means that there are 0 real solutions in this equation.

4 0

4 years ago

Simplify this expression. Write your answer using positive exponents.<br>

storchak [24]

What expression ? There isn’t one

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3 years ago

a shipping box will be filled with two different machine parts. one type of machine part weights 0.15 ponds and the other weight

gulaghasi [49]

$\frac{x}{2} =0.15$ hope so this helps

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3 years ago

A fair die is cast four times. Calculate

svetlana [45]

Step-by-step explanation:

<h2><em><u>You can solve this using the binomial probability formula.</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows:</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: </u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) </u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)</u></em></h2>

8 0

3 years ago