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2 years ago

Need help with this math problem

Mathematics

1 answer:

Dmitrij [34]2 years ago

3 0

Answer:

So the value of u is $24$ degree.

Step-by-step explanation:

Given;

Three angle $2u$ , $(u+18)$ degree and $90$ degree in a Triangle.

We know;

Addition of three angle in a triangle is equal $180$ degree

$2u+u+18+90=180$

$3u=180-90-18$

$3u=72$

$u=\frac{72}{3}$

$u=24$

∴ The value of u is [tex]24[\tex] degree.

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Graph a system of equations to solve log (−5.6x + 1.3) = −1 − x. Round to the nearest tenth.

Neporo4naja [7]

Answer:

Step-by-step explanation:

-2.1 and 0.2

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3 years ago

According to Masterfoods, the company that manufactures M&M’s, 12% of peanut M&M’s are brown, 15% are yellow, 12% are re

Anna35 [415]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

P(brown) = 12% = 0.12

P(Yellow) = 15% = 0.15

P(Red) = 12% = 0.12

P(blue) = 23% = 0.23

P(orange) = 23% = 0.23

P(green) = 15% = 0.15

A.) Compute the probability that a randomly selected peanut M&M is not yellow.

P(not yellow) = P(Yellow)' = 1 - P(Yellow) = 1 - 0.15 = 0.85

B.) Compute the probability that a randomly selected peanut M&M is brown or red.

P(Brown) or P(Red) :

0.12 + 0.12 = 0.24

C.) Compute the probability that three randomly selected peanut M&M’s are all brown.

P(brown) * P(brown) * P(brown)

0.12 * 0.12 * 0.12 =0.001728

D.) If you randomly select three peanut M&M’s, compute that probability that none of them are blue.

P(3 blue)' = 1 - P(3 blue)

P(3 blue) = 0.23 * 0.23 * 0.23 = 0.012167

1 - P(3 blue) = 1 - 0.012167 = 0.987833

If you randomly select three peanut M&M’s, compute that probability that at least one of them is blue.

P(1 blue) + p(2 blue) + p(3 blue)

(0.23) + (0.23*0.23) + (0.23*0.23*0.23)

0.23 + 0.0529 + 0.012167

= 0.295067

5 0

3 years ago

Square root of 2tanxcosx-tanx=0

kobusy [5.1K]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3242555

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Solve the trigonometric equation:

$\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}$

Restriction for the solution:

$\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.$

Square both sides of (i):

$\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}$

$\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}$

Let

$\mathsf{sin\,x=t\qquad (0\le t$

So the equation becomes

$\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}$

Solving the quadratic equation:

$\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.$

$\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}$

$\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}$

You can discard the negative value for t. So the solution for (ii) is

$\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}$

Substitute back for t = sin x. Remember the restriction for x:

$\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}$

where k is an integer.

I hope this helps. =)

3 0

3 years ago

Hey whats 2+2? i need to know ASAP! IM IN PRESCHOOL-

Karo-lina-s [1.5K]

Answer:

lol Its 4

Step-by-step explanation:

Dont worry you got this

4 0

2 years ago

Read 2 more answers

Please help me 20 points