Boxers in the nineteenth century used their bare fists. In modern boxing,

fighters wear padded gloves. How do gloves protect the brain of the boxer from

injury?​

the answer is [tex]x^{2} \sqrt{x} \leq \geq \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \pi x^{2} \sqrt{x} \frac{x}{y} \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left \{ {{y=2} \atop {x=2}} \right.[/tex]random but marter than einstein

Answer:

32 Years, 7 Months, 2 Days old

Step-by-step explanation:

Answer:

2590.33

Step-by-step explanation:

...............

Answer:

option D

Step-by-step explanation:

18 1/5 + 22 2/5 +40 1/5

it wouldnt let me type the answer-

Answer:

[tex]\frac{2}{9}[/tex]

Step-by-step explanation:

Since both fractions have a different denominator, we need to look for a common denominator.

[tex]\frac{1}{3}=\frac{3}{9}[/tex]   Multiply 3 in both numerator and denominator

[tex]\frac{5}{9} -\frac{3}{9} =\frac{2}{9}[/tex]    

Hence, 2/9 is the answer

Answer:

Directions: Write the new coordinates for each point or figure.

Sample Question:

Point F has the coordinates (-9,6). What are the coordinates of its image point after the translation of six units down and three units to the left?

Answer:

The coordinates of its image point after the translation are (-12, 0).

Explanation: When you look at an ordered pair the x number comes first, followed by the y number (x,y). Moving a point left or right changes the x number. Moving a point up or down changes the y number. Left and down require subtraction, while right and up require addition. In the sample, we had to go 3 units to the left, -9 - 3 = -12. We also had to go 6 units down, so 6 - 6 = 0.

Point Q has the coordinates (9,5). What are the coordinates of its image point after a translation of six units up and three units to the right?

Point P has the coordinates (3,4). What are the coordinates of its image point after a translation of 8 units down and 5 units to the left?

Point Z has the coordinates (10,7). What are the coordinates of its image point after a translation of six units down and three units to the right?

Point Y has the coordinates (0,-4). What are the coordinates of its image point after a translation of 10 units up and 3 units to the left?

Point X has the coordinates (3,-8). What are the coordinates of its image points after a translation of 11 units down and 10 units to the right?

Answer:

(-10,5)

Step-by-step explanation:

9514 1404 393

Answer:

  $80

Step-by-step explanation:

You are given the formula ...

  P = I/(rt)

and the values I = 20, r = 0.05, t = 5.

Put the numbers in place of the corresponding variables and do the arithmetic.

  P = 20/(0.05·5) = 20/0.25 = 80

The amount $80 will generate $20 in interest over 5 years at 5% per year.

Answer:

option C

Step-by-step explanation:

[tex]distance = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}[/tex]

              [tex]=\sqrt{(-4-2)^2+(15-7)^2} \\\\=\sqrt{(-6)^2 + 8^2}\\\\=\sqrt{36 + 64 }\\\\= \sqrt{100} \\\\= 10[/tex]

Answer:

A

Step-by-step explanation:

Answer:

[tex]x < - 2[/tex]

Step-by-step explanation:

Hope it is helpful...

Answer:

2nd option:

8x+2y=32

x+y=7

Step-by-step explanation:

Answer:

It is in fact the second one.

Step-by-step explanation:

It says that he bought a total of 7 tickets and it cost $32.

x represents the number of adult tickets and y represents the number of chid tickets.

So:

x + y = 7 because the adding the number of adult and child tickets should equal the total number bought.

and

8x + 2y = 32 because each adult ticket was 8 dollars and each child was 2 and the total should add up to 32

Answer:

[tex]A_{\Delta ABP}=\sin \theta[/tex]

Step-by-step explanation:

Let the center of the circle be O.

Recall that the area of a triangle can be given by:

[tex]\displaystyle A=\frac{1}{2} ab\sin C[/tex]

Where C is the angle between the two sides.

ΔABP is equal to the sum of ΔAPO and ΔBOP.

Let a = OP and b = OB. Since this is the unit circle and PO and BO are radii, they both equal one. C will be θ. Hence, the area of ΔBOP is:

[tex]\displaystyle A_{\Delta BOP}=\frac{1}{2}(1)(1)\sin \theta=\frac{1}{2}\sin\theta[/tex]

For ΔAPO, we can use the two sides OP and OA. Again, they are the radii of the unit circle, so they equal one. The angle in this case will be π - θ radians. Hence:

[tex]\displaystyle A_{\Delta APO}=\frac{1}{2}(1)(1)\sin\left(\pi -\theta\right)=\frac{1}{2}\sin\left(\pi -\theta\right)[/tex]

However, note that sin(π - θ) = sin(θ). Hence:

[tex]\displaystyle A_{\Delta APO}=\frac{1}{2}\sin\left(\theta\right)[/tex]

Hence, the area of ΔABP is:

[tex]\displaystyle A_{\Delta ABP}=\frac{1}{2}\sin \theta+\frac{1}{2}\sin\theta =\sin \theta[/tex]

Answer:

NEDD POITS SERY

Step-by-step explanation:

Answer:

X = -6 - 3y

Y = -1

Step-by-step explanation: Solve for X first using basic algebra. Plug in X vaalue to original equation and solve. -6 - 3y + 3y = -6. Solve by adding 6 to both sides of the equation and dividing -3y by 3y.

Answer:

See Explaination

Step-by-step explanation:

According To Question

(Diagram, Please Find In Attachment)

Question 38 . Cos ∅ = 12/13   ∅ in Quadrant-4(Where only Cos & Sec are Positive all other Functions Are Negative in this Quadrant)

Now ∅ Represent a Right Angle Triangle with Base=12 Perpendicular=5 & Hypothenese=13

Sin∅ = -5/13 , Cosec∅=-13/5 , Tan∅=-5/12 , Cot∅=-12/5 & Sec∅=13/12

Question 39 . SinX=-3/4   X Lies in Quadrant-3(Where only Tan & Cot are Positive all other Functions Are Negative in this Quadrant)

Find By Using Basic Formulas

[tex]SinX^{2}+CosX^{2}=1[/tex] ↔ Put Value of Sin∅=-3/4

We get CosX= ±[tex]\frac{\sqrt{5} }{4}[/tex] (but X Lies in Quadrant-3 Where Cosine Is Negative , So Neglect CosX= +[tex]\frac{\sqrt{5} }{4}[/tex])

So CosX= -[tex]\frac{\sqrt{5} }{4}[/tex] ...... now you get the point . All Questions Will Solve like This .

Answer:

22.5

Step-by-step explanation:

Inscribed Angle =1/2 Intercepted Arc

F = 1/2 (45)

<F = 22.5

[tex]0.375[/tex] ✅

Step-by-step explanation:

[tex] - 1 \frac{7}{8} + 2 \frac{1}{4} \\ = - \frac{15}{8} + \frac{9}{4} \\ = - \frac{15}{8} + \frac{9 \times 2}{4 \times 2} \\ = \frac{ - 15 + 18}{8} \\ = \frac{3}{8} \\ = 0.375[/tex]

[tex]\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}[/tex]

Answer:

01:04

Step-by-step explanation:

add 7 minutes to 22:57 to get 23:04. then add 2 hours to get to 01:04.

Answer:

1

Step-by-step explanation:

Answer:

She didn't put the sign's in the proper place, it should be ( x + 3 ) ( x - 7 )

Step-by-step explanation:

( x - 3 ) ( x + 7 ) = x² + 7x - 3x - 21 = x² + 4x - 21

( x + 3 ) ( x - 7 ) = x² - 7x + 3x - 21 = x² - 4x - 21

Answer:

See Below.

Step-by-step explanation:

Problem A)

We have:

[tex]\displaystyle \csc^2\theta \tan^2\theta -1=\tan^2\theta[/tex]

When in doubt, convert all reciprocal trig functions and tangent into terms of sine and cosine.

So, let cscθ = 1/sinθ and tanθ = sinθ/cosθ. Hence:

[tex]\displaystyle \left(\frac{1}{\sin^2\theta}\right)\left(\frac{\sin^2\theta}{\cos^2\theta}\right)-1=\tan^2\theta[/tex]

Cancel:

[tex]\displaystyle \frac{1}{\cos^2\theta}-1=\tan^2\theta[/tex]

Let 1/cosθ = secθ:

[tex]\sec^2\theta -1=\tan^2\theta[/tex]

From the Pythagorean Identity, we know that tan²θ + 1 = sec²θ. Hence, sec²θ - 1 = tan²θ:

[tex]\tan^2\theta =\tan^2\theta[/tex]

Problem B)

We have:

[tex]\sin^3x=\sin x-\sin x \cos^2 x[/tex]

Factor out a sine:

[tex]\sin x(\sin^2 x)=\sin x-\sin x\cos^2 x[/tex]

From the Pythagorean Identity, sin²θ + cos²θ = 1. Hence, sin²θ = 1 - cos²θ:

[tex]\sin x(1-\cos^2 x)=\sin x-\sin x\cos^2x[/tex]

Distribute:

[tex]\sin x- \sin x \cos^2 x=\sin x-\sin x\cos^2 x[/tex]

Problem C)

We have:

[tex]\displaystyle \frac{\cos 2x+1}{\sin 2x}=\cot x[/tex]

Recall that cos2θ = cos²θ - sin²θ and that sin2θ = 2sinθcosθ. Hence:

[tex]\displaystyle \frac{\cos^2 x-\sin^2 x+1}{2\sin x\cos x}=\cot x[/tex]

From the Pythagorean Identity, sin²θ + cos²θ = 1 so cos²θ = 1 - sin²θ:

[tex]\displaystyle \frac{2\cos^2 x}{2\sin x\cos x}=\cot x[/tex]

Cancel:

[tex]\displaystyle \frac{\cos x}{\sin x}=\cot x[/tex]

By definition:

[tex]\cot x = \cot x[/tex]

Answer:

16

Step-by-step explanation:

8/1/2

8 divided by 1/2

8 x 2/1 = 16

Answer:

Step-by-step explanation:

P=8/V

1/2=8/V

do cross multiplication

2*8=V*1

16=V

Answer:

C: .75

Step-by-step explanation:

I really just subtracted

Answer:

0,4

Step-by-step explanation:

See image below:)

Answer:

[tex]x = 70^o[/tex]

Step-by-step explanation:

Given

See attachment for triangle

Required

Find x

To solve for x, we make use of:

[tex]x + 50^o + 60^o = 180^o[/tex] --- angles in a triangle

[tex]x + 110^o = 180^o[/tex]

Collect like terms

[tex]x = 180^o-110^o[/tex]

[tex]x = 70^o[/tex]

Step-by-step explanation:

P(red) = 12/(12+10+5+4+29)

= 12/60

= 1/5

= 0.2

= 20%

Answer:

[tex] \cos(x) = \frac{24}{30} [/tex]

OPTION C