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3 years ago

Please help me with his geometry question

Mathematics

1 answer:

Reil [10]3 years ago

6 0

Is this <ABD?

if so, all lines tangent to a circle, its point of tangency is perpendicular to the center of the circle. So <ABD = 90 degrees

You might be interested in

Find the limit as x approaches 0 of x/sin3x

frozen [14]

$\lim_{x \to 0} \frac{x}{sin 3x} = \\ =\frac{0}{0}= \\ = \lim_{x \to 0} \frac{3x}{3sin3x} =$
= 1/3 ( because the limit as x approaches 0 of t/sin t = 1 ).

5 0

3 years ago

The members of a high school band asked a number of students whether they would like blue, gold, or both for the uniforms for th

MaRussiya [10]

The values of a and b are a = 33% and b = 43% ⇒ last answer

Step-by-step explanation:

The venn-diagram contains:

A circle labeled blue 32
A circle labeled gold 25
The two circles are overlap with label 12 on the overlap

The table:

→ Band : blue : N.b : total

→ gold : 16% : a : 49%

→ N.g : b : 8% : 51%

→ Total : 59% : 41% : 100%

We need to find the values of a and b

From the table

16% represents the percentage of blue and gold uniforms (2nd row with 2nd column)

From venn-diagram

The common part of blue and gold label with 12 (overlap of two circles)

∴ 16% of the total number of students = 12

∵ 16% × x = 12

∴ $\frac{16}{100}$ × x = 12

∴ 0.16 x = 12

- Divide both sides by 0.16

∴ x = 75

∵ x represents the total number of students

∴ The total number of students were asked is 75

From the table

b represents the percentage of blue but not gold uniforms (3rd row and 2nd column)

From venn-diagram

The part that has blue and not gold labeled with 32 (the part of the blue circle not in the gold circle)

∴ 32 of the total number of students like the blue but not gold

- Divide 32 by the total 75 , then multiply the quotient by 100%

to find the percentage of blue but not gold

∴ b = $\frac{32}{75}$ × 100% = 42.6666%

∴ b ≅ 43%

From the table

a represents the percentage of gold nut not blue uniforms (2nd row and 3rd column)

From venn-diagram

The part that has gold and not blue labeled with 25 (the part of the gold circle not in the blue circle)

∴ 25 of the total number of students like the gold but not blue

- Divide 25 by the total 75 , then multiply the quotient by 100%

to find the percentage of gold but not blue

∴ a = $\frac{25}{75}$ × 100% = 33.3333%

∴ a ≅ 33%

The values of a and b are a = 33% and b = 43%

To check your answer complete the table you will find the total of it is 100% ( for the columns 16% + 43% = 59% , 33% + 8% = 41% , then 59% + 41% = 100%, for the rows 16% + 33% = 49% , 43% + 8% = 51% , then 49% + 51% = 100%)

Learn more:

You can learn more about the probability in brainly.com/question/3756853

#LearnwithBrainly

4 0

3 years ago

Read 2 more answers

HELP Please<br>I really need the answer

SpyIntel [72]

Step-by-step explanation:

a full circle has 360°.

so, since each individual result is represented by an amount of degrees, the total sum of all results must be 360°.

labour = 360 - 136 - 34 - 54 - 14 = 122°

the fraction 122/360 represents the 183 votes for labour.

122/360 = 183

the fraction 1/1 = 1 then represents the total number of votes in the whole system.

61/180 = 183

1/180 = 3

1 = 3×180 = 540

540 pupils voted altogether.

3 0

2 years ago

3x – 2y+3<br> 3y - 2x<br><br> X=7 Y=6

faust18 [17]

<h2><em>Answer:</em></h2><h2><em>3×7-2×6+3</em></h2><h2 /><h2><em>=21-12+3=24-12</em><em>=</em><em>1</em><em>2</em></h2><h2 />

8 0

3 years ago

Help with 5b please . thank you.

Allushta [10]

Answer:

See explanation

Step-by-step explanation:

We are given f(x)=ln(1+x)-x+(1/2)x^2.

We are first ask to differentiate this.

We will need chain rule for first term and power rule for all three terms.

f'(x)=(1+x)'/(1+x)-(1)+(1/2)×2x

f'(x)=(0+1)/(1+x)-(1)+x

f'(x)=1/(1+x)-(1)+x

We are then ask to prove if x is positive then f is positive.

I'm thinking they want us to use the derivative part in our answer.

Let's look at the critical numbers.

f' is undefined at x=-1 and it also makes f undefined.

Let's see if we can find when expression is 0.

1/(1+x)-(1)+x=0

Find common denominator:

1/(1+x)-(1+x)/(1+x)+x(1+x)/(1+x)=0

(1-1-x+x+x^2)/(1+x)=0

A fraction can only be zero when it's numerator is.

Simplify numerator equal 0:

x^2=0

This happens at x=0.

This means the expression,f, is increasing or decreasing after x=0. Let's found out what's happening there. f'(1)=1/(1+1)-(1)+1=1/2 which means after x=0, f is increasing since f'>0 after x=0.

So we should see increasing values of f when we up the value for x after 0.

Plugging in 0 gives: f(0)=ln(1+0)-0+(1/2)0^2=0.

So any value f, after this x=0, should be higher than 0 since f(0)=0 and f' told us f in increasing after x equals 0.

8 0

2 years ago