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3 years ago

~Only answer if you know for sure~ -Meaning 100% positive on the answer- Thank you

Mathematics

2 answers:

patriot [66]3 years ago

8 0

All you have to do is find (3/10) of 60, and find the triangle with that number
for its label.

The exact same kind of problem was answered correctly by 2 people for the question you posted 9 minutes earlier, and one of them even explained how
he did it. If you're getting anything more than answers from Brainly ... I mean
if you're actually getting HELP here ... then you ought to be able to answer
this one on your own.

xxMikexx [17]3 years ago

7 0

The question is asking three tenth of the actual figure, resulting in having to do 60 x 3/10= 180/10= 18. So, your answer us eighteen centimeters. I hope this helped u

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The drama and math club scenario is represented by the system of equations in which x represents the drama club members and y re

shtirl [24]

Answer:

-8, 40, 32.

Step-by-step explanation:

On edge owo <3~

6 0

3 years ago

A baseball player gets 16 hits in 40 at bats. What percent of his at bats did not result

Katarina [22]

Answer:

40 percent

Step-by-step explanation:

Now percentage formula

Whatever

- - - - - - - - - - x 100

Total whatever

In this case your whatever is the hits. according to the question, if i said he had 40 hits in 40, that's a 100percent. So... Your whatever is 16, total whatever is 40

- - - - - - x 100

= 40 percent

6 0

2 years ago

let X = {a,b,c}, Y = {a,c,b}, Z = {a,b,b,c,c,c}. what are the elements of X, Y, and Z? how are X,Y, and Z related?

zhannawk [14.2K]

They're equal.

$X=Y=Z$

7 0

3 years ago

Read 2 more answers

The sum of squares of two

Elanso [62]

Answer:

the product of the 2 numbers is 22

Step-by-step explanation:

x² + y² = 80

(x - y)² = 36

x - y = 6

or y - x = 6

let's start with the first one x-y=6

x = 6 + y

(6+y)² + y² = 80

y² + 6y +6y + 36 + y² = 80

2y² + 12y + 36 = 80

2y² + 12y - 44 = 0

y² + 6y - 22 = 0

the solution of a quadratic equation

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case here we have an squadron in y.

a=1

b=6

c=-22

(-6 ± sqrt(36 + 4×22))/2 = (-6 ± sqrt(36+88))/2 =

= (-6 ± sqrt(124))/2 = (-6 ± sqrt(4×31))/2 =

= (-6 ± 2×sqrt(31))/2 = -3 ± sqrt(31)

y1 = -3 + sqrt(31)

y2 = -3 - sqrt(31)

x1 = 6 + -3 + sqrt(31) = 3 + sqrt(31)

x2 = 3 - sqrt(31)

control :

(-3 + sqrt(31))² + (3 + sqrt(31))² = 80

9 - 3 sqrt(31) - 3 sqrt(31) + 31 + 9 +3 sqrt(31) + 3 sqrt(31) + 31 = 80

9 + 31 + 9 +31 = 80

18 + 62 = 80

80 = 80 correct

solving now for y-x=6

delivers exactly the same calculations, just with x and y trading places.

so, the resulting 2 number pairs are the same.

the product of the 2 numbers :

(3 + sqrt(31))(-3 + sqrt(31)) = -9 - 3 sqrt(31) + 3 sqrt(31) + 31 =

= -9 + 31 = 22

(3 - sqrt(31))(-3 - sqrt(31)) = -9 + 3 sqrt(31) - 3 sqrt(31) + 31 =

= 22

so, the product is the same in both cases.

6 0

3 years ago

Can you please answer the question?

Roman55 [17]

The trigonometric expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the trigonometric expression $\sec \alpha \cdot \csc \alpha + 1$ .

<h3>How to prove a trigonometric equivalence</h3>

In this problem we must prove that one side of the equality is equal to the expression of the other side, requiring the use of algebraic and trigonometric properties. Now we proceed to present the corresponding procedure:

$\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} - \frac{\frac{1 }{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\frac{\tan^{3}\alpha - 1}{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\tan^{3}\alpha - 1}{\tan \alpha \cdot (\tan \alpha - 1)}$

$\frac{(\tan \alpha - 1)\cdot (\tan^{2} \alpha + \tan \alpha + 1)}{\tan \alpha\cdot (\tan \alpha - 1)}$

$\frac{\tan^{2}\alpha + \tan \alpha + 1}{\tan \alpha}$

$\tan \alpha + 1 + \cot \alpha$

$\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} + 1$

$\frac{\sin^{2}\alpha + \cos^{2}\alpha}{\cos \alpha \cdot \sin \alpha} + 1$

$\frac{1}{\cos \alpha \cdot \sin \alpha} + 1$

$\sec \alpha \cdot \csc \alpha + 1$

To learn more on trigonometric expressions: brainly.com/question/10083069

#SPJ1

6 0

2 years ago

~Only answer if you know for sure~ -Meaning 100% positive on the answer- *Thank you*

~Only answer if you know for sure~ -Meaning 100% positive on the answer- Thank you