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2 years ago

Select the lengths that would form a right triangle.

Mathematics

1 answer:

allsm [11]2 years ago

8 0

Hello

The three different lengths are 39 as a leg 80 as a leg and 89 as the hypotneuse becasue of the Pythagoran therom. a^2+b^2=c^2
So your answers would be 39, 80, and 89

Hope this helps
Plz mark me as Brainliest

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3x + 8 < 4(x - 3)Solve the inequality and graph the solution set.

solmaris [256]

Step-by-step explanation:

3x+8<4x−12

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−x+8<−12

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−x<−20

x>20

Interval Notation:(20,∞)

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2 years ago

F(x) = 5x^3+ 2x^2 - 90x - 36. <br><br> finding all zeros by factoring and explaining the steps

ella [17]

Answer:

f(x)=5x^3-2x^2-90x-36=0

=x^2(5x-2)-18(5x-2)=(x^2-18)(5x-2)=0

x^2-18=0/5x-2=0

x^2=18=x=9√2

5x-2=0

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8 0

2 years ago

If -2 is added to a number and the sum is doubled, the result is -15 less than the number.

VMariaS [17]

The required number is 19 given that -2 is added to a number, the sum is doubled and the result is -15 less than the number. This can be obtained by assuming the number as x, converting the given conditions to algebraic expression, forming algebraic equation and solving for x.

<h3>Find the required number:</h3>

Here in the question it is given that,

-2 is added to a number

The sum is doubled

The result is -15 less than the number

We have to find the required number.

Let the required number be x.

⇒ -2 is added to the number ⇒ -2 + x

⇒ the sum is doubled ⇒ 2(-2 + x)

⇒ the result is -15 less than the number ⇒ 2(-2 + x) = x -(-15)

2(-2 + x) = x + 15

⇒ - 4 + 2x = x + 15

⇒ x = 19

Hence the required number is 19 given that -2 is added to a number, the sum is doubled and the result is -15 less than the number.

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1 year ago

Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

$sin^2(\theta)+ cos^2(\theta)=1$

we have

$cos(\theta)=-\frac{\sqrt{2}}{3}$

substitute

$sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1$

$sin^2(\theta)+ \frac{2}{9}=1$

$sin^2(\theta)=1- \frac{2}{9}$

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

$tan^2(\beta)+1= sec^2(\beta)$

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

$cos(\beta)=\frac{3}{5}$

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

$cos(\beta)=\frac{3}{5}$

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

$sin(\beta)=\frac{4}{5}$

step 4

Find sin(θ+β)

we know that

$sin(A + B) = sin A cos B + cos A sin B$

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

$cos(\theta)=-\frac{\sqrt{2}}{3}$

$sin(\beta)=\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

8 0

3 years ago

Dave is helping his grandmother make trail mix. His grandmother asks him to add 1/5 cup of fruit for every 1/3 cup of nuts. To s

erma4kov [3.2K]

3/5 cup of fruit because 1/3 +1/3 +1/3 = 3/3 or one. 1/5 +1/5 +1/5 = 3/5

3 0

2 years ago