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3 years ago

What is the height of a rectangle whose base is 40 inches and whose diagonal is 41 inches?

Mathematics

2 answers:

maw [93]3 years ago

8 0

Answer: 9 inches

Step-by-step explanation:

romanna [79]3 years ago

7 0

Answer:

You must draw a right triangle and label the hypotenuse as 41 and the longest leg (horizontal in this case, since we are to determine the height) as 40. We do not know the height so call it h.

Now the Pythagorean (sp /) Theorem states that the square of the hyp. is equal to the sum of the squares of the two legs.

so, 41 ^ 2 = 40 ^ 2 + h ^ 2

41 ^ 2 = 1681 40 ^ 2 = 1600 Therefore h ^2 must be equal to 81. This means that h = 9.

The height is 9 inches.

Step-by-step explanation:

mark me as brainliest pls :)

~Alex

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How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$

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2 years ago

aylah plans to make small cheese pizzas to sell at a school fundraiser. She has nine bars of cheese. How many pizzas can she mak

Serhud [2]

Answer:

27 pizzas

Step-by-step explanation:

Each brick of cheese can make 3 pizzas since there are 3 thirds in 1. We multiply 3 by the total number of bricks to get 27.

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2 years ago

Which polynomial is prime

Vikentia [17]

We need more information. You don't give us enough information....? Provide an image please. Or at least name the polynomials...

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3 years ago

HURRY!!!!!!!!!!!

andrey2020 [161]

Answer:

2.5

Step-by-step explanation:

Conversion a mixed number 2 1/

to a improper fraction: 2 1/2 = 2 1/

= 2 · 2 + 1/

= 4 + 1/

= 5/

To find new numerator:

a) Multiply the whole number 2 by the denominator 2. Whole number 2 equally 2 * 2/

= 4/

b) Add the answer from previous step 4 to the numerator 1. New numerator is 4 + 1 = 5

c) Write a previous answer (new numerator 5) over the denominator 2.

Two and one half is five halfs

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2 years ago

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How do I solve this?

Ivenika [448]

$\overline{XS}\cong\overline{YT}\Rightarrow 3m+7=4.2m+5\ \ \ |-7\\\\3m=4.2m-2\ \ \ \ |-4.2m\\\\-1.2m=-2\ \ \ \ |:(-1.2)\\\\m=\dfrac{2}{1.2}\\\\m=\dfrac{20}{12}\\\\m=\dfrac{5}{3}\to m=1\dfrac{2}{3}$

$\overline{YS}\cong\overline{XT}\Rightarrow3\dfrac{1}{2}r+2=2r+5\ \ \ \ |-2\\\\3\dfrac{1}{2}r=2r+3\ \ \ \ |-2r\\\\1\dfrac{1}{2}r=3\ \ \ |\cdot2\\\\3r=6\ \ \ \ |:3\\\\r=2$

7 0

3 years ago