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3 years ago

15

Find the value of x in the isosceles triangle shown

1 answer:

Nat2105 [25]3 years ago

8 0

Answer:

x = 8√5

Step-by-step explanation:

The large triangle is divided into two smaller ones. The right one has one leg of length 8, one leg of length x/2, and hypotenuse of length √80.

Appling the Pythagorean Theorem, we can find x:

8² + (x/2)² = (√80)²,

or: x^2

64 + x^2/4 = 80, or ------- = 80

4

Then x^2 = 320 = 64*5. Taking the square root of both sides, we get:

x = 8√5

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Cousin Edie drank all of Clark's egg nog.Edie finds a coupon for 50 cents off that can only be use at a local grocery store.Clar

Rudik [331]

Answer:

Edie has a better deal since his final price is lower

Step-by-step explanation:

Cousin Edie drank all of Clark's egg nog . Edie finds a coupon for 50 cents off that can only be use at a local grocery store.Clark normally buys his egg nog at the supermarket.

The tax rate at both stores is 2.25%.

Clark buys the egg nog at $5.95 plus tax .

Edie buys the egg nog at $6.35 plus coupon plus tax.

After Edie applies the coupon , the final price is $( 6.35 - 0.50 ) = $5.85

<u>Since the percentage of tax applied is the same, the deal with lower final price is better.</u>

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3 years ago

Which property is shown? 4/5+(-4/5)=0

Aliun [14]

Answer:

Inverse Property

Step-by-step explanation:

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8 0

3 years ago

Directions: Find the indicated trigonometric ratio as a fraction in simplest form. 1. Sin L =

Rainbow [258]

Answer:

$\sin L = 0.60$

$tan\ N = 1.33$

$\cos L = 0.80$

$\sin N = 0.80$

Step-by-step explanation:

Given

See attachment

From the attachment, we have:

$MN = 6$

$LN = 10$

First, we need to calculate length LM,

Using Pythagoras theorem:

$LN^2 = MN^2 + LM^2$

$10^2 = 6^2 + LM^2$

$100 = 36 + LM^2$

Collect Like Terms

$LM^2 = 100 - 36$

$LM^2 = 64$

$LM = 8$

Solving (a): $\sin L$

$\sin L = \frac{Opposite}{Hypotenuse}$

$\sin L = \frac{MN}{LN}$

Substitute values for MN and LN

$\sin L = \frac{6}{10}$

$\sin L = 0.60$

Solving (b): $tan\ N$

$tan\ N = \frac{Opposite}{Adjacent}$

$tan\ N = \frac{LM}{MN}$

Substitute values for LM and MN

$tan\ N = \frac{8}{6}$

$tan\ N = 1.33$

Solving (c): $\cos L$

$\cos L = \frac{Adjacent}{Hypotenuse}$

$\cos L = \frac{LM}{LN}$

Substitute values for LN and LM

$\cos L = \frac{8}{10}$

$\cos L = 0.80$

Solving (d): $\sin N$

$\sin N = \frac{Opposite}{Hypotenuse}$

$\sin N = \frac{LM}{LN}$

Substitute values for LM and LN

$\sin N = \frac{8}{10}$

$\sin N = 0.80$

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3 years ago

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3 years ago

A rectangular garden is fenced on all sides with 248 feet of fencing. The garden is 6 feet longer than it is wide. Fine the leng

Shalnov [3]

The total length of the fencing is 248 ft; we call this the "perimeter" of the garden. Since P = 2W + 2L in general, here:

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Subtracting 12 from both sides of this equation: 236 ft = 4W. Then W = 59 ft, which means that L = 65 ft.

8 0

3 years ago