A tree casts a shadow of 18 feet. A horse standing 9 feet tall casts a 6-foot shadow. Determine

Trigonometry

schepotkina [342]

The area of the triangle ABC is 207.5 square units.

Explanation:

The measurements of the sides of the triangle are $AB=19$ , $AC=24$ and $m\angle A=65^{\circ}$

We need to determine the area of the triangle ABC.

<u>Area of the triangle:</u>

The area of the triangle can be determined using the formula,

${Area}=\frac{1}{2} b c \sin A$

where $b=19$ , $c=24$ and $m\angle A=65^{\circ}$

Substituting these values in the above formula, we get,

${Area}=\frac{1}{2}(19)(24) \sin 65^{\circ}$

Simplifying the values, we get,

${Area}=\frac{1}{2}(456) (0.91)$

${Area}=\frac{1}{2}(414.96)$

${Area}=207.48$

Rounding off to the nearest tenth, we get,

${Area}=207.5$

Thus, the area of the triangle ABC is 207.5 square units.

5 0

3 years ago

Read 2 more answers

I will give brainliest to anyone who gives me a correct answer

Pachacha [2.7K]

Answer:

C

Step-by-step explanation:

Let's check the first condition [y=110, t=0] for all the choices and eliminate:

A)

$y = 110e^{7t} - 80\\y = 110e^{0} - 80\\y=30$

NO

B)

$y = 110 - 80e^{-5t}\\y = 110 - 80e^{0}\\y=30$

NO

C)

$y = 30 + 80e^{-2t}\\y = 30 + 80e^{0}\\y=110$

YES

D)

$y = 110 - 80e^{2t}\\y = 110 - 80e^{0}\\y=30$

NO

E)

$y = 30 - 80e^{-4t}\\y = 30 - 80e^{0}\\y=-50$

NO

Note: e^0 = 1

Only C is satisfied. But let's make sure the 2nd condition applies as well.

$y = 30 + 80e^{-2t}\\y = 30 + 80e^{-\infty}\\y=30+\frac{80}{e^{\infty}}\\y>>30$

Hence, y approaches 30 as t goes to infinity. 2nd condition is satisfied as well.

Answer is C

4 0

4 years ago

Can someone please answer this ?

hjlf

We need to know the density of Marble City, hence we know the density of Pebble City.
MC density = population/area
MC density = 52,348/125
MC density = 418.784 people/miles

From the question, we know that the two cities have the same population density.
PC density = MC density
PC density = 418.784 people/miles

Use the density formula to figure out the dimension of PC
PC density = population/area
418.784 = 85,312/(l × w)
(l × w) = 85,312/418.784
11.8 w = 85,312/418.784
11.8 w = 203.7
w = 203.7/11.8
w = 17.2

The closest value of x is 17 miles

6 0

4 years ago

Hey can anyone help me with this question please and thank you !

Inessa [10]

Answer:

(6x-12)+(8x-4)=180

14x-16=180

14x=196

x=14

Step-by-step explanation:

3 0

3 years ago

Consider a paint-drying situation in which drying time for a test specimen is normally distributed with s = 6. The hypotheses H0

adell [148]

Answer:

There is no statistical evidence to support the claim that the mean is elss than 74.

Step-by-step explanation:

Given that a paint-drying situation in which drying time for a test specimen is normally distributed with s = 6.

The hypotheses H0: µ = 74 and Ha: µ < 74

are to be tested using a random sample of n = 25 observations

a) Z = $\frac{72.3-74}{\frac{6}{\sqrt{25} } } \\=-1.417$

ie. -1.417 std deviations to the left of mean.

b) p value = 0.078241

Since p > 0.01 , our alpha, accept H0

There is no statistical evidence to support the claim that the mean is elss than 74.

6 0

4 years ago