The most common size of a sedan tank is 18.5 gallons. So if it is 75% full, it contains 13.875 gallons of gas. If it recorded a mileage of 15,121 miles, mileage per gallon, therefore, is 1090 rounded off to the nearest whole number.
Given: Gas full tank of a sedan – 18.5 gallons
<span> 75% full = 13.875 gallons</span>
<span> Mileage using 13.875 gallons of fuel = 15,121 miles</span>
The mileage per gallon computation using the ratio equation:
Ratio is 15121:13.875 or
<span> X/1= 15121/13.875</span>
<span> 13.875x = 15121</span>
<span> X = 15121/13.875</span>
<span> X =1089.80 or 1090 miles per gallon (rounded off)</span>
By definition we have to:
Acute triangle - It is a triangle that has all acute angles. An acute angle is one whose degree of measurement is less than 90.
Therefore, Charlene's definition is correct.
Charlene said:
triangle is a triangle whose three interior angles have measures less than 90 °
Answer:
C.Yes, because this definition fits all acute triangles and does not fit triangles that are not acute.
Answer:
Maria owes more than 30
Step-by-step explanation:
Maria's Credit card is less than $-30, which would mean Maria owes more than $30
Answer:
D
Step-by-step explanation:
Using the law of sines in ΔABC, then
= ![\frac{a}{sinA}](https://tex.z-dn.net/?f=%5Cfrac%7Ba%7D%7BsinA%7D)
Substituting in values gives
=
( cross- multiply )
c × sin38° = 25 × sin31.6° ( divide both sides by sin38° )
c =
≈ 21.28 → D
Answer:
Height of stone face is : <em>56.7 ft</em>
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Step-by-step explanation:
Kindly refer to the attached image for the diagram of the given conditions and values.
Let C be the base of mountain.
D be the point from where two sightings are taken.
AB be the stone face.
Angle of elevations:
![\angle BDC =35^\circ\\\angle ADC =38^\circ](https://tex.z-dn.net/?f=%5Cangle%20BDC%20%3D35%5E%5Ccirc%5C%5C%5Cangle%20ADC%20%3D38%5E%5Ccirc)
To find:
Height of stone face = ?
AB = ?
Solution:
We can use trigonometric function of tangent here in two triangles
:
![In\ \triangle BCD :](https://tex.z-dn.net/?f=In%5C%20%5Ctriangle%20BCD%20%3A)
![tan(\angle BDC) = \dfrac{Perpendicular}{Base} = \dfrac{BC}{CD}\\\Rightarrow BC = 700 \times tan35 ..... (1)](https://tex.z-dn.net/?f=tan%28%5Cangle%20BDC%29%20%3D%20%5Cdfrac%7BPerpendicular%7D%7BBase%7D%20%3D%20%5Cdfrac%7BBC%7D%7BCD%7D%5C%5C%5CRightarrow%20BC%20%3D%20700%20%5Ctimes%20tan35%20.....%20%281%29)
![In\ \triangle ACD :](https://tex.z-dn.net/?f=In%5C%20%5Ctriangle%20ACD%20%3A)
![tan(\angle ADC) = \dfrac{Perpendicular}{Base} = \dfrac{AC}{CD}\\\Rightarrow AC = 700 \times tan38\\\Rightarrow AB +BC = 700 \times tan38\\\\\text{Using equation (1):}\\\Rightarrow AB + 700 \times tan 35 = 700 \times tan 38\\\Rightarrow AB = 700 \times tan 38-700 \times tan35\\\Rightarrow AB = 700 \times (tan 38-tan35)\\\Rightarrow AB = 700 \times 0.081\\\Rightarrow AB = \bold{56.7}\ ft](https://tex.z-dn.net/?f=tan%28%5Cangle%20ADC%29%20%3D%20%5Cdfrac%7BPerpendicular%7D%7BBase%7D%20%3D%20%5Cdfrac%7BAC%7D%7BCD%7D%5C%5C%5CRightarrow%20AC%20%3D%20700%20%5Ctimes%20tan38%5C%5C%5CRightarrow%20AB%20%2BBC%20%3D%20700%20%5Ctimes%20tan38%5C%5C%5C%5C%5Ctext%7BUsing%20equation%20%281%29%3A%7D%5C%5C%5CRightarrow%20AB%20%2B%20700%20%5Ctimes%20tan%2035%20%3D%20700%20%5Ctimes%20tan%2038%5C%5C%5CRightarrow%20AB%20%20%3D%20700%20%5Ctimes%20tan%2038-700%20%5Ctimes%20tan35%5C%5C%5CRightarrow%20AB%20%20%3D%20700%20%5Ctimes%20%28tan%2038-tan35%29%5C%5C%5CRightarrow%20AB%20%20%3D%20700%20%5Ctimes%200.081%5C%5C%5CRightarrow%20AB%20%20%3D%20%5Cbold%7B56.7%7D%5C%20ft)
So, Height of stone face is : <em>56.7 ft</em>