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yarga [219]
3 years ago
14

One angle in an obtuse isosceles triangle measures 40°. What are the measures of the other two angles?

Mathematics
2 answers:
Ronch [10]3 years ago
7 0

Answer:

C.40° and 100°

Step-by-step explanation:

Damm [24]3 years ago
5 0
An obtuse triangle has one obtuse angle, or an angle that measures more than 90° but less than 180°. Options 1 and 2 are not correct, because these are acute angles.
So, one of the angles must be 40°, and the other may be 100° or 120°.
The measures of the angles in a triangle add up to 180°.
40^\circ + 40^\circ +x = 180^\circ \\
80^\circ + x=180^\circ \\
x=180^\circ - 80^\circ \\
x=100^\circ
The other two angles are 40° and 100°.
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6 0
3 years ago
what do i write for this " Write a problem that requires adding 1 to the quotient when interpreting the remainder."
Mariana [72]
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3 years ago
The length of one leg of a right triangle is 3 feet more than the other leg. If the hypotenuse is 15 feet find the length of eac
melamori03 [73]

Answer:

Leg1 = 9 and leg2 = 12

Step-by-step explanation:

<u>Points to remember</u>

<u>Pythagorean theorem</u>

Hypotenuse² = Base² + Height²

<u>To find the length of each leg</u>

It is given that, hypotenuse = 15 feet

Let base = x the height = x +3

By using Pythagorean theorem we can write,

Hypotenuse² = Base² + Height²

15²  = x² + (x + 3)²

225 = x²  + x²  + 6x + 9

2x²  + 6x + 216 = 0

x²  + 3x + 108 = 0

By solving we get x = 9 and x -12

Therefore one leg = 9 and other leg = 9 + 3 = 12

5 0
3 years ago
Read 2 more answers
F(t)=t^2+19t+60<br> What are the zeros of the function and what is the vertex of the parabola?
SVEN [57.7K]

Answer: -4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)

Step-by-step explanation:

Given

Function is F(t)=t^2+19+60

Zeroes of the function are

t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15

Using completing the square method

y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)

4 0
3 years ago
Simplify this please
aliina [53]

Answer:

2a +5

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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