The distance from the center to where the foci are located exists 8 units.
<h3>How to determine the distance from the center?</h3>
The formula associated with the focus of an ellipse exists given as;
c² = a² − b²
Where c exists the distance from the focus to the center.
a exists the distance from the center to a vertex,
the major axis exists 10 units.
b exists the distance from the center to a co-vertex, the minor axis exists 6 units
c² = a² − b²
c² = 10² - 6²
c² = 100 - 36
c² = 64
c = 8
Therefore, the distance from the center to where the foci are located exists 8 units.
To learn more about the Pythagorean theorem here:
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Hello there!
x² + x = 7/4
x² + - 7/4 = 0
Now we gonna use the quadratic formula to find x
a= 1
b=1
c = -1.75
x = -b+/-√b² -4ac all of them divide by 2a
x = -(1)+/-√(1)² - (4)(1)(-1.75) all of them divide by 2(1)
x= -1+/-√8 all of them divide by 2
x = -1/2 + √2 or x = -1/2 - √2
The correct option is option C
I hope that helps!
A is the answer and i am in 6th grade
1. Angles ADC and CDB are supplementary, thus
m∠ADC+m∠CDB=180°.
Since m∠ADC=115°, you have that m∠CDB=180°-115°=65°.
2. Triangle BCD is isosceles triangle, because it has two congruent sides CB and CD. The base of this triangle is segment BD. Angles that are adjacent to the base of isosceles triangle are congruent, then
m∠CDB=m∠CBD=65°.
The sum of the measures of interior angles of triangle is 180°, therefore,
m∠CDB+m∠CBD+m∠BCD=180° and
m∠BCD=180°-65°-65°=50°.
3. Triangle ABC is isosceles, with base BC. Then
m∠ABC=m∠ACB.
From the previous you have that m∠ABC=65° (angle ABC is exactly angle CBD). So
m∠ACB=65°.
4. Angles BCD and DCA together form angle ACB. This gives you
m∠ACB=m∠ACD+m∠BCD,
m∠ACD=65°-50°=15°.
Answer: 15°.
