A number with the same value but written differently
1 answer:
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Answer:
UV=29
Step-by-step explanation:
In right triangles AQB and AVB,
∠AQB = ∠AVB ...(i) {Right angles}
∠QBA = ∠VBA ...(ii) {Given that they are equal}
We know that sum of all three angles in a triangle is equal to 180 degree. So wee can write sum equation for each triangle
∠AQB+∠QBA+∠BAQ=180 ...(iii)
∠AVB+∠VBA+∠BAV=180 ...(iv)
using (iii) and (iv)
∠AQB+∠QBA+∠BAQ=∠AVB+∠VBA+∠BAV
∠AVB+∠VBA+∠BAQ=∠AVB+∠VBA+∠BAV (using (i) and (ii))
∠BAQ=∠BAV...(v)
Now consider triangles AQB and AVB;
∠BAQ=∠BAV {from (v)}
∠QBA = ∠VBA {from (ii)}
AB=AB {common side}
So using ASA, triangles AQB and AVB are congruent.
We know that corresponding sides of congruent triangles are equal.
Hence
AQ=AV
5x+9=7x+1
9-1=7x-5x
8=2x
divide both sides by 2
4=x
Now plug value of x=4 into UV=7x+1
UV=7*4+1=28+1=29
<u>Hence UV=29 is final answer.</u>
Answer:
- 0
Step-by-step explanation:
<u>For each odd i the term is:</u>
- (-1)
(5!*5/4!)
= -
<u>For each even i the term is:</u>
- (-1)
So the sum of the first 100 terms is zero
ANSWER
B.Yes, f is continuous on [1, 7] and differentiable on (1, 7).
EXPLANATION
The given
The hypotheses are
1. The function is continuous on [1, 7].
2. The function is differentiable on (1, 7).
3. There is a c, such that:
This implies that;
Since the function is continuous on [1, 7] and differentiable on (1, 7) it satisfies the mean value theorem.
B.Yes, f is continuous on [1, 7] and differentiable on (1, 7).
EXPLANATION
The given
The hypotheses are
1. The function is continuous on [1, 7].
2. The function is differentiable on (1, 7).
3. There is a c, such that:
This implies that;
Since the function is continuous on [1, 7] and differentiable on (1, 7) it satisfies the mean value theorem.