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Eduardwww [97]
3 years ago
6

I don’t know how to do this:(

Mathematics
1 answer:
Wewaii [24]3 years ago
4 0

Answer:

me niether

Step-by-step explanation:

lol

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"In the 45°-45°-90° triangle above, r = 10. Find the other two lengths.
Basile [38]

We are given a right triangle that has angles of 45°-45°-90°. This would indicate that the triangle is an isosceles type. We can use some trigonemetric functions to solve for the other legs. We do as follows:

 

sin 45 = p / 10

p = 5√2

 

cos 45 = q /10

q = 5√2

 

<span>Hope this answers the question. Have a nice day.</span>

5 0
3 years ago
What is the simple way of solving quadratic equation ​
aleksandrvk [35]

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factorization

completing square

quadratic formula

but factorization is the simplest

4 0
3 years ago
Work out the area of this quarter circle of radius 8 cm.<br> Give your answer in terms of pie
Bess [88]

16pi

Step-by-step explanation:

pi x r² = ans ÷ 4 = Answer for area

Pi×8²=64pi ÷ 4 = 16pi

3 0
3 years ago
Prove 2√(x) + 1/√(x + 1) &lt;= 2√(x+1) for all x in [0,inf)
EleoNora [17]
Start by multiplying each side of the inequality by \sqrt{x + 1} and simplifying:

2 \sqrt x + \frac{1}{\sqrt{x+1}}  \leq  2 \sqrt{x + 1}
(2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})
2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)
2 \sqrt{x^2 + x} + 1 \leq 2x + 2
2 \sqrt{x^2 + x} \leq 2x + 1
\sqrt{x^2 + x} \leq x + \frac{1}{2}

From here, we can square both sides to get

x^2 + x \leq (x + \frac{1}{2})^2
x^2 + x \leq x^2 + x + \frac{1}{4}
0  \leq \frac{1}{4}, which is always true.
3 0
3 years ago
Jack earns $50 for washing cars on the weekend. He makes an average of $4.25in tips per hour. Write the function of jack's earni
jarptica [38.1K]
F(x) = 4.25x + 50

for 6 hrs....x = 6
f(6) = 4.25(6) + 50
f(6) = 25.50 + 50
f(6) = 75.50 <==

4 0
3 years ago
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