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4 years ago

What is the length of the side opposite the 45 degree angle?

2 answers:

7 0

Pythagorean theorem.
a^2 + b^2 = c^2 a^2 = b^2 because this is an isosceles right triangle. c = 10 -- given a^2 + a^2 = 100 2a^2 = 100 a^2 = 50 a = sqrt (50) = 5 sqrt (2) ~ 7.07

yulyashka [42]4 years ago

6 0

Sin 45° = opposite side length/hypotenuse length

Cross Multiply and you get
opposite side length = hypotenuse length x Sin 45°

Now all You need to know is the hypotenuse length. It will help you find the opposite side length.

HOPE THIS WILL HELP.

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Need some help here.

Brut [27]

Answer:

Step-by-step explanation:

you have the right answer.

6 0

3 years ago

Read 2 more answers

Simplify the expression in part C, and explain what the number means with regard to the problem.

alexdok [17]

Answer: |55.75 – (-15.8)| = |55.75 + 15.8| = |71.55| = 71.55

Step-by-step explanation: The distance between two positions with longitudes A and B is given by ||A| - |B|| if the two positions are at the same side of the meridian (0 degrees longitude) and |A| + |B| if both positions are at different sides of the meridian.

4 0

3 years ago

The product of (3 + 2i) and a complex number is (17 + 7i)

Tatiana [17]

<span> Of course you're looking for the ratio: (17+7i)/(3+2i). To get a clear number, you need to get a real number at the denominator -- and it so happens that the product of two complex conjugates is a real number. So, you multiply the denominator and the numerator by the conjugate of 3+2i. The denominator becomes 13 and the numerator becomes 51 + 7i. Hence, the multiplier is 51/13 + 7/13 i. </span>

<span>As a side note, while you probably knew how to factor a difference of squares, I just showed how to factor a sum of squares: a^2 + b^2 = (a + bi)(a - bi). That holds because i^2 = -1.

Hope this's useful/helpful !!!^_~!!!</span>

8 0

3 years ago

ohn wants to buy a new bike that costs $337 and he already has saved $103. Estimate to find about how much more John needs to sa

Sholpan [36]

Your answer is C because:

337 rounded is 300
103 rounded is 100
300-100=200

200 is your answer

8 0

3 years ago

Read 2 more answers

A one-parameter family of solutions of the DE P' = P( 1 - P) is given below. P = c1et/1 + c1et Does any solution curve pass thro

topjm [15]

Answer:

a. The curve $P(t) = -\frac{6e^t}{5-6e^t}$ passes through the point (0, 6)

b. No solution of the curve $P(t)$ passes through the point (0, 1)

Step-by-step explanation:

Consider the family of the solution of DE $P' = P(1 - P)$ is $P = \frac{c_1e^t}{1 + c_1e^t}$

a. If any solution passes through the point (0, 6), then there is $c_1$ such that the point (0, 6) satisfies the solution $P = \frac{c_1e^t}{1 + c_1e^t}$

Substitute $t = 0, P = 6$ in $P = \frac{c_1e^t}{1 + c_1e^t}$ and then solve the equation to obtain $c_1$

$P(t) = \frac{c_1e^t}{1 + c_1e^t}\\P(0) = \frac{c_1e^0}{1+c_1e^0}\\ 6 = \frac{c_1}{1 + c_1}\\ c_1 = -\frac{6}{5}$

Therefore, the curve $P(t) = -\frac{6e^t}{5 - 6e^t}$ passes through the point (0, 6)

b. If any solution passes through the point(0, 1), then there is $c_1$ such that the point (0, 1) satisfies the solution $P = \frac{c_1e^t}{1+c_1e^t}$

$P(t) = \frac{c_1e^t}{1 + c_1e^t}\\ P(0) = \frac{c_1e^0}{1 + c_1e^0}\\ 1 = \frac{c_1}{1+c_1} \\1 + c_1 = c_1$

this is not possible

Hence, there is no curve $P(t)$ that exists which passes through the point (0, 1)

4 0

3 years ago