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3 years ago

Help please I might have a quiz soon!

Mathematics

1 answer:

Thepotemich [5.8K]3 years ago

5 0

Answer:

We are given that the two triangles are approximately equal or similar thus

angle M=angle Z = 124

angle L = angle X = 33

angle N= angle Y = ?

The total angles is 180, thus angle N or Y is 180- (124+33) = 23

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How do you do this question?

mihalych1998 [28]

Answer:

C) π/6

Step-by-step explanation:

The area under the curve from x=-π/2 to x=k is 3 times the area under the curve from x=k to x=π/2.

$\int\limits^k_{-\frac{\pi }{2}} {cos\ x} \, dx = 3\int\limits^{\frac{\pi }{2}}_k {cos\ x} \, dx\\sin\ x\ |^{k}_{-\frac{\pi }{2}} = 3\ sin\ x\ |^{\frac{\pi }{2}}_k\\(sin\ k - (-1)) = 3\ (1 - sin\ k)\\sin\ k + 1 = 3 - 3\ sin\ k\\4\ sin\ k = 2\\sin\ k = \frac{1}{2}\\k = \frac{\pi}{6}$

Graph: desmos.com/calculator/mezlen9hb4

3 0

3 years ago

9 ten thousands divided by 10

MA_775_DIABLO [31]

The answer is9000 thats the answer

5 0

2 years ago

Read 2 more answers

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago

The moon is a sphere with radius of 959 km. Determine an equation for the ellipse if the distance of the satellite from the surf

Sergeeva-Olga [200]

Answer:

$\frac{x^2}{1316^2}+\frac{y^2}{1669^2}=1$

Step-by-step explanation:

An ellipse is the locus of a point such that its distances from two fixed points, called foci, have a sum that is equal to a positive constant.

The equation of an ellipse with a center at the origin and the x axis as the minor axis is given by:

$\frac{x^2}{b^2}+\frac{y^2}{a^2} =1 \\\\where\ a>b$

Since the distance of the satellite from the surface of the moon varies from 357 km to 710 km, hence:

b = 357 km + 959 km = 1316 km

a = 710 km + 959 km = 1669 km

Therefore the equation of the ellipse is:

$\frac{x^2}{1316^2}+\frac{y^2}{1669^2}=1$

5 0

3 years ago

Suppose the graph of a cubic polynomial function has the same zeroes and passes through the coordinate (0, –5).

Alexeev081 [22]

Zeros are the x values which make the function equal to zero. Set it up as you would for a binomial with a constant multiplier "k" to account for the y-intercept (0, -5) given.

f(x) = k(x-2)(x-3)(x-5)

Use the y-intercept (0,-5) to solve for k.

-5 = k(0-2)(0-3)(0-5)
-5 = -30k
-5/-30 = k
1/6 = k

The cubic polynomial function is then ..

f(x) = (1/6)(x-2)(x-3)(x-5)

Linear factors are the linear (line) expressions you can factor out of the polynomial. They are (x-2), (x-3) and (x-5).

5 0

2 years ago

Read 2 more answers