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3 years ago

Joan was asked to draw a right triangle how many right angles are in a right triangle

Mathematics

2 answers:

Trava [24]3 years ago

6 0

A right triangle only has one right angle

Temka [501]3 years ago

4 0

Full Answer.
If the triangle is an equilateral triangle, then all three angles are exactly 60 degrees each. A right-angled triangle has one angle of 90 degrees, which is the right angle, with two more angles that total 90 degrees when added together.

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How does the measure of the pink inscribed angle compare with the measure of the blue intercepted arc? How do you know?

Goryan [66]

Answer:

The pink angle is half the angle subtended by the blue arc.

Step-by-step explanation:

Given

See attachment for circle

Required

Compare the pink angle to the blue arc

Let the pink angle be x and the angle subtended by the blue arc be y.

So, we have:

$x \to$ angle at circumference

$y \to$ angle subtended by arc

The relationship between x and y is:

$y =2x$

Make x the subject

$x = \frac{1}{2}y$

7 0

3 years ago

Given the function s(x) = 7x - 17, find s(-8).

Scrat [10]

Answer:

-73

Step-by-step explanation:

substitue the -8 in for x. then multiply -8 by 7 you get -56. then subtract 17.

8 0

3 years ago

Mrs. Nguyen used 1.48 meters of netting to make 4 identical mini hockey goals. How much netting did she use per goal?

snow_tiger [21]

0.37 meters of netting

<h3>Further explanation</h3>

<u>Given:</u>

Mrs. Nguyen used 1.48 meters of netting to make 4 identical mini hockey goals.

<u>Question:</u>

How much netting did she use per goal?

<u>The Process:</u>

We will solve the problem of conversion or in other words the ratio or proportion.

Option A: the conversion

Mrs. Nguyen used 1.48 meters of netting to make 4 identical mini hockey goals.

Then for every 1 goal, she needs 0.37 meters of netting.

Because $\boxed{ \ 1 \ goal \times \frac{1.48 \ meters}{4 \ goals} = 0.37 \ meters \ of \ netting.}$

Option B: the ratio (or proportion)

In the ratio, we process as follows:

$\boxed{ \ 4 \ goals : 1.48 \ meters \ of \ netting \ }$

Both are divided by 4.

$\boxed{ \ 1 \ goals : 0.37 \ meters \ of \ netting \ }$

Thus, she uses 0.37 meters of netting per goal.

Alternative problem

How much netting did she use for 8 goals?

$\boxed{ \ 8 \ goal \times \frac{1.48 \ meters}{4 \ goals} = \ ? \ }$ meters of netting.

We cross out eight and four because they can be divided.

$\boxed{ \ 2 \times 1.48 \ meters = 2.96 \ meters \ of \ netting. \ }$

She uses 2.96 meters of netting to make 8 identical mini hockey goals.

<h3>Learn more</h3>

About the engine displacement of Kawasaki Ninja brainly.com/question/5009365
The clothing maker brainly.com/question/1594110
The unit conversion brainly.com/question/5416146

Keywords: Mrs. Nguyen, 1.48 meters of netting, to make 4 identical mini hockey goals, how much netting, she uses per goal, conversion, the ratio, proportion

8 0

3 years ago

A hot air balloon holds 1,592 cubic meters of helium. The density of helium is 0.1785 kilograms per cubic

yuradex [85]

Answer: 285.2 kg

Step-by-step explanation:

Given

The volume of helium in the balloon is $V=1592\ m^3$

The density of the helium is $\rho =0.1785\ kg/m^3$

Mass of the helium gas is given by the product of density and volume

$\Rightarrow M=\rho \cdot V\\\Rightarrow M=0.1785\times 1592\\\Rightarrow M=285.2\ kg$

Thus, the mass of gas is 285.2 kg

3 0

3 years ago

Use this list of Basic Taylor Series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the Taylor Series for f(x) = sin2(3x) based

notsponge [240]

Answer:

The Taylor series for $sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$ , the first three non-zero terms are $9x^{2} -27x^{4}+\frac{162}{5}x^{6}$ and the interval of convergence is $( -\infty, \infty )$

Step-by-step explanation:

<u>These are the steps to find the Taylor series for the function</u> $sin^2(3 x)$

Use the trigonometric identity:

$sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$

2. The Taylor series of $cos(x)$

$cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}$

Substituting y=6x we have:

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

3. Find the Taylor series for $sin^2(3x)$

$sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$ (1)

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$ (2)

Substituting (2) in (1) we have:

$\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

Bring the factor $\frac{1}{2}$ inside the sum

$\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )$

$\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

Extract the term for n=0 from the sum:

$\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

<u>To find the first three non-zero terms you need to replace n=3 into the sum</u>

$sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}$

<u>To find the interval on which the series converges you need to use the Ratio Test that says</u>

For the power series centered at x=a

$P(x)=C_{0}+C_{1}(x-a)+C_{2}(x-a)^{2}+...+ C_{n}(x-a)^{n}+...,$

suppose that $\lim_{n \to \infty} |\frac{C_{n}}{C_{n+1}}| = R.$ . Then

If $R=\infty,$ the the series converges for all x
If $0$ then the series converges for all $|x-a|$
If R=0, the the series converges only for x=a

So we need to evaluate this limit:

$\lim_{n \to \infty} |\frac{\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}}{\frac{-9^{n+1}2^{2*(n+1)-1}x^{2*(n+1)}}{(2*(2n+1))!}} |$

Simplifying we have:

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |$

Next we need to evaluate the limit

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |\\ \frac{1}{18x^{2} } \lim_{n \to \infty} |-(n+1)(2n+1)}|}$

-(n+1)(2n+1) is negative when n -> ∞. Therefore $|-(n+1)(2n+1)}|=2n^{2}+3n+1$

You can use this infinity property $\lim_{x \to \infty} (ax^{n}+...+bx+c) = \infty$ when a>0 and n is even. So

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } | \\ \frac{1}{18x^{2}} \lim_{n \to \infty} 2n^{2}+3n+1=\infty$

Because this limit is ∞ the radius of converge is ∞ and the interval of converge is $( -\infty, \infty )$ .

6 0

3 years ago