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3 years ago

Compare the triangles and determine whether they can be proven congruent by SSS, SAS, ASA, AAS, or HL. If not, type "NONE".

Mathematics

1 answer:

Komok [63]3 years ago

6 0

Answer:

SAS

Step-by-step explanation:

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A drawer contains 3 white balls and 6 black balls. Two balls are drawn out of the box at random. What is the probability that th

Rudik [331]

You have a 2/9 chance of it being white. It is how many of that color or number over the total, so you have a 2/9 chance of it being white.

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3 years ago

Ben uses a compass and a straightedge to bisect angle PQR, as shown below: The figure shows two rays QP and QR with a common end

maria [59]

Answer:

The correct option is;

∠AQS ≅ ∠BQS when AS = BS

Step-by-step explanation:

Given that AQ is equal to BQ. When AS is drawn congruent to BS, we have;

QS is congruent to SQ by reflective property

Therefore;

The three sides of triangle QAS are congruent to the three sides of triangle QBS, from which we have;

∠AQS and ∠BQS are corresponding angles, therefore;

∠AQS ≅∠BQS because corresponding angles of congruent triangles are also congruent.

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3 years ago

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I need help finding the degree of rotation for number 5 and 6

stiks02 [169]

Answer:

Step-by-step explanation:

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3 years ago

Which function can represent the height, y, of the railing in inches according to the horizontal distance in inches, x, from the

hodyreva [135]

The function representing the height is y = 36 - $\frac{3}{4} x$ .

<h3>What is function?</h3>

A formula, rule, or legislation that specifies how one variable (the independent variable) and another variable are related (the dependent variable). In mathematics, functions exist everywhere, and they are crucial for constructing physical links in the sciences.

<h3>What is fraction?</h3>

Any number of equal parts is represented by a fraction, which also represents a portion of a whole. A fraction, such as one-half, eight-fifths, or three-quarters, indicates how many components of a particular size there are when stated in ordinary English.

Given that,

Height of rails h is 36in.

The rate of decrement is 9 inches for every 12 horizontal inches,

We represent the rate (r) as a fraction.

r = $-\frac{9}{12}$ (negative, because it decreases),

r = $-\frac{3}{4}$ ,

So, the function is

Y = h + r × x

Where,

X is distance from top

So, y = 36 $-\frac{3}{4}x$

To know more about function, visit:

brainly.com/question/14418346

#SPJ4

4 0

1 year ago

Suppose a > 0 is constant and consider the parameteric surface sigma given by r(phi, theta) = a sin(phi) cos(theta)i + a sin(

Gnom [1K]

$\Sigma$ should have parameterization

$\vec r(\varphi,\theta)=a\sin\varphi\cos\theta\,\vec\imath+a\sin\varphi\sin\theta\,\vec\jmath+a\cos\varphi\,\vec k$

if it's supposed to capture the sphere of radius $a$ centered at the origin. ( $\sin\theta$ is missing from the second component)

a. You should substitute $x=a\sin\varphi\cos\theta$ (missing $\cos\theta$ this time...). Then

$x^2+y^2+z^2=(a\sin\varphi\cos\theta)^2+(a\sin\varphi\sin\theta)^2+(a\cos\varphi)^2$

$x^2+y^2+z^2=a^2\left(\sin^2\varphi\cos^2\theta+\sin^2\varphi\sin^2\theta+\cos^2\varphi\right)$

$x^2+y^2+z^2=a^2\left(\sin^2\varphi\left(\cos^2\theta+\sin^2\theta\right)+\cos^2\varphi\right)$

$x^2+y^2+z^2=a^2\left(\sin^2\varphi+\cos^2\varphi\right)$

$x^2+y^2+z^2=a^2$

as required.

b. We have

$\vec r_\varphi=a\cos\varphi\cos\theta\,\vec\imath+a\cos\varphi\sin\theta\,\vec\jmath-a\sin\varphi\,\vec k$

$\vec r_\theta=-a\sin\varphi\sin\theta\,\vec\imath+a\sin\varphi\cos\theta\,\vec\jmath$

$\vec r_\varphi\times\vec r_\theta=a^2\sin^2\varphi\cos\theta\,\vec\imath+a^2\sin^2\varphi\sin\theta\,\vec\jmath+a^2\cos\varphi\sin\varphi\,\vec k$

$\|\vec r_\varphi\times\vec r_\theta\|=a^2\sin\varphi$

c. The surface area of $\Sigma$ is

$\displaystyle\iint_\Sigma\mathrm dS=a^2\int_0^\pi\int_0^{2\pi}\sin\varphi\,\mathrm d\theta\,\mathrm d\varphi$

You don't need a substitution to compute this. The integration limits are constant, so you can separate the variables to get two integrals. You'd end up with

$\displaystyle\iint_\Sigma\mathrm dS=4\pi a^2$

# # #

Looks like there's an altogether different question being asked now. Parameterize $\Sigma$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k$

with $\sqrt2\le u\le\sqrt6$ and $0\le v\le2\pi$ . Then

$\|\vec s_u\times\vec s_v\|=u\sqrt{1+4u^2}$

The surface area of $\Sigma$ is

$\displaystyle\iint_\Sigma\mathrm dS=\int_0^{2\pi}\int_{\sqrt2}^{\sqrt6}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$

The integrand doesn't depend on $v$ , so integration with respect to $v$ contributes a factor of $2\pi$ . Substitute $w=1+4u^2$ to get $\mathrm dw=8u\,\mathrm du$ . Then

$\displaystyle\iint_\Sigma\mathrm dS=\frac\pi4\int_9^{25}\sqrt w\,\mathrm dw=\frac{49\pi}3$

# # #

Looks like yet another different question. No figure was included in your post, so I'll assume the normal vector points outward from the surface, away from the origin.

Parameterize $\Sigma$ by

$\vec t(u,v)=u\,\vec\imath+u^2\,\vec\jmath+v\,\vec k$

with $-1\le u\le1$ and $0\le v\le 2$ . Take the normal vector to $\Sigma$ to be

$\vec t_u\times\vec t_v=2u\,\vec\imath-\vec\jmath$

Then the flux of $\vec F$ across $\Sigma$ is

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=\int_0^2\int_{-1}^1(u^2\,\vec\jmath-uv\,\vec k)\cdot(2u\,\vec\imath-\vec\jmath)\,\mathrm du\,\mathrm dv$

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=-\int_0^2\int_{-1}^1u^2\,\mathrm du\,\mathrm dv$

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=-2\int_{-1}^1u^2\,\mathrm du=-\frac43$

If instead the direction is toward the origin, the flux would be positive.

8 0

4 years ago