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2 years ago

ABCD is an isosceles trapezoid.

Mathematics

2 answers:

timurjin [86]2 years ago

7 0

Answer:

127

Step-by-step explanation:

Eduardwww [97]2 years ago

6 0

Answer:

∡b = 53°

∡d = 127°

∡c = 127°

Step-by-step explanation:

in trapezoids, the two top angles are equal and the two bottom are equal

also, there are 360 degrees in any 4-sided polygon

53 + 53 + x + x = 360

2x + 106 = 360

2x = 254

x = 127

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Leokris [45]

The answer to your problem is C

7 0

2 years ago

What is 39 in base two

ser-zykov [4K]

Answer: 15

Step-by-step explanation:

39 in base two

3 x 2^1 + 9 x 2^0

3 x 2 + 9 x 1

6 + 9

Note : any number raised to power zero is given as 1

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2 years ago

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BRAINLIEST

krok68 [10]

Answer:

ΔABC ≅ ΔEDF by SSS

Step-by-step explanation:

Its not a right triangle so it cant be the 2nd or last one.

ABC is equal to EDF, if you trace where the lines go

it is not equal to fde

7 0

2 years ago

Read 2 more answers

Compute the matrix of partial derivatives of the following functions.

s344n2d4d5 [400]

For a vector-valued function

$\mathbf f(\mathbf x)=\mathbf f(x_1,x_2,\ldots,x_n)=(f_1(x_1,x_2,\ldots,x_n),\ldots,f_m(x_1,x_2,\ldots,x_n))$

the matrix of partial derivatives (a.k.a. the Jacobian) is the $m\times n$ matrix in which the $(i,j)$ -th entry is the derivative of $f_i$ with respect to $x_j$ :

$D\mathbf f(\mathbf x)=\begin{bmatrix}\dfrac{\partial f_1}{\partial x_1}&\dfrac{\partial f_1}{\partial x_2}&\cdots&\dfrac{\partial f_1}{\partial x_n}\\\dfrac{\partial f_2}{\partial x_1}&\dfrac{\partial f_2}{\partial x_2}&\cdots&\dfrac{\partial f_2}{\partial x_n}\\\vdots&\vdots&\ddots&\vdots\\\dfrac{\partial f_m}{\partial x_1}&\dfrac{\partial f_m}{\partial x_2}&\cdots&\dfrac{\partial f_n}{\partial x_n}\end{bmatrix}$

So we have

(a)

$D f(x,y)=\begin{bmatrix}\dfrac{\partial(e^x)}{\partial x}&\dfrac{\partial(e^x)}{\partial y}\\\dfrac{\partial(\sin(xy))}{\partial x}&\dfrac{\partial(\sin(xy))}{\partial y}\end{bmatrix}=\begin{bmatrix}e^x&0\\y\cos(xy)&x\cos(xy)\end{bmatrix}$

(b)

$D f(x,y,z)=\begin{bmatrix}\dfrac{\partial(x-y)}{\partial x}&\dfrac{\partial(x-y)}{\partial y}&\dfrac{\partial(x-y)}{\partial z}\\\dfrac{\partial(y+z)}{\partial x}&\dfrac{\partial(y+z)}{\partial y}&\dfrac{\partial(y+z)}{\partial z}\end{bmatrix}=\begin{bmatrix}1&-1&0\\0&1&1\end{bmatrix}$

(c)

$Df(x,y)=\begin{bmatrix}y&x\\1&-1\\y&x\end{bmatrix}$

(d)

$Df(x,y,z)=\begin{bmatrix}1&0&1\\0&1&0\\1&-1&0\end{bmatrix}$

5 0

2 years ago

1) A group of 100 students was surveyed about their interest in a new international studies program. interest was measured in te

liubo4ka [24]

Answer:

1) 0.30

2) 0.308

Step-by-step explanation:

Interest Frequency

High 30

Medium 40

Low 30

Total frequency=30+40+30=100.

The relative frequency of student with high interest can be computed by dividing the frequency of high interest to the total frequency.

R.f of high interest=30/100=0.30

Monthly Commission Frequencies

600-800 3

800-1000 7

1000-1200 11

1200-1400 22

1400-1600 40

1600-1800 24

1800-2000 9

2000-2200 4

Total frequency=3+7+11+22+40+24+9+4=120

The relative frequency of earning 1600 or more computed by dividing the frequency of earning 1600 or more to the total frequency.

R.f of earning more than 1600=24+9+4/120=37/120=0.308

7 0

2 years ago