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3 years ago

Which equation is a linear function? Please answer within 5 minutes

Mathematics

2 answers:

Fynjy0 [20]3 years ago

8 0

Answer:

y= x/2-5

Step-by-step explanation:

notka56 [123]3 years ago

6 0

Answer:

y=x²+7

Step-by-step explanation:

I'm pretty sure this is the answer

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Dallas was told to run around the perimeter of the high

Zina [86]

Answer:

Option (4)

Step-by-step explanation:

Perimeter of a polygon is the sum of measures of all sides of the polygon.

From the picture attached,

Distance run by Dallas = Perimeter of the ground

= 48 + 50 + 9 + 30 + 30 + 30 + 9 + 50

= 256 yards

Total warm up run by Dallas = 256 yards

Therefore, Option (4) will be the correct option.

7 0

3 years ago

Read 2 more answers

Pls help it is easy math

Serjik [45]

Answer:

all I know is 40,000

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Andria wrote the following statements:

mezya [45]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

The geometry term does each statement above represents is Statement 1 is postulate because you have to prove that corresponding angles are congruent.

6 0

4 years ago

Use the Chain Rule to find the indicated partial derivatives. u = x2 + yz, x = pr cos(θ), y = pr sin(θ), z = p + r; (partial u)/

devlian [24]

$u(x,y,z)=x^2+yz$

$\begin{cases}x(p,r,\theta)=pr\cos\theta\\y(p,r,\theta)=pr\sin\theta\\z(p,r,\theta)=p+r\end{cases}$

At the point $(p,r,\theta)=(2,2,0)$ , we have

$\begin{cases}x(2,2,0)=4\\y(2,2,0)=0\\z(2,2,0)=4\end{cases}$

Denote by $f_x:=\dfrac{\partial f}{\partial x}$ the partial derivative of a function $f$ with respect to the variable $x$ . We have

$\begin{cases}u_x=2x\\u_y=z\\u_z=y\end{cases}$

The Jacobian is

$\begin{bmatrix}x_p&x_r&x_\theta\\y_p&y_r&y_\theta\\z_p&z_r&z_\theta\end{bmatrix}=\begin{bmatrix}r\cos\theta&p\cos\theta&-pr\sin\theta\\r\sin\theta&p\sin\theta&pr\cos\theta\\1&1&0\end{bmatrix}$

By the chain rule,

$u_p=u_xx_p+u_yy_p+u_zz_p=2xr\cos\theta+zr\sin\theta+y$

$u_p(2,2,0)=2\cdot4\cdot2\cos0+4\cdot2\sin0+0\implies\boxed{u_p(2,2,0)=16}$

$u_r=u_xx_r+u_yy_r+u_zz_r=2xp\cos\theta+zp\sin\theta+y$

$u_r(2,2,0)=2\cdot4\cdot2\cos0+4\cdot2\sin0+0\implies\boxed{u_r(2,2,0)=16}$

$u_\theta=u_xx_\theta+u_yy_\theta+u_zz_\theta=-2xpr\sin\theta+zpr\cos\theta$

$u_\theta(2,2,0)=-2\cdot4\cdot2\cdot2\sin0+4\cdot2\cdot2\cos0\implies\boxed{u_\theta(2,2,0)=16}$

7 0

3 years ago

HELP PLEASE! BRAINLIEST

V125BC [204]

Since AED forms a straight line, all involved angles sum to 180°. Therefore AEB + BED = 180. Also, since EC bisects BED, BEC = CED, and BED = 2× CED. Now to substitute the first equation:
AEB + BED = 180
AEB + 2×CED = 180
11x-12 + 2(4x+1) = 180
11x-12+8x+2 = 180
19x-10 = 180
19x-10+10 = 180+10
19x = 190
x = 10
So what is m<AEC?? It is the sum of AEB + BEC, and since BEC = CED we can say that:
AEC = AEB + CED
AEC = 11x-12 + 4x+1 = 15x-11 = 15(10)-11 = 150-11
m<AEC = 139°

3 0

4 years ago