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sasho [114]
3 years ago
12

What is the relationship between ∠a and∠b? NO LINKS ! if you give me a link you're getting reported. Also the last option says n

one of the above.

Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
3 0
The answer would be B. As you can see Angle A and angle B are connected by that line and together they would equal 90 degrees like on the other side of line AB.
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1.What is the y-intercept of the plane whose equation is 4x + 5y – z = 20?
Katena32 [7]

In case of plane, for y intercept, x and z are o .

So we will get

4(0)+5y-0=20

5y=20

Dividing both sides by 5

y = 4

So the y intercept is (0,4,0).

In second question, x intercept is 1, y intercept is -1 and z intercept is 2 .

So the correct option is the third equation , that is x-y+2z=4 .

8 0
3 years ago
Read 2 more answers
the runner's time to complete the marathon is 45 minutes. The time to complete the last marathon was 49.5 minutes. What is the p
Lana71 [14]

Answer:

9.09%

Step-by-step explanation:

To find percent decrease you have to use this equation:

Percent Decrease = \frac{Initial -New}{Initial} x 100

Initial = 49.5

New = 45

Plug in and solve:

PD = \frac{49.5-45}{49.5} x 100\\\\PD = \frac{4.5}{49.5} x 100 \\\\PD = 0.0909 x 100\\\\PD = 9.09

The percent decrease is 9.09%

<em>Hope this helps!!</em>

<em>- Kay :)</em>

6 0
3 years ago
Solve the for the inequality ​
Naddika [18.5K]

Answer:

k ≥ -18

Step-by-step explanation:

-8\leq \frac{2}{5}(k-2)\\\\-8(5)\leq (5)\frac{2}{5}(k-2)\\\\-40\leq 2(k-2)\\\\\frac{-40}{2}\leq \frac{2(k-2)}{2}\\\\-20\leq k-2\\\\-20+2\leq k-2+2\\\\-18\leq k

5 0
3 years ago
Read 2 more answers
The perimeter of a shape will always be greater in value than the area of the shape (ignoring the units) True or False
Xelga [282]

Answer:

false

Step-by-step explanation:

the are of a shape will always be greater than the perimeter

5 0
3 years ago
Evaluate the integral e^xy w region d xy=1, xy=4, x/y=1, x/y=2
LUCKY_DIMON [66]
Make a change of coordinates:

u(x,y)=xy
v(x,y)=\dfrac xy

The Jacobian for this transformation is

\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}

and has a determinant of

\det\mathbf J=-\dfrac{2x}y

Note that we need to use the Jacobian in the other direction; that is, we've computed

\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}

but we need the Jacobian determinant for the reverse transformation (from (x,y) to (u,v). To do this, notice that

\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}

we need to take the reciprocal of the Jacobian above.

The integral then changes to

\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv
=\displaystyle\frac12\int_{v=}^{v=}\int_{u=}^{u=}\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\ln2}2
8 0
3 years ago
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