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Natali [406]
2 years ago
13

In exercises 21 and 22, find the indicated angle measure. (See photo)

Mathematics
2 answers:
joja [24]2 years ago
7 0
Measure of angle rsv is 42 and that of lhk is 101
Fudgin [204]2 years ago
5 0

Answer:

42°, 101°

Step-by-step explanation:

These answers come from the angle subtraction postulate.

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Please help me! Please please please help me!
photoshop1234 [79]
Graph C is the correct option.

If you drew a straight line (mostly) connecting the points, it would be linear. A relationship is linear if one variable, in this case, x, increases by approximately the same rate as the other variables, in this case y.

If you drew lines connecting points on the other graphs, they wouldn’t resemble a straight line, therefore, they aren’t linear.
5 0
3 years ago
Which of the following points lies on the graph of y=x^2-5?
grandymaker [24]

Answer:

(3, 4)

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p
Juli2301 [7.4K]
I'm reading this as

\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy

with \nabla f=(2xe^{-y},2y-x^2e^{-y}).

The value of the integral will be independent of the path if we can find a function f(x,y) that satisfies the gradient equation above.

You have

\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}

Integrate \dfrac{\partial f}{\partial x} with respect to x. You get

\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx
f=x^2e^{-y}+g(y)

Differentiate with respect to y. You get

\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]
2y-x^2e^{-y}=-x^2e^{-y}+g'(y)
2y=g'(y)

Integrate both sides with respect to y to arrive at

\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy
y^2=g(y)+C
g(y)=y^2+C

So you have

f(x,y)=x^2e^{-y}+y^2+C

The gradient is continuous for all x,y, so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e
8 0
3 years ago
Mrs. Woods prepared bags of baked cookies. The graph shows the number of bags and the number of cookies she used.
larisa86 [58]

Answer: 4 cookies per bag

Step-by-step explanation

3 0
2 years ago
Write the slope intercept form of the r question of the line parallel to the graph of 2x+y=5 that passes through (0,1)
Fofino [41]

Step-by-step explanation:

first let's get the original line in slope intercept form.

the vegetal slope intercept form is

y = ax + b

a is the slope, b is the y-intercept.

2x + y = 5

y = -2x + 5

there !

a line parallel to this line must have the same slope (-2).

but it will intersect with the y-axis at a different point.

the y-intersect is the y value when x = 0.

the given point (0, 1) is already that y-intersect (because x=0).

so, the desired parallel line function is

y = -2x + 1

if we had a different point of the line, e.g. (4, -7), we would go back to the general equation

y = ax + b

put in the slope we know (-2)

y = -2x + b

and then put in the x and y cakes if the point and calculate b

-7 = -2×4 + b

-7 = -8 + b

b = 1

and then we get again

y = -2x + 1

4 0
2 years ago
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