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posledela
3 years ago
7

Questions that are circled with work please

Mathematics
1 answer:
Ymorist [56]3 years ago
3 0
Longer sides have longer angles. So the side length corresponds with the angle size.
 For example in 16 (that is the only one I can see properly), angle C is the largest. That means that the side facing it will be the largest. So side AB will be the largest, then BC and AC in that order because angle A is greater than angle B.
Hope this helps. 
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a lighthouse casts a 128 footshadow a nearby lampost that measures 5 feet 3inches cast an 8 Foot Shadow
k0ka [10]
5 times 3 times 8 is the correct answer, because it's like finding the volume.<span />
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If you change the y-intercept in the equation y = -2x + 5 to y = -2x + 1, how will its graph change? A. It will increase at a fa
FinnZ [79.3K]
The starting value will decrease.
6 0
3 years ago
Based on the line plot, how many recipes call for more than 1/4 tsp of salt?
Studentka2010 [4]
It would be 6 haha I just took a test and got it right
5 0
3 years ago
Read 2 more answers
5+3=4d please help me
ad-work [718]
So the object of this problem is to find out the value of d.

To do this, first think about 5 + 3. This equals 8, right? So the value of d times 4 has to equal 8.

No other number besides 2 would make sense. 2 would equal d because 4 x 2 is 8, and 5 + 3 is 8.

So you would write the answer as:

5+3=4(2)
4 0
3 years ago
Read 2 more answers
Assuming that the equations define x and y implicitly as differentiable functions x=f(t),y=g(t) find the slope of the curve x=f(
Mumz [18]
The given equations are
x(t+1)-4t \sqrt{x} =9            (1)
2y+4y^{3/2}=t^{3}+t           (2)

When t=0, obtain
x=9 \\ 2y+4y^{3/2}=0 \,\,=\ \textgreater \ \, y(1+2 \sqrt{y} )=0 \,=\ \textgreater \ \,y=0

Obtain derivatives of (1) and find x'(0).
x' (t+1) + x - 4√x - 4t*[(1/2)*1/√x = 0
x' (t+1) + x - 4√x -27/√x = 0
When t=0, obtain
x'(0) + x(0) - 4√x(0) = 0
x'(0) + 9 - 4*3 = 0
x'(0) = 3
Here, x' means \frac{dx}{dt}.

Obtain the derivative of (2) and find y'(0).
2y' + 4*(3/2)*(√y)*(y') = 3t² + 1
When t=0, obtain
2y'(0) +6√y(0) * y'(0) = 1
2y'(0) = 1 
y'(0) = 1/2.
Here, y' means \frac{dy}{dt}.

Because \frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt}, obtain
\frac{dy}{dx} |_{t=0}\, =  \frac{1/2}{3}= \frac{1}{6}

Answer:
The slope of the curve at t=0 is 1/6.



3 0
3 years ago
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