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3 years ago

Caleb and his classmates are making felt dog toys for a local animal shelter.

Mathematics

1 answer:

TEA [102]3 years ago

3 0

Not really sure how to answer this hope someone helps you!

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In the diagram of circle R, m∠FGH is 50°. What is m? 130° 230° 260° 310°

omeli [17]

we know that

The measurement of the exterior angle is the semi-difference of the arcs which comprises

In this problem

∠FGH is the exterior angle

∠FGH= $50\°$

∠FGH= $\frac{1}{2}(arc\ FEH-arc\ FH)$

$50\°=\frac{1}{2}(arc\ FEH-arc\ FH)$

$100\°=(arc\ FEH-arc\ FH)$ -----> equation A

$arc\ FEH+arc\ FH=360\°$

$arc\ FH=360\°-arc\ FEH$ --------> equation B

Substitute equation B in equation A

$100\°=(arc\ FEH-[360\°-arc\ FEH])$

$100\°=2arc\ FEH-360\°$

$2arc\ FEH=360\°+100\°$

$arc\ FEH=230\°$

therefore

The answer is

The measure of arc FEH is equal to $230\°$

6 0

3 years ago

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This model is composed of two shapes. What is the area of this figure?

mote1985 [20]

Measurements of 30 I believe yw

7 0

2 years ago

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Please Help I’m in a test!!!!!

Semenov [28]

Answer:

undefined

Step-by-step explanation:

The x coordinates didn't change so the line is vertical and vertical lines have undefined slope

5 0

3 years ago

If L=√(x^2+y^2), dx/dt =-4, dy/dt=3, find dL/dt when x=4 and y=3

yulyashka [42]

Hello,

$L=\sqrt{x^2+y^2} \\\\\dfrac{dx}{dt}=-4\\\\\dfrac{dy}{dt}=3\\\\x=4, y=3\\\\\dfrac{\partial L} {\partial x} =\dfrac{x}{\sqrt{x^2+y^2}} \\\\\dfrac{\partial L} {\partial y} =\dfrac{y}{\sqrt{x^2+y^2}} \\\\\dfrac{dL}{dt} =\dfrac{\partial L} {\partial x}*\dfrac{dx}{dt}+\dfrac{\partial L} {\partial y}*\dfrac{dy}{dt}\\\\=-4*\frac{-4}{\sqrt{4^2+3^2}} +3*\frac{3}{\sqrt{4^2+3^2}}\\\\=\dfrac{16}{5} +\dfrac{9}{5} \\\\=5\\$

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3 years ago

State the domain of the function (-3,-1,1,2,4) (-3,-1,1,2,3) {0, 1,3,4,5) {0, 1.4)

dmitriy555 [2]

Answer:

-4/4

Step-by-step explanation:

bc divison

3 0

3 years ago