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nasty-shy [4]
3 years ago
12

Find the measure of each numbered angle for each figure

Mathematics
1 answer:
cricket20 [7]3 years ago
6 0

Answer:

Angle 1 is 58. Angle 2 is 32.

Step-by-step explanation:

The measure of Angle ACB is 90 because C is on the circle and A and B connect to form a diameter of the circle. So, Angle 1 and Angle 2 add up to 90 (total degrees in triangle - 90). Now you can add the expressions the question gave for Angle 1 and Angle 2, and you get 7x + 6. So you have the equation 7x + 6 = 90. Solve the equation and you get x = 12. Now you can plug in that value for x into the expressions for Angles 1 and 2 to find their measures.

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Marisa bought four bottles of water.Each bottle of water was 95 cents. Write an equation with the same product as the total cost
meriva

Answer:

5\times 0.76= 3.8

Step-by-step explanation:

Given: Marisa bought four bottles

          Price of each bottle is $0.95.

Solving with the factors as given to find the total cost paid by Marisa.

Total cost= 4\ bottles\times \$ 0.95= \$ 3.8

∴ Total cost paid by Marisa is $3.8.

As given to write an equation with the same product as the total cost but different factors.

∴ Lets assume the number of bottles bought by Marisa be "5" and cost of each bottle be "x"

We already find the total cost she paid is $3.8.

Now, writing an equation with different factors, as total cost remain same.

5\times x= 3.8

Dividing both side by 5

we get x= $ 0.76

∴ The required equation is 5\times 0.76= 3.8

6 0
3 years ago
What is bigger 5/6 or 11/12?
Maurinko [17]
First you find the common deoninator.
Step 1: Reduce (simplify) entered fractions to lowest terms, if the case:Fraction: 5 / 6 it's already reduced to lowest terms
Fraction: 11 / 12 it's already reduced to lowest terms
Step 2: Calculate LCM (lowest common multiple) of the reduced fractions' denominators, it will be the common denominator of the compared fractions:Denominator 6, factored = 2 * 3
Denominator 12, factored = 22<span> * 3</span>
LCM (6, 12) = 22<span> * 3 = 12</span>Step 3: Calculate each fraction's expanding number (LCM divided by each fraction's denominator):For fraction: 5 / 6 is 12 : 6 = (22<span> * 3) : 6 = 2</span>
For fraction: 11 / 12 is 12 : 12 = (22<span> * 3) : 12 = 1</span>
Step 4: Expand fractions to bring them to the common denominator (LCM):5 / 6 = (2 * 5) / (2 * 6) = 10 / 12
<span>11 / 12 = (1 * 11) / (1 * 12) = 11 / 12</span>
4 0
2 years ago
For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f
butalik [34]

\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k

Let

\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k

The curl is

\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)

where \partial_\xi denotes the partial derivative operator with respect to \xi. Recall that

\vec\imath\times\vec\jmath=\vec k

\vec\jmath\times\vec k=\vec i

\vec k\times\vec\imath=\vec\jmath

and that for any two vectors \vec a and \vec b, \vec a\times\vec b=-\vec b\times\vec a, and \vec a\times\vec a=\vec0.

The cross product reduces to

\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

\nabla\times\vec f=\vec0

which means \vec f is indeed conservative and we can find f.

Integrate both sides of

\dfrac{\partial f}{\partial y}=2xze^{2xyz}

with respect to y and

\implies f(x,y,z)=e^{2xyz}+g(x,z)

Differentiate both sides with respect to x and

\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}

2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}

4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}

\implies g(x,z)=4\sin(xz^2)+h(z)

Now

f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)

and differentiating with respect to z gives

\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}

2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}

\dfrac{\mathrm dh}{\mathrm dz}=0

\implies h(z)=C

for some constant C. So

f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C

3 0
3 years ago
This is the last one!
elixir [45]
I think is C or D, but I'm not to sure
7 0
3 years ago
What is the product of the rational expression?? CORRECT ANSWER'S ONLY
padilas [110]
The answer is B.

Breaking down the operation:

NUmerators: x × x = x² and +3 × -3 = -9

Denumerators: x × x = x² and +2 × -2 = -4

Answer:

\frac{x+3}{x+2} × \frac{x-3}{x-2}

Hope it helped,

Happy homework/ study/ exam!

7 0
2 years ago
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