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neonofarm [45]
3 years ago
10

PLEASE HELP ANYONE I WILL MAKE YOU BRAINLIST PLEASE

Mathematics
2 answers:
MariettaO [177]3 years ago
8 0

Answer:

T'(2,-2) ,  C'(2,-5) ,  Z'(5, -4) , F'(5, 0)

Step-by-step explanation:

5) For reflection across the x-axis

Step 1)  the preimage coordinate (x,y) will change to image coordinates (x,-y).

T(2,2)  =  T'(2,-2)

C(2,5)  =  C'(2,-5)

Z(5, 4)  =  Z'(5, -4)

F(5, 0)  =  F'(5, 0)

4vir4ik [10]3 years ago
3 0

Answer:

Step-by-step explanation: simply make a number line across the sides and the point them accordingly like it says in the directions above

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Given data set:
denpristay [2]

Answer:

see below

Step-by-step explanation:

6, 2, 7, 12, 6, 1, 7, 1, 9, 7, 8

First put the numbers in order from smallest to largest

1,1, 2,6,6,  7, 7, 7, 8, 9, 12

The mean is adding all the numbers and dividing by the number of numbers

( 1+1+ 2+6+6+  7+ 7+ 7+ 8+ 9+ 12 )/11

66/1 =6

The mean is 6

The median is the middle

There are 11 terms

11/5 = 5.5  so 6 is the middle number

1,1, 2,6,6,    7,      7, 7, 8, 9, 12

7 is the median

Mode is the most often

7 appears three times so it is the mode

6 0
3 years ago
When is the additive relationship used? Select all that apply.
Mrac [35]

Answer:

your answer is for sure D but I think C is also the answer

Step-by-step explanation:

Additive relationship are two quantities can be expressed as related to each other through addition. It can be written as y = x + a, where y is related to x through the addition of a constant, a. The value for a may be positive or negative.

7 0
3 years ago
Read 2 more answers
Mason compares two plans for an online book club.
horsena [70]

Answer:

Book clubs in the classroom: 10 tips for success

4/19/201746 Comments

Picture

Whether you’re considering a classroom book club for the first time, or are already guiding your students through their third book this year, here’s a list of benefits and some tips for success you can employ right away!

NYC’s District 75 published their Middle School Units of Study, Developing Autonomy when Engaging with Literature, online. In it they list some of the benefits of book clubs in the classroom:

Book clubs:

Promote a love for literature and a positive attitude towards reading;

Reflect a student-centered model of literacy (employing the Gradual

Release of Responsibility);

Encourage extensive and intensive reading;

Invite natural discussions that lead to student inquiry and critical

Thinking;

Support diverse responses to text;

Foster interaction, cooperation and collaboration;

Provide choice and encourage responsibility;

Expose students to literature from multiple perspectives; and

Nurture reflection and self-evaluation.

Thoughtfully planned book clubs position learning in the hands of the students and provide discussion tools students can use as they work out their responses to the book.

We provide approaches to professional development that bridge theory and practice to generate solutions that work for a multitude of schools. With that bridge in mind, enjoy the following as you think through your classroom’s book club possibilities!

Step-by-step explanation:

pls brainlieste

6 0
2 years ago
Question 6<br> Express this decimal as a fraction.<br> ?<br> 1.21 =<br> ^ it has a line over the 21.
saul85 [17]

Answer: 40/33

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
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
4 years ago
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