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Rina8888 [55]
3 years ago
9

Again(sorry) did anyone know 96 and 97

Mathematics
2 answers:
vladimir2022 [97]3 years ago
7 0
96: 

sugar: 1,2,3,4,5,6
flour:2,4,6,8,10,12

97:

yellow:3,6,9,12,15,18
blue:1,2,3,4,5,6
Dahasolnce [82]3 years ago
4 0
  #96:     sugar : flour 
1:2           2:4            3:6          4:8          5:10
  
  #97:       yellow : blue 
3:1           6:2}           9:3        12:4        15:5
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What is the slope of the line that passes through the points (2, 1) and (17, -17)?
Nastasia [14]

Answer:

-6/5

Step-by-step explanation:

-17-1 = -18=-6

17-2= 15=5

7 0
3 years ago
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Determine whether the triangles are congruent by SSS or SAS. If not congruent, write "not congruent." b) If congruent, write a c
Ulleksa [173]

Answer:

sss is all sides are the same length, in the same order, and if they both have those side they are congruent. sas is two sides that are connected by an angle, and if the sides connected by that angle match the other triangle then they will be congruent

Step-by-step explanation:

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3 years ago
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PLEASE HELP!! Jennifer is wrapping presents. One of the rectangular gift boxes has dimensions 1 in. • 7 in. • 3 in. and the othe
cluponka [151]

Answer:

252 inches

Step-by-step explanation:

Alright, so we've got to find the surface area of both of the rectangular prisms and add them together.

We can use the equation 2(2l+hl+hw)

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Do the same for the second, you should get 190.

Now we just need to add the two together. 190+62= 252

5 0
4 years ago
find an equation of the tangent plane to the given parametric surface at the specified point. x = u v, y = 2u2, z = u − v; (2, 2
Alexxandr [17]

The surface is parameterized by

\vec s(u,v) = x(u, v) \, \vec\imath + y(u, v) \, \vec\jmath + z(u, v)) \, \vec k

and the normal to the surface is given by the cross product of the partial derivatives of \vec s :

\vec n = \dfrac{\partial \vec s}{\partial u} \times \dfrac{\partial \vec s}{\partial v}

It looks like you're given

\begin{cases}x(u, v) = u + v\\y(u, v) = 2u^2\\z(u, v) = u - v\end{cases}

Then the normal vector is

\vec n = \left(\vec\imath + 4u \, \vec\jmath + \vec k\right) \times \left(\vec \imath - \vec k\right) = -4u\,\vec\imath + 2 \,\vec\jmath - 4u\,\vec k

Now, the point (2, 2, 0) corresponds to u and v such that

\begin{cases}u + v = 2\\2u^2 = 2\\u - v = 0\end{cases}

and solving gives u = v = 1, so the normal vector at the point we care about is

\vec n = -4\,\vec\imath+2\,\vec\jmath-4\,\vec k

Then the equation of the tangent plane is

\left(-4\,\vec\imath + 2\,\vec\jmath - 4\,\vec k\right) \cdot \left((x-2)\,\vec\imath + (y-2)\,\vec\jmath +  (z-0)\,\vec k\right) = 0

-4(x-2) + 2(y-2) - 4z = 0

\boxed{2x - y + 2z = 2}

6 0
2 years ago
The sum of four times a number and negative three is twelve. What is the number?
neonofarm [45]
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x=3 and 3/4

answer is 3 and 3/4
3 0
3 years ago
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