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VARVARA [1.3K]
3 years ago
12

Someone pls just comment something​

Mathematics
2 answers:
hichkok12 [17]3 years ago
4 0

Answer:

Eleventeen

Step-by-step explanation:

solmaris [256]3 years ago
4 0

Answer:

THIRTEEN

Step-by-step explanation:

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Put the equation in standard form with a positive x coefficient y - 1 = 5/6 (x-4) ​
Anon25 [30]

Answer:

5x - 6y = 14

Step-by-step explanation:

The equation of a line in standard form is

Ax + By = C ( A is a positive integer and B, C are integers )

given

y - 1 = \frac{5}{6}(x - 4)

Multiply through by 6 to eliminate the fraction

6y - 6 = 5(x - 4) ← distribute

6y - 6 = 5x - 20 ( add 20 to both sides )

6y + 14 = 5x ( subtract 6y from both sides )

14 = 5x - 6y , thus

5x - 6y = 14 ← in standard form

3 0
3 years ago
On four exams, Wallace’s grades were 79,93,91, and 68. What grade must he obtain on his fifth exam to have an 80 average?
neonofarm [45]
(79 + 93 + 91 + 68 + x) / 5 = 80
(331 + x) / 5 = 80.....multiply both sides by 5
331 + x = 80 * 5
331 + x = 400
x = 400 - 331
x = 69 <=== he would have to make at least a 69
7 0
3 years ago
Read 2 more answers
Badria, Khalid, and Yousif have a total of 95 in their wallets. Yousif has 5 less than Badria. Khalid has 3 times what Yousif ha
schepotkina [342]

Answer:

<u>Amount in Badria's wallet:</u> 18

<u />

<u />

<u>Amount in Khalid's wallet:</u> 54

<u />

<u>Amount in Yousif's wallet:</u> 23

Step-by-step explanation:

6 0
2 years ago
How do larger multiplicities affect the<br> graph of the polynomial
valentina_108 [34]
-multiplicity: the number of times a particular number is a zero for a given polynomial
-affects shape of graph
-odd multiplicity: crosses x axis at root
-even multiplicity: touches x axis at root but doesn't cross it
7 0
4 years ago
15. The equation below defines z as a differentiable function of x and y. Find the value of dz/dy at the point (1, 1, 1).
MariettaO [177]

Isolate the term with z^2.

x^2 - 5y^2 + xyz^2 = y - 4 \implies xyz^2 = -x^2 + 5y^2 + y - 4

Differentiate both sides with respect to y.

\dfrac{\partial(xyz^2)}{\partial y} = \dfrac{\partial(-x^2 + 5y^2 + y - 4)}{\partial y}

By the product and chain rules,

xz^2 + 2xyz \dfrac{\partial z}{\partial y} = 10y + 1

Solve for the partial derivative, then evaluate at (x,y,z) = (1,1,1).

\dfrac{\partial z}{\partial y} = \dfrac{10y + 1 - xz^2}{2xyz}

\dfrac{\partial z}{\partial y} \bigg|_{x=1,y=1,z=1} = \dfrac{10 + 1 - 1}{2} = \boxed{5}

4 0
2 years ago
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