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3 years ago

PLEAS HELP ME I BEG ONYL 4 QUESTIONS PLEASE PLEASE

1 answer:

3 0

It works well to put both equations in the same form. Here, we use "general form". Any common factors are removed from the numbers. The leading coefficient is positive.

The equations are inconsistent if the x and y coefficients have the same ratio but the constants are different. They are consistent otherwise.

1. 8x +y +4 = 0
.. 8x +y +4 = 0
"consistent" . . . . . and also "dependent"

2. 7x +2y +6 = 0
.. 7x +2y +1 = 0
"inconsistent"

3. x +y +3 = 0
.. x +y +3 = 0
"consistent" . . . . . and also "dependent"

4. x -2y -7 = 0
.. x +3y -7 = 0
"consistent" . . . . . (x, y) = (7, 0)

5. 2x -y +5 = 0
.. 2x -y -2 = 0
"inconsistent"

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What is the slope of the line?

kherson [118]

-8/5 is the slope of the line

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2 years ago

−3(y−8)=33 What is the value of y?

Julli [10]

Answer:

Step-by-step explanation:

-3(y-8)= 33

+3 +3

y - 8 = 36

+ 8 + 8

y = 44

3 0

3 years ago

ValentinkaMS [17]

Answer:

$P = 58$

Step-by-step explanation:

The image if the circle and triangle can be found online

Given

$JA = 8$

$AL = 8$

$CK = 10$

Required

The perimeter of JKL

Since the mid-points of the tangents are A, B and C respectively.

$JK = JB + BK$

Where

$JB = JA = 8$

$BK = CK = 10$

So:

$JK =8+10 =18$

$JL = JA + AL$

$JL = 8+11 = 19$

$KL = CL + CK$

Where

$CL =AL = 11$

So:

$KL = 11+10 = 21$

So, the perimeter (P) is:

$P = JK+ KL +JL$

$P = 18 + 19 + 21$

$P = 58$

4 0

3 years ago

Differentiating Exponential Functions In Exercise, find the derivative of the function. y = x2ex

I am Lyosha [343]

Answer:

$y' = xe^{x}(2 + x)$

Step-by-step explanation:

If we have a product function y, in the following format

$y = f(x)*g(x)$

This function has the following derivative

$y' = f'(x)*g(x) + g'(x)*f(x)$

In this problem, we have that:

$y = x^{2}e^{x}$

$f(x) = x^{2}, f'(x) = 2x, g(x) = e^{x}, g'(x) = e^{x}$

The derivative of the function is:

$y' = f'(x)*g(x) + g'(x)*f(x)$

$y' = 2xe^{x} + x^{2}e^{x}$

$y' = e^{x}(2x + x^{2})$

$y' = xe^{x}(2 + x)$

8 0

3 years ago

A cook used all of a two-pound chunk of cheese in making three recipes. Recipe I used one-third as much cheese as Recipe II did,

skelet666 [1.2K]

Answer:

Recipe I uses 10% of the whole chunk of cheese.

Recipe II uses 30% of the whole chunk of cheese.

Recipe II uses 60% of the whole chunk of cheese.

Recipe I uses 0.2 pounds of cheese.

Recipe II uses 0.6 pounds of cheese.

Recipe II uses 1.2 pounds of cheese.

Step-by-step explanation:

Total amount of cheese = 2 pounds

Recipe I used one-third as much cheese as Recipe II did, and Recipe II used one-half as much cheese as Recipe III.

Let the amount of cheese used by recipe I, II and III be x, y and z respectively.

Total amount of cheese = 2 pounds

x + y + z = 2 (eqn 1)

Recipe I used one-third as much cheese as Recipe II did

x = (1/3)y

y = 3x

Recipe II used one-half as much cheese as Recipe III.

y = (1/2)z

z = 2y

z = 2y = 2(3x) = 6x

Substituting all of these into eqn 1

x + y + z = 2

x + 3x + 6x = 2

10x = 2

x = 0.2 pounds of cheese

y = 3x = 0.6 pounds of cheese

z = 6x = 1.2 pounds of cheese.

Recipe I uses (0.2/2) = 10% of the whole chunk of cheese.

Recipe II uses (0.6/2) = 30% of the whole chunk of cheese.

Recipe II uses (1.2/2) = 60% of the whole chunk of cheese.

Hope this Helps!!!

5 0

3 years ago