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Helen [10]
2 years ago
15

X + y=-3 can someone help me sketching the graph of each line infinite pre-algebra

Mathematics
1 answer:
Leni [432]2 years ago
7 0
Plot these points and then draw the line
X = 0
Y = -3

X = -1
Y = -2

X= -2
Y = -1

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Solve the equation and check the solution.
myrzilka [38]

A - 2 1/2 = 1 1/2

Solve for A by adding 2 1/2 to both sides:

A = 1 1/2 + 2 1/2

A = 4

The answer is c. A = 4

Check: 4 - 2 1/2 = 1 1/2

8 0
3 years ago
(a) The function k is defined by k(x)=f(x)g(x). Find k′(0).
Brut [27]

Answer:

(a) k'(0) = f'(0)g(0) + f(0)g'(0)

(b) m'(5) = \frac{f'(5)g(5) - f(5)g'(5)}{2g^{2}(5) }

Step-by-step explanation:

(a) Since k(x) is a function of two functions f(x) and g(x) [ k(x)=f(x)g(x) ], so for differentiating k(x) we need to use <u>product rule</u>,i.e., \frac{\mathrm{d} [f(x)\times g(x)]}{\mathrm{d} x}=\frac{\mathrm{d} f(x)}{\mathrm{d} x}\times g(x) + f(x)\times\frac{\mathrm{d} g(x)}{\mathrm{d} x}

this will give <em>k'(x)=f'(x)g(x) + f(x)g'(x)</em>

on substituting the value x=0, we will get the value of k'(0)

{for expressing the value in terms of numbers first we need to know the value of f(0), g(0), f'(0) and g'(0) in terms of numbers}{If f(0)=0 and g(0)=0, and f'(0) and g'(0) exists then k'(0)=0}

(b) m(x) is a function of two functions f(x) and g(x) [ m(x)=\frac{1}{2}\times\frac{f(x)}{g(x)} ]. Since m(x) has a function g(x) in the denominator so we need to use <u>division rule</u> to differentiate m(x). Division rule is as follows : \frac{\mathrm{d} \frac{f(x)}{g(x)}}{\mathrm{d} x}=\frac{\frac{\mathrm{d} f(x)}{\mathrm{d} x}\times g(x) + f(x)\times\frac{\mathrm{d} g(x)}{\mathrm{d} x}}{g^{2}(x)}

this will give <em>m'(x) = \frac{1}{2}\times\frac{f'(x)g(x) - f(x)g'(x)}{g^{2}(x) }</em>

on substituting the value x=5, we will get the value of m'(5).

{for expressing the value in terms of numbers first we need to know the value of f(5), g(5), f'(5) and g'(5) in terms of numbers}

{NOTE : in m(x), g(x) ≠ 0 for all x in domain to make m(x) defined and even m'(x) }

{ NOTE : \frac{\mathrm{d} f(x)}{\mathrm{d} x}=f'(x) }

4 0
3 years ago
A box contains four equal sized cards labeled 1, 3,5, and 7. Tim will select one card from the bow. What is the probability that
Zielflug [23.3K]

Answer: The probability is zero.

Step-by-step explanation:

The labels on the 4 cards are 1, 3, 5 and 7. Because the cards are equal, the probability of dragging at random each one of them is equal to 1/4 = 0.25.

Now, the probability that Tim selects a card labeled with a 4 is equal to the number of cards labeled with a 4, divided the total number of cards.

We do not have cards labeled with a 4, so the probability is:

p = 0/4 = 0

because we do not have any card labeled with a 4, the probability of selecting one at random is equal to zero.

8 0
3 years ago
Help me out asap!please and thank you
irga5000 [103]

Answer:

c. (5, -2)

Step-by-step explanation:

as we can clearly see, the 4th vertex has to be on the positive side of x.

therefore, all answer options with negative x are out.

(2, -5) is almost at D (-1, 5). that cannot be right for a rectangle.

that leaves only c as right answer.

FYI - how to get this without predefined answer portions ?

the x- difference of B to A is 4 (from -5 to -1). the x-difference from D to C must be the same (4). 1 + 4 = 5.

so, x of D must be 5.

the y- difference of B to A is 3 (from 3 to 6). the y-difference from D to C must be the same (3). -5 + 3 = -2.

so, y of D must be -2.

3 0
2 years ago
A The length of a rectangle is 4 m more
Lady bird [3.3K]

Given :

  • The length of a rectangle is 4m more than the width.
  • The area of the rectangle is 45m²

⠀

To Find :

  • The length and width of the rectangle.

⠀

Solution :

We know that,

\qquad { \pmb{ \bf{Length \times Width = Area_{(rectangle)}}}}\:

So,

Let's assume the length of the rectangle as x and the width will be (x – 4).

⠀

Now, Substituting the given values in the formula :

\qquad \sf \: { \dashrightarrow x  \times  (x - 4) = 45 }

\qquad \sf \: { \dashrightarrow {x}^{2}  - 4x = 45 }

\qquad \sf \: { \dashrightarrow {x}^{2}  - 4x - 45 = 0 }

\qquad \sf \: { \dashrightarrow {x}^{2}    - 9x+ 5 x - 45 = 0 }

\qquad \sf \: { \dashrightarrow x(x - 9) + 5(x - 9) = 0 }

\qquad \sf \: { \dashrightarrow (x  - 9) (x  + 5) = 0 }

\qquad \sf \: { \dashrightarrow x = 9, \: \: x =  - 5}

⠀

Since, The length can't be negative, so the length will be 9 which is positive.

⠀

\qquad { \pmb{ \bf{ Length _{(rectangle)} = 9\:m}}}\:

\qquad { \pmb{ \bf{ Width _{(rectangle)} = 9 - 4=5\: m}}}\:

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

8 0
2 years ago
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