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Masteriza [31]
2 years ago
6

Use slope and y intercept to graph a line. y=2x-5

Mathematics
2 answers:
Solnce55 [7]2 years ago
8 0
On your graph:

-- Mark a dot on the y-axis, at y = -5 .

-- From there, move 1 unit to the right and 2 units up and make a mark.  If this
is too tiny for you, then you can move 7 units to the right and 14 units up, or
11 units to the right and 22 units up ... any way you want to do it, as long as
the distance 'up' is double the distance to the right, because the slope is 2 to 1.
Wherever you wind up, mark a dot.

-- Using your pencil and your ruler, draw a straight line between the two dots you have
marked.  You may extend it as far as you wish in either or both directions.
Naily [24]2 years ago
6 0
M=2,b=-5 where m is the slope and b is the y-intercept
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100 equals 15 minus x divided by 5
zloy xaker [14]
The answer is -485. 

To find this we first need to get rid of the 5 in the denominator. 

100 = \frac{15 - x}{5}

So we multiply each side by 5. 

 500 = 15 - x

Now we need to get rid of the 15. So we subtract 15 from both sides. 

485 = -x

Then since our x is negative, we need to divide by -1. 

-485 = x
7 0
3 years ago
Solve for d<br> 3(d+12)=8-4d<br> ?????
zysi [14]

Answer:

d=-4

Step-by-step explanation:

the answer is negative 4 have a nice day :)

5 0
3 years ago
Find the value of x.
nikdorinn [45]

Answer:

X=11

Step-by-step explanation:

(10x+30)=140

-30 both sides

10x=110

Divide 10 both sides

X=11

8 0
3 years ago
Read 2 more answers
PRECAL:<br> Having trouble on this review, need some help.
ra1l [238]

1. As you can tell from the function definition and plot, there's a discontinuity at x = -2. But in the limit from either side of x = -2, f(x) is approaching the value at the empty circle:

\displaystyle \lim_{x\to-2}f(x) = \lim_{x\to-2}(x-2) = -2-2 = \boxed{-4}

Basically, since x is approaching -2, we are talking about values of x such x ≠ 2. Then we can compute the limit by taking the expression from the definition of f(x) using that x ≠ 2.

2. f(x) is continuous at x = -1, so the limit can be computed directly again:

\displaystyle \lim_{x\to-1} f(x) = \lim_{x\to-1}(x-2) = -1-2=\boxed{-3}

3. Using the same reasoning as in (1), the limit would be the value of f(x) at the empty circle in the graph. So

\displaystyle \lim_{x\to-2}f(x) = \boxed{-1}

4. Your answer is correct; the limit doesn't exist because there is a jump discontinuity. f(x) approaches two different values depending on which direction x is approaching 2.

5. It's a bit difficult to see, but it looks like x is approaching 2 from above/from the right, in which case

\displaystyle \lim_{x\to2^+}f(x) = \boxed{0}

When x approaches 2 from above, we assume x > 2. And according to the plot, we have f(x) = 0 whenever x > 2.

6. It should be rather clear from the plot that

\displaystyle \lim_{x\to0}f(x) = \lim_{x\to0}(\sin(x)+3) = \sin(0) + 3 = \boxed{3}

because sin(x) + 3 is continuous at x = 0. On the other hand, the limit at infinity doesn't exist because sin(x) oscillates between -1 and 1 forever, never landing on a single finite value.

For 7-8, divide through each term by the largest power of x in the expression:

7. Divide through by x². Every remaining rational term will converge to 0.

\displaystyle \lim_{x\to\infty}\frac{x^2+x-12}{2x^2-5x-3} = \lim_{x\to\infty}\frac{1+\frac1x-\frac{12}{x^2}}{2-\frac5x-\frac3{x^2}}=\boxed{\frac12}

8. Divide through by x² again:

\displaystyle \lim_{x\to-\infty}\frac{x+3}{x^2+x-12} = \lim_{x\to-\infty}\frac{\frac1x+\frac3{x^2}}{1+\frac1x-\frac{12}{x^2}} = \frac01 = \boxed{0}

9. Factorize the numerator and denominator. Then bearing in mind that "x is approaching 6" means x ≠ 6, we can cancel a factor of x - 6:

\displaystyle \lim_{x\to6}\frac{2x^2-12x}{x^2-4x-12}=\lim_{x\to6}\frac{2x(x-6)}{(x+2)(x-6)} = \lim_{x\to6}\frac{2x}{x+2} = \frac{2\times6}{6+2}=\boxed{\frac32}

10. Factorize the numerator and simplify:

\dfrac{-2x^2+2}{x+1} = -2 \times \dfrac{x^2-1}{x+1} = -2 \times \dfrac{(x+1)(x-1)}{x+1} = -2(x-1) = -2x+2

where the last equality holds because x is approaching +∞, so we can assume x ≠ -1. Then the limit is

\displaystyle \lim_{x\to\infty} \frac{-2x^2+2}{x+1} = \lim_{x\to\infty} (-2x+2) = \boxed{-\infty}

6 0
2 years ago
Allie already owns 33 necklaces, and additional necklaces are priced at 1 for a dollar. How much money does Allie need to spend
ziro4ka [17]

Answer:

75-33=42

Step-by-step explanation:

total (75) minus what she already owns (33) leaves you with how much she still needs since they're each a dollar

6 0
2 years ago
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