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alekssr [168]
3 years ago
11

Find the slope and the y-intercept of the equation y − 3(x − 1) = 0.

Mathematics
2 answers:
nekit [7.7K]3 years ago
6 0
In this problem, we need to isolate the y variable. To do so, we can first distribute the -3 into the (x-1) to get y - 3x + 1 = 0.
Then, we can add 3x - 1 to both sides of the equation, making it:
y = 3x - 1
This is a linear equation in the form of y = mx + b. The m, 3 in this equation, is the slope. The b, -1 in this equation, is the y-intercept.
The slope is 3.
The y-intercept is -1.
Nikolay [14]3 years ago
3 0
<span>y − 3(x − 1) = 0.
y - 3x + 3 = 0
y = 3x - 3

slope m = 3
y-intercept (0, -3)</span>
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I really need help for this! Please!
Serhud [2]

Answer: -16

Step-by-step explanation:

-3 is less than -2 so use the first one

Plug into equation

5(-3)-1

-15-1= -16

8 0
3 years ago
2 please answer this question
Fittoniya [83]
B. End of the reconstruction. 
4 0
4 years ago
Read 2 more answers
What is the value of k in the equation below
Dima020 [189]

Answer:

<h2>4</h2>

Step-by-step explanation:

5k-2k=12

3k=12

k=4

5 0
3 years ago
1. Two triangles have the following congruence statement: ACGI = AMPR Name all 6 pairs of corresponding congruent parts. 1.​
rodikova [14]

Answer:

\angle C \cong \angle M

\angle G \cong \angle P

\angle I \cong \angle R

\overline{CG} \cong \overline{MP}

\overline{GI} \cong \overline{PR}

\overline{CI} \cong \overline{MR}

Step-by-step explanation:

Given the congruence statement ∆CGI \cong ∆MPR, it follows that the corresponding sides of both ∆s are equal, as well as the corresponding vertices or angles. It implies that ∆CGI and ∆MPR are of the same shape and size.

✅Thus, the 6 pairs of the corresponding congruent parts of ∆CGI and ∆MPR are:

\angle C \cong \angle M

\angle G \cong \angle P

\angle I \cong \angle R

\overline{CG} \cong \overline{MP}

\overline{GI} \cong \overline{PR}

\overline{CI} \cong \overline{MR}

3 0
3 years ago
To two decimal places, find the value of k that will make the function f(x) continuous everywhere. f of x equals the quantity 3x
Fiesta28 [93]

Given function is

f(x)=\left\{\begin{matrix}3x+k & x\leq -4 \\ kx^2-5 & x> -4\end{matrix}\right.

now we need to find the value of k such that function f(x) continuous everywhere.

We know that any function f(x) is continuous at point x=a if left hand limit and right hand limits at the point x=a are equal.

So we just need to find both left and right hand limits then set equal to each other to find the value of k

To find the left hand limit (LHD) we plug x=-4 into 3x+k

so LHD= 3(-4)+k

To find the Right hand limit (RHD) we plug x=-4 into

kx^2-5

so RHD= k(-4)^2-5

Now set both equal

k(-4)^2-5=3(-4)+k

16k-5=-12+k

16k-k=-12+5

15k=-7

k=-\frac{7}{15}

k=-0.47

<u>Hence final answer is -0.47.</u>




7 0
3 years ago
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