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Blizzard [7]
4 years ago
6

Y=×^2+4x+1 and y=x+1 Solve a linear/quadratic system by graphing and substitution.

Mathematics
1 answer:
LiRa [457]4 years ago
8 0
Graphing is used when you have paper

I will not graph this equation because it takes time, sorry

but I will solve

subsitute y=x^2+4x+1
subsitute x+1 for y
x+1=x^2+4x+1
subtract x from bot sides
1=x^2+3x+1
subtract 1 from both sides
0=x^2+3x
factor
0=(x)(x+3)
if xy=0 then assume that x and y=0
so
set them to zero
x=0

x+3=0
subtract 3 from both sides
x=-3

subsitue
y=x+1
y=0+1
y=1

y=-3+1
y=-2

so the answers could be
(x,y)
(-3,-2)
(0,1)
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To subtract integers you…<br> Pls, tell me the steps to subtract integers IN YOUR OWN WORDS PLS.
Aleks04 [339]

Answer:

Steps on How to Subtract Integers!

1. Keep the first number!

2. Change the operation from subtraction to addition!

3. Get the opposite sign of the second number!

4. Proceed with the regular addition of integers!

Step-by-step explanation:

8 0
3 years ago
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Which quotient matches the description of solving with partial quotients for 240÷15
Illusion [34]

Answer:

C. 16

Step-by-step explanation:

240/15

Step one: 10x15= 150

              240-150= 90  

Partial quotient: 10

Step two: 6x15= 90

                 90-90= 0

Partial quotient: 6

Step 3: 10+6= 16

Answer check: 240/15= 16

3 0
3 years ago
What is the minimum number of the data set shown on the box-and-whisker plot?
Dmitriy789 [7]
Hello there, and thank you for posting your question here on brainly.

Short answer: C. 9

Why?

Box and whisker plots are confusing, the line in the middle is the median, but we don't need to do that now. The dots on the lines outside of the box show the largest and the shortest number. The plotted point on the way left is your answer, the line plotted is 9.

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3 0
4 years ago
What is 2 wholes minus 1/4
bonufazy [111]
2-1/4
2/1=8/4       Multiply top and bottom by 4
8/4-1/4
7/4
1 3/4
3 0
4 years ago
Read 2 more answers
Prove that if {x1x2.......xk}isany
Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where n\geq 2. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set X= (x_1, x_2,....,x_n) is linearly dependent, so then by definition we have scalars c_1,c_2,....,c_n in C such that:

c_1 x_1 +c_2 x_2 +.....+c_n x_n =0

And not all the scalars c_1,c_2,....,c_n are equal to 0.

Since at least one constant is non zero we can assume for example that c_1 \neq 0, and we have this:

c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n

We can divide by c1 since we assume that c_1 \neq 0 and we have this:

v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n

And as we can see the vector v_1 can be written a a linear combination of the remaining vectors v_2,v_3,...,v_n. We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select v_1 and we have this:

v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n

For scalars defined c_2,c_3,...,c_n in C. So then we have this:

v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0

So then we can conclude that the set X is linearly dependent.

And that complet the proof for this case.

5 0
3 years ago
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