All categories

3 years ago

How many solutions does this system have? y = -2x + 4 y = x^2 - 2x +4

Mathematics

1 answer:

AVprozaik [17]3 years ago

6 0

Answer:

One Solution

Step-by-step explanation:

You might be interested in

Drag a statement or reason to each box to complete the proof.

astra-53 [7]

Answer:

The answers of the proof are the following:

2. Division Property of Equality

3. x+ 1 = -4

4. Subraction Property of Equality

5. x = -5

5 0

3 years ago

Simplify. 3/5 +(− 1/4 ) 7/10 Enter your answer, in simplest form, in the boxes.

vitfil [10]

3/5+(-7/40)

=3/5-7/40

=24/40-7/40

=17/40

8 0

3 years ago

Read 2 more answers

Which could be the first step in solving this system of equations by substitution?

Ainat [17]

Answer:

The first step is to replace the y in y-x=15 with 7x (we can do this because the first equation tells us that they're equal)

solve and get

(2.5,17.5)

or x= 2.5 y=17.5

Step-by-step explanation:

The first step is to recognize that y=7x which means that we can just replace the y in the second equation with 7x

7x-x=15

6x=15

x=2.5

Then we can solve for y

y=7(2.5)

y=17.5

3 0

2 years ago

Grey ran 2.4 times as many miles as Emory. If Emory ran 2.08 miles, how many miles did Grey run

MrRissso [65]

2.08 x 2.4= 4.992
Round answer=5 miles

6 0

3 years ago

The quadratic function g(x) = a.ca + bx+c has the

Mumz [18]

<em>The value of b is 14 and the value of c is 65</em>

<h2>Explanation:</h2>

The quadratic function is a function of the form:

$f(x)=ax^2+bx+c$

Here we know that the leading coefficient $a=1$ so we reduce our equation to:

$g(x)=x^2+bx+c$

The roots are those values at which $g(x)=0$

So:

$x^2+bx+c=0 \\ \\ First \ root: \\ \\ (-7+4i)^2+b(-7+4i)+c=0 \\ \\ (-7)^2-2(7)(4i)+(4i)^2-7b+4bi+c=0 \\ \\ 49-56i+16i^2-7b+4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49-56i+16(-1)-7b+4bi+c=0 \\ \\ 49-56i-16-7b+4bi+c=0 \\ \\ 33-56i-7b+4bi+c=0 \\ \\ \\$

$Second \ root: \\ \\ (-7-4i)^2+b(-7-4i)+c=0 \\ \\ (-1)^2(7+4i)^2+b(-7-4i)+c=0 \\ \\ (7)^2+2(7)(4i)+(4i)^2-7b-4bi+c=0 \\ \\ 49+56i+16i^2-7b-4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49+56i+16(-1)-7b-4bi+c=0 \\ \\ 49+56i-16-7b-4bi+c=0 \\ \\ 33+56i-7b-4bi+c=0$

So we have:

$(1) \ 33-56i-7b+4bi+c=0 \\ \\ (2) \ 33+56i-7b-4bi+c=0 \\ \\ \\ Subtract \ 2 \ from: \\ \\ 33-56i-7b+4bi+c-(33+56i-7b-4bi+c)=0 \\ \\ 33-56i-7b+4bi+c-33-56i+7b+4bi-c=0 \\ \\ \\ Combine \ like \ terms: \\ \\ 33-33-56i-56i-7b+7b+4bi+4bi+c-c=0 \\ \\ -112i+8bi=0 \\ \\ Isolating \ b: \\ \\ b=\frac{112i}{8i} \\ \\ \boxed{b=14}$

Finding c from (1):

$33-56i-7b+4bi+c=0 \\ \\ \\ Substituting \ b: \\ \\ 33-56i-7(14)+4(14)i+c=0 \\ \\ 33-56i-98+56i+c=0 \\ \\ -65+c=0 \\ \\ \boxed{c=65}$

<h2>Learn more:</h2>

Complex conjugate: brainly.com/question/2137496

#LearnWithBrainly

5 0

2 years ago