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Alex17521 [72]
3 years ago
6

Look at the graph of this system of equations: y = - x2 + 1 and y = x2. At which approximate points are the two equations equal?

Mathematics
2 answers:
Alexandra [31]3 years ago
8 0
Y=-x^2+1 and y=x^2

When the two curves intersect their coordinates will be equal, so we can say y=y whenever a solution exists, so:

x^2=-x^2+1  add x^2 to both sides

2x^2=1  divide both sides by 2

x^2=1/2  take the square root of both sides

x=±√(1/2), so there are two solutions, we can use x^2 to find the corresponding y values.

y=x^2, y=1/2 in each instance, so the two point where the curves intersect are:

(-√(1/2), 1/2) and (√(1/2), 1/2)  or if you want approximations....

(-0.7, 0.5) and (0.7, 0.5)
NemiM [27]3 years ago
6 0

Answer: they are correct....

Step-by-step explanation: I know this cause I just took the mastery test and got it right.

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3x+2(x-5)=25 show your work
Zanzabum

Answer:

x = 7

Step-by-step explanation:

Expand 3x+2(x-5)=25 which looks like

3x+2x-10=25

Add similar elements: ( 3x+2x = 5x)

5x-10=25

Add 10 to both sides

5x-10+10 = 25 + 10

Simplfy

5x = 35

Divide both sides by 5

5x ÷ 5 = 35 ÷ 5  

Simplfy:

x = 7

have a wonderful day

8 0
3 years ago
Can someone help me please
kherson [118]

Answer:

c = 72

Step-by-step explanation:

multiply both sides of the equation by 8 to eliminate the fraction

c = 8 × 9 = 72

note that \frac{72}{8} = 9


5 0
3 years ago
When a denominator is greater then numerator does the cutting fraction stop there?or it can still be reduced
omeli [17]

Answer:

When the denominator is greater than the numerator, the cutting fraction continues and it can still be reduced

Step-by-step explanation:

Pls mark brainliest

4 0
1 year ago
Read 2 more answers
Given right triangle ABC with altitude BD drawn to hypotenuse AC. If AC =16 and DC=5 what is the length of BC in the simplest ra
Nana76 [90]

The length of BC is 4 \sqrt{5}.

Solution:

Given ABC is a right triangle.

AC is the hypotenuse and BD is the altitude.

AB and BC are legs of the triangle ABC.

AC = 16 and DC = 5

<u>Leg rule of geometric mean theorem:</u>

$\frac{\text { hypotenuse }}{\text { leg }}=\frac{\text { leg }}{\text { part }}$

$\Rightarrow \frac{AC}{BC}=\frac{BC}{DC}$

$\Rightarrow \frac{16}{x}=\frac{x}{5}$

Do cross multiplication.

\Rightarrow  16\times 5 = x\times x

\Rightarrow  80= x^2

\Rightarrow  16\times 5= x^2

Taking square root on both sides.

\Rightarrow  \sqrt{16\times 5} = \sqrt{x^2}

\Rightarrow  \sqrt{4^2\times 5} = \sqrt{x^2}

square and square roots get canceled, we get

\Rightarrow  4\sqrt{ 5} = x

The length of BC is 4 \sqrt{5}.

7 0
3 years ago
Given that f.x 3x-2 over x+1 g[x] x +5 evaluate f[-4] and gf [-2]
Jobisdone [24]

The value of f[ -4 ] and g°f[-2] are \frac{14}{3} and 13 respectively.

<h3>What is the value of f[-4] and g°f[-2]?</h3>

Given the function;

  • f(x) = \frac{3x-2}{x+1}
  • g(x)=x+5
  • f[ -4 ] = ?
  • g°f[ -2 ] = ?

For f[ -4 ], we substitute -4 for every variable x in the function.

f(x) = \frac{3x-2}{x+1}\\\\f(-4) = \frac{3(-4)-2}{(-4)+1}\\\\f(-4) = \frac{-12-2}{-4+1}\\\\f(-4) = \frac{-14}{-3}\\\\f(-4) = \frac{14}{3}

For g°f[-2]

g°f[-2] is expressed as g(f(-2))

g(\frac{3x-2}{x+1}) =  (\frac{3x-2}{x+1}) + 5\\\\g(\frac{3x-2}{x+1}) =  \frac{3x-2}{x+1} + \frac{5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{3x-2+5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{8x+3}{x+1}\\\\We\ substitute \ in \ [-2] \\\\g(\frac{3x-2}{x+1}) =  \frac{8(-2)+3}{(-2)+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{-16+3}{-2+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{-13}{-1}\\\\g(\frac{3x-2}{x+1}) =  13

Therefore, the value of f[ -4 ] and g°f[-2] are \frac{14}{3} and 13 respectively.

Learn more about composite functions here: brainly.com/question/20379727

#SPJ1

6 0
1 year ago
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