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3 years ago

Solve x^2—y=—2 —x+y=4 using the elimination method

1 answer:

5 0

X² - y = -2 .................. equation 1
-x + y = 4 .................. equation 2 from the y's we can obtain y - y = 0
the new equation will be
x² - x = 2 to for a quadratic equation make the equation equals to zero
x² - x - 2 = 0
x² - 2x + x - 2 = 0 using factorization method
x(x - 2) + 1(x - 2) = 0
(x + 1)(x - 2) = 0
x + 1=0 and x - 2=0
x = -1 or 2
subtitute x= -1 in equation 2 subtitute x= 2 in equation 2
-(-1) + y = 4 -(2) + y = 4
1 + y = 4 -2 + y = 4
y= 4 - 1 y= 4 + 2
y= 3 y= 6

the values of x are -1 and 2 . The values of y are 3 and 6
when x is -1, y is 3
when x is 2, y is 6...

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Solve step by step solution then only i can do it plxx

Anuta_ua [19.1K]

Answer:

<h3>Let's understand the concept:-</h3>

Here angle B is 90°

So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle

So we use Pythagoras thereon for solution

<h3>Required Answer:-</h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

3 0

3 years ago

Geometry: fill in the blanks (ASAP!! It’s urgent)

stepladder [879]

5. m∠C = 95°

6. m∠C = 70°

7. The other acute angle in the right triangle = 70°

8. m∠C = 70°

9. m∠C = 60° [equilateral triangle]

10. Measure of the exterior angle at ∠C = 110°

11. m∠B = 70°

12. m∠Z = 70°

<h3>What are Triangles?</h3>

A triangle is a 3-sided polygon with three sides and three angles. The sum of all its interior angles is 180 degrees. Some special triangles are:

Isosceles triangle: has 2 equal base angles.
Equilateral triangle: has three equal angles, each measuring 60 degrees.
Right Triangle: Has one of its angles as 90 degrees, while the other two are acute angles.

5. m∠C = 180 - 50 - 35 [triangle sum theorem]

m∠C = 95°

6. m∠C = 180 - 25 - 85 [triangle sum theorem]

m∠C = 70°

7. The other acute angle in the right triangle = 180 - 90 - 25 [triangle sum theorem]

The other acute angle = 70°

8. m∠C = 180 - 55 - 55 [isosceles triangle]

m∠C = 70°

9. m∠C = 60° [equilateral triangle]

10. Measure of the exterior angle at ∠C = 50 + 60

Measure of the exterior angle at ∠C = 110°

11. m∠B = 115 - 45