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2 years ago

Match the number with its opposite. (4 points)

Mathematics

1 answer:

Svetach [21]2 years ago

8 0

Answer:

This should definetly b more points

Step-by-step explanation:

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WILL GIVE BRAINLIEST!!! Question in pic!!!!! Help pls!!

aalyn [17]

Answer:

x=26

y=11

Step-by-step explanation:

y is equal to the one on the opposite side, so y=11

since 24 and the other side are equal, and the other side is x-2, you do 24+2=26

3 0

2 years ago

Read 2 more answers

What are the solutions of x^2-2x+5=0

sweet [91]

The solution of $x^{2}-2 x+5=0$ are 1 + 2i and 1 – 2i

Solution:

Given, equation is $x^{2}-2 x+5=0$

We have to find the roots of the given quadratic equation

Now, let us use the quadratic formula

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ --- (1)

Let us determine the nature of roots:

Here in $x^{2}-2 x+5=0$ a = 1 ; b = -2 ; c = 5

$b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16$

Since $b^2 - 4ac < 0$ , the roots obtained will be complex conjugates.

Now plug in values in eqn 1, we get,

$x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 1 \times 5}}{2 \times 1}$

On solving we get,

$x=\frac{2 \pm \sqrt{4-20}}{2}$

$x=\frac{2 \pm \sqrt{-16}}{2}$

$x=\frac{2 \pm \sqrt{16} \times \sqrt{-1}}{2}$

we know that square root of -1 is "i" which is a complex number

$\begin{array}{l}{\mathrm{x}=\frac{2 \pm 4 i}{2}} \\\\ {\mathrm{x}=1 \pm 2 i}\end{array}$

Hence, the roots of the given quadratic equation are 1 + 2i and 1 – 2i

6 0

2 years ago

Write the equation of circle b with center B(-2,3) that passes through (1,2)

professor190 [17]

Answer:

$(x+2)^2 + (y-3)^2 = 10$

Step-by-step explanation:

The standard equation for circle is

$(x-a)^2 + (y-b)^2 = r^2$

where point (a,b) is coordinate of center of circle and r is the radius.

______________________________________________________

Given

center of circle =((-2,3)

let r be the radius of circle

Plugging in this value of center in standard equation for circle given above we have

$(x-a)^2 + (y-b)^2 = r^2 \ substitute (a,b) \ with (-2,3) \\=>(x-(-2))^2 + (y-3)^2 = r^2 \\=>(x+2)^2 + (y-3)^2 = r^2 (1)$

Given that point (1,2 ) passes through circle. Hence this point will satisfy the above equation of circle.

Plugging in the point (1,2 ) in equation 1 we have

$\\=>(x+2)^2 + (y-3)^2 = r^2 \\=> (1+2)^2 + (2-3)^2 = r^2\\=> 3^2 + (-1)^2 = r^2\\=> 9 + 1 = r^2\\=> 10 = r^2\\=> r^2 = 10\\$

now we have value of r^2 = 10, substituting this in equation 1 we have

Thus complete equation of circle is $=>(x+2)^2 + (y-3)^2 = r^2\\=>(x+2)^2 + (y-3)^2 = 10$

7 0

3 years ago

Prove that the value of the expression does not depend on a: a−(6a−(5a−8))

djyliett [7]

Step-by-step explanation:

a−(6a−(5a−8))

a-(6a-5a+8)

a-6a+5a-8

6a-6a-8

0-8

=-8

so the answer doesn't depend on a

3 0

2 years ago

The quotient of n and 6

Shalnov [3]

Answer:

The quotient of n and 6 is n/6, as quotient means divide.

for example, if it was "the quotient of 24 and 6", it would be 24/6, which is 4.

5 0

2 years ago