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3 years ago

Solve: x − 2 3 = 61 3 A) 42 9 B) 52 3 C) 6 D) 7

Mathematics

1 answer:

Zepler [3.9K]3 years ago

8 0

Answer:

Step-by-step explanation:

$x-\dfrac{2}{3}=6\dfrac{1}{3}\qquad\text{add}\ \dfrac{2}{3}\ \text{to both sides}\\\\x-\dfrac{2}{3}+\dfrac{2}{3}=6\dfrac{1}{3}+\dfrac{2}{3}\\\\x=6\dfrac{1+2}{3}\\\\x=6\dfrac{3}{3}\\\\x=7$

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3 years ago

Can anyone help? It’s not 14, I’ll give Brainly + 15 pts

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Short side length = x

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2 years ago

Use rhombus TQRS below for questions what is the value of x? what is the value of y?

sladkih [1.3K]

The value of x is 1.

The value of y is 4.

Solution:

Given TQRS is a rhombus.

Property of rhombus:

Diagonals bisect each other.

In diagonal TR

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⇒ 3x – y = –1 – – – – (1)

In diagonal QS

⇒ x + 3 = y

⇒ x – y = –3 – – – – (2)

Solve (1) and (2) by subtracting

⇒ 3x – y – (x – y) = –1 – (–3)

⇒ 3x – y – x + y = –1 + 3

⇒ 2x = 2

⇒ x = 1

Substitute x = 1 in equation (2), we get

⇒ 1 – y = –3

⇒ –y = –3 – 1

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⇒ y = 4

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The value of y is 4.

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3 years ago

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Nataly [62]

Step-by-step explanation:

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2 years ago

Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)

Sergeeva-Olga [200]

Answer:

The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

$x^{2} +y^{2}+4x-1=0$

Step-by-step explanation:

Explanation:-

Step 1:-

The equation of the circle having center and radius is

$(x-h)^2+(y-k)^2=r^2$

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

The diameter of the circle is passing through the center of the circle

so center of the circle = midpoint of two end points

$(\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2} )$

$(-2,0)$

therefore center (h,k) = (-2,0)

Step 2:-

we have to find the radius of the circle

the radius of the circle = the distance from center to the one end point

i.e., C P = r

Given one end point is P(-3,2) and center C(-2,0)

The distance formula of two points are

$\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}$

$r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}$

$r=\sqrt{5}$

Step 3:-

center (h,k) = (-2,0) and

radius $r=\sqrt{5}$

The standard form of circle equation

$(x-h)^2+(y-k)^2=r^2$

$(x-(-2))^2+(y-0)^2=\sqrt{5} ^2$

on simplification is

$x^{2} +y^{2}+4 x-1=0$

5 0

2 years ago