All categories

2 years ago

Im stuck on these 2 questions

Mathematics

2 answers:

Kryger [21]2 years ago

8 0

Answer:

In the first question it is the 2nd table. On the second it is : No, the xost of the 9 shirts is incorrect

Goryan [66]2 years ago

3 0

Answer:In the first question it is the 2nd table

Step-by-step explanation:

You might be interested in

11Alexandr11 [23.1K]

It is 6 units because if you divide 9 by 3 you get 3 and you multiple that by 2 to make it 6:9 which is equal to 2:3

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%7B80%29%5C%5C" id="TexFormula1" title="\sqrt{80)\\" alt="\sqrt{80)\\" align="absmiddle

Naddika [18.5K]

Answer:

8.944271909999

Step-by-step explanation:

hope you happy

6 0

2 years ago

Read 2 more answers

raph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the ci

andreyandreev [35.5K]

Answer:

Equation:

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

The point (0,-5), (0,7), (5,0) and (-7,0)also lie on this circle.

Step-by-step explanation:

We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).

The center of this circle is the midpoint of (-3, 4) and (5, -2).

We use the midpoint formula:

$( \frac{x_1+x_2}{2}, \frac{y_1+y_2,}{2} )$

Plug in the points to get:

$( \frac{ - 3+5}{2}, \frac{ - 2+4}{2} )$

$( \frac{ -2}{2}, \frac{ 2}{2} )$

$( - 1, 1)$

We find the radius of the circle using the center (-1,1) and the point (5,-2) on the circle using the distance formula:

$r = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

$r = \sqrt{ {(5 - - 1)}^{2} + {( - 2- - 1)}^{2} }$

$r = \sqrt{ {(6)}^{2} + {( - 1)}^{2} }$

$r = \sqrt{ 36+ 1 } = \sqrt{37}$

The equation of the circle is given by:

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

Where (h,k)=(-1,1) and r=√37 is the radius

We plug in the values to get:

$(x- - 1)^2 + (y-1)^2 = {( \sqrt{37}) }^{2}$

$(x + 1)^2 + (y - 1)^2 = 37$

We expand to get:

${x}^{2} + 2x + 1 + {y}^{2} - 2y + 1 = 37$

${x}^{2} + {y}^{2} + 2x - 2y +2 - 37= 0$

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

We want to find at least four points on this circle.

We can choose any point for x and solve for y or vice-versa

When y=0,

${x}^{2} + {0}^{2} + 2x - 2(0) - 35= 0$

${x}^{2} +2x - 35= 0$

$(x - 5)(x + 7) = 0$

$x = 5 \: or \: x = - 7$

The point (5,0) and (-7,0) lies on the circle.

When x=0

${0}^{2} + {y}^{2} + 2(0) - 2y - 35= 0$

${y}^{2} - 2y - 35= 0$

$(y - 7)(y + 5) = 0$

$y = 7 \: or \: y = - 5$

The point (0,-5) and (0,7) lie on this circle.

3 0

3 years ago

2x + 12 = 1/3 (7/3x+4/3)

gladu [14]

Answer:

Step-by-step explanation:

2x + 12 = 1/3 (7/3x+4/3)

$2x+12=\frac{1}{3}*\frac{7x}{3}+\frac{1}{3}*\frac{4}{3}\\\\2x+12=\frac{7x}{9}+\frac{4}{9}\\\\2x-\frac{7x}{9}=\frac{4}{9}-12\\\\\frac{2x*9}{1*9}-\frac{7x}{9}=\frac{4}{9}-\frac{12*9}{1*9}\\\\\frac{18x}{9}-\frac{7x}{9}=\frac{4}{9}-\frac{108}{9}\\\\\frac{18x-7x}{9}=\frac{4-108}{9}\\\\\frac{11x}{9}=\frac{-104}{9}\\\\\\x=\frac{-104*9}{11*9}=\frac{-104}{11}=-9\frac{5}{11}$

8 0

3 years ago

Please help me on this

ozzi

Answer:

q=371

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers