All categories

2 years ago

7(x-3)+6=5x-9 please solve the equation fully

Mathematics

1 answer:

TiliK225 [7]2 years ago

4 0

Answer:

I love algebra anyways

The ans is x=3 steps is in the picture with the ans

(hope this helps can i plz have brainlist :D hehe)

Step-by-step explanation:

You might be interested in

Due 4/29/22. add explanation to please if you can, so that i can understand better

vagabundo [1.1K]

Answer:

240 ÷ 5 = 48

Please Mark Brainliest If This Helped!

8 0

2 years ago

Greg is designing a clock face. Using the center of the clock face as the origin, he keeps its diameter at 10 units. Match the p

Nat2105 [25]

The center of the clock is taken as the origin.The clock is a circle with a diameter 10 units.Radius is half the diameter .Radius = 10 ÷2= 5 units.

The clock is divided in four quadrants .On x axis y=0 and on y axis x=0.

When it is 12 o'clock the hour hand is on positive of y axis.Coordinates of the point at 12 o'clock=(0,5)

When it is 3 o 'clock the hour hand is on positive of x axis .Coordinate of the point at 3o'clock is (5,0)

When it is 6 o'clock the hour hand is on negative of y axis .The coordinates of the point at 6o'clock is (0,-5)

At 9o'clock the hour hand is on negative of x axis .The coordinate of the point at 6o'clock is(-5,0)

3 0

3 years ago

Read 2 more answers

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

HELP ME PLEASE<br><br> SJSJSJJSJSJSJSJ

djverab [1.8K]

Answer:

24.12-9.8=24.32

Step-by-step explanation:

hope this helps, sorry I'm bad at explaining, but just trust me it is this im 100% positive.

7 0

2 years ago

Read 2 more answers

The diagonals of an isosceles trapezoid are 3x + 7 and 5x – 11. What is the value of x?

Alja [10]

In a trapazoid the diagonals are the same so therefore the two equations equal echother

3x+7=5x-11 solve for x by combining like terms

18=2x then divide to unto multiplication

9=x

9 is the value of x

I hope I've helped!

7 0

3 years ago

Read 2 more answers