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sweet [91]
3 years ago
10

%5B3%5D%7B125%7D%20%20%2B%20%20%7B9%7D%5E%7B3%7D%20%20%3D%20....%7D%20" id="TexFormula1" title=" \bf{ {15}^{2} \times {3}^{2} \div \sqrt[3]{125} + {9}^{3} = ....} " alt=" \bf{ {15}^{2} \times {3}^{2} \div \sqrt[3]{125} + {9}^{3} = ....} " align="absmiddle" class="latex-formula">​
Mathematics
1 answer:
34kurt3 years ago
7 0

\\ \tt\Rrightarrow 15^2\times 3^2\div \sqrt[3]{125}+9^3

\\ \tt\Rrightarrow 3^25^2\times 3^2\div 5^1+3^33^3

\\ \tt\Rrightarrow 3^45^1\div 3^6

\\ \tt\Rrightarrow 3^{-2}5

\\ \tt\Rrightarrow \dfrac{1}{3^2}5

\\ \tt\Rrightarrow \dfrac{5}{9}

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Y=|x-2|-4<br> Can someone help me
aleksley [76]
Y= - x + 2 - 4
y = - x - 2
-y = x + 2
8 0
3 years ago
Does anybody know how to classify and graph the quadrilateral points L(1,2) M(3,3) N(5,2) P(3,1), I'm unsure how to graph these
iVinArrow [24]

Answer:

Step-by-step explanation:

Yeah it’s pretty easy

4 0
3 years ago
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Please help will mark brainliest! (random answers will be reported)
Akimi4 [234]

Answer:

value of x =

\sqrt{216}

Step-by-step explanation:

here's the solution,

the given triangle is a right angled triangle so, by applying pythagoras theorem,

=》

perpendicular {}^{2}  =  hypotenuse {}^{2}   - base {}^{2}

=》

{x}^{2}  =  {21}^{2}   -  {15}^{2}

=》

{x}^{2}  = 441 - 225

=》

{x}^{2}  = 216

=》

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5 0
3 years ago
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Graph the line and the circle (x-1)²+(y-2)²=2² and y=2x+2.
Temka [501]

Answer:

The graph of the line and the circle is in the attachment.

Step-by-step explanation:

In order to graph the circle, you have to use the circle equation formula to obtain the center and the radius.

(x-xo)² + (y-yo)² = r²

where the point (xo,yo) is the center and r is the radius.

Using the given formula, the center is:

xo=1, yo=2

(1,2)

And the radius is:

r=2

Now you have to graph the point (1,2) and draw a circle centered at that point, with a distance of 2 from the center.

To obtain the exact values of the intersection of the circle with the x-axis you have to replace y=0 in the equation and solve it for x. For the points of intersection with the y-axis you have to replace x=0 in the equation and solve it for y.

-For x=0, solving for y:

(0-1)²+(y-2)²=2²

1+(y-2)²=4

(y-2)²= 3

Applying the squareroot both sides

y-2 = ± √3

y1= -√3 +2

y2= √3 +2

P1(0, -√3 +2) and P2(0, √3 +2)

-For y=0

(x-1)²+(0-2)²=2²

(x-1)² + 4 = 4

(x-1)² = 0

Applying the squareroot both sides:

x-1 = √0

x-1=0

x=1

P3 (1,0)

In order to graph the linear equation, you have to obtain the intersection with the x-axis replacing y=0 in the equation and the intersection with the y-axis replacing x=0 in the equation. That way, you obtain two point of the line and then you have to trace the line containing those points.

-For x=0

y=2(0)+2 =2

P1 (0,2)

-For y=0

0= 2x+2

Adding -2 both sides

2x= -2

Dividing by 2 both sides

x= -1

P2(-1,0)

5 0
3 years ago
Answer ASAP <br> ————————-
11111nata11111 [884]

Answer:

it should be 680

Step-by-step explanation:

1 kilogram is 1000 grams

720*6=4320

5000-4320=680

8 0
3 years ago
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