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wel
3 years ago
14

Find the value of given expression

tle=" \sqrt{35 + 1} " alt=" \sqrt{35 + 1} " align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
aliya0001 [1]3 years ago
6 0
Answer is 6 if you do 35+1 that gives you 36, the square root of 36 is 6 .6•6=36 so that’s your answer 6
Simora [160]3 years ago
4 0

Answer and Step-by-step explanation:

35 + 1 = 36

\sqrt{36} (The square root of 36) is equal to 6.

6 is the answer.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

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4 0
3 years ago
Read 2 more answers
Please don't answer wrong.​
Tamiku [17]

Answer:

<h3><u>Part (a)</u></h3>

<u />

<u>Equation of a circle</u>

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

where:

  • (a, b) is the center
  • r is the radius

Given equation:  x^2+y^2=6.25

Comparing the given equation with the general equation of a circle, the given equation is a <u>circle</u> with:

  • center = (0, 0)
  • radius = \sqrt{6.25}=2.5

To draw the circle, place the point of a compass on the origin. Make the width of the compass 2.5 units, then draw a circle about the origin.

<h3><u>Part (b)</u></h3>

Given equation:  x+y=1.5

Rearrange the given equation to make y the subject:  y=-x+1.5

Find two points on the line:

x=-2 \implies -(-2)+1.5=3.5\implies (-2,3.5)

x=2 \implies -(2)+1.5\implies-0.5\implies (2,-0.5)

Plot the found points and draw a straight line through them.

The <u>points of intersection</u> of the circle and the straight line are the solutions to the equation.

To solve this algebraically, substitute  y=-x+1.5  into the equation of the circle to create a quadratic:

\implies x^2+(-x+1.5)^2=6.25

\implies x^2+x^2-3x+2.25=6.25

\implies 2x^2-3x-4=0

Now use the quadratic formula to solve for x:

x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0

\implies x=\dfrac{-(-3) \pm \sqrt{(-3)^2-4(2)(-4)}}{2(2)}

\implies x=\dfrac{3 \pm \sqrt{41}}{4}

To find the coordinates of the points of intersection, substitute the found values of x into y=-x+1.5

\implies y=-\left(\dfrac{3 + \sqrt{41}}{4}\right)+1.5=\dfrac{3-\sqrt{41}}{4}

\implies y=-\left(\dfrac{3 - \sqrt{41}}{4}\right)+1.5=\dfrac{3+\sqrt{41}}{4}

Therefore, the two points of intersection are:

\left(\dfrac{3 + \sqrt{41}}{4},\dfrac{3-\sqrt{41}}{4}\right) \textsf{ and }\left(\dfrac{3 - \sqrt{41}}{4},\dfrac{3+\sqrt{41}}{4}\right)

Or as decimals to 2 d.p.:

(2.35, -0.85) and (-0.85, 2.35)

6 0
2 years ago
What is 14 1/6 minus 7 1/3
Levart [38]
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6 0
3 years ago
What is the standard form of 5.93-5
vredina [299]
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4 0
3 years ago
Consider U = {x\x is a real number}
zysi [14]

Answer:

The fourth pair of statement is true.

9∈A, and 9∈B.

Step-by-step explanation:

Given that,

U={x|  x is real number}

A={x| x∈ U and x+2>10}

B={x| x∈ U and 2x>10}

  • 5 ∈ A , 5 ∈ B

If 5∈ A, Then it will be satisfies x+2>10 , but 5+2<10.

Similarly, If 5∈ B, Then it will be satisfies 2x>10 , but 2.5=10.

So, 5∉A, and 5∉B.

  • 6 ∈ A , 6 ∈ B

If 6∈ A, Then it will be satisfies x+2>10 , but 6+2<10.

Similarly, If 6∈ B, Then it will be satisfies 2x>10 , and 2.6=12>10.

So, 6∉A, and 6∈B.

  • 8 ∈ A , 8 ∈ B

If 8∈ A, Then it will be satisfies x+2>10 , but 8+2=10.

Similarly, If 8∈ B, Then it will be satisfies 2x>10.  2.8=16>10.

So, 8∉A, and 8∈B.

  • 9 ∈ A , 9 ∈ B

If 9∈ A, Then it will be satisfies x+2>10 , but 9+2=11>10.

Similarly, If 9∈ B, Then it will be satisfies 2x>10.  2.9=18>10.

So, 9∈A, and 9∈B.

3 0
3 years ago
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