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3 years ago

Solve (x-5)^2=3 apex!!!

Mathematics

2 answers:

slega [8]3 years ago

8 0

The answer would be 6.5

Anna71 [15]3 years ago

7 0

Answer:

x = 5 + √3 and x = 5 - √3

Step-by-step explanation:

First take the square root of both sides...

x - 5 = ±√3

Now add 5 to both sides....

x = 5 ± √3

so x = 5 + √3 and x = 5 - √3

You might be interested in

3.3^(2x+1)-103^x+1=0 need value of x

photoshop1234 [79]

Answer:

The value of $x$ is approximately -1.531.

Step-by-step explanation:

Let $3.3^{2\cdot x + 1}-103^{x+1} = 0$ , we proceed to solve this expression by algebraic means:

1) $3.3^{2\cdot x + 1}-103^{x+1} = 0$ Given

2) $3.3^{2\cdot x}\cdot 3.3 -103^{x}\cdot 103 = 0$ $a^{b}\cdot a^{c} = a^{b+c}$

3) $(3.3^{x})^{2}\cdot 3.3 -\left[\left( \sqrt{103} \right)^{2}\right]^{x}\cdot 103 = 0$ $(a^{b})^{c} = a^{b\cdot c}$

4) $(3.3^{x})^{2}\cdot 3.3 - \left[\left(\sqrt{103}\right)^{x}\right]^{2}\cdot 103 = 0$ $(a^{b})^{c} = a^{b\cdot c}$ /Commutative property

5) $\left[\left(\frac{3.3}{\sqrt{103}}\right)^{x}\right] ^{2}-\frac{103}{3.3} = 0$ Existence of multiplicative inverse/Definition of division/Modulative property/ $a^{b}\cdot a^{c} = a^{b+c}$

6) $\left(\frac{3.3}{\sqrt{103}} \right)^{2\cdot x}=\frac{103}{3.3}$ Existence of additive inverse/Modulative property/ $(a^{b})^{c} = a^{b\cdot c}$

7) $\log \left(\frac{3.3}{\sqrt{103}} \right)^{2\cdot x}=\log \frac{103}{3.3}$ Definition of logarithm.

8) $2\cdot x\cdot \log \left(\frac{3.3}{\sqrt{103}} \right)= \log \frac{103}{3.3}$ $\log_{b} a^{c} = c\cdot \log_{b} a$

9) $2\cdot x \cdot [\log 3.3-\log \sqrt{103}] = \log 103 - \log 3.3$ $\log_{b} \frac{a}{d}$

10) $x\cdot (2\cdot \log 3.3-\log 103) = \log 103 - \log 3.3$ $\log_{b} a^{c} = c\cdot \log_{b} a$ /Associative property

11) $x = \frac{\log 103-\log 3.3}{2\cdot \log 3.3-\log 103}$ Existence of multiplicative inverse/Definition of division/Modulative property

12) $x \approx -1.531$ Result

The value of $x$ is approximately -1.531.

4 0

3 years ago

Can someone please solve FAST I have 5 minutes

Hunter-Best [27]

Answer:

no time left sorry

Step-by-step explanation:

7 0

3 years ago

The perimeter of the base of a square pyramid is 16 feet and it stands 20 ft tall. Calculate the volume.

Vikentia [17]

Answer: $1706.67 feet$

Step-by-step explanation:

Formula: Volume = $\frac{1}{3} \alpha^{2}h$

a is the length of the base's edge and h is the height. Since 16 is the length we would square it as it is the length, then we would multiply it by the height which is 20 ft.

<u>Steps: Multiply</u>

Volume = $\frac{1}{3} \alpha^{2}h$

= $\frac{1}{3} * 16^{2} * 20$