QuestionIf $(a+2,63)=\left(-1, b^{3}-1\right)$ , find $\sqrt{a^{2}+b^{2}}=\ldots$ (a) 25 (b) 7 (c) 5 (d) $\pm 5$

Studdy Solution

STEP 1

Assumptions1. The given coordinates are $(a+,63)$ and $(-1, b^{3}-1)$ . We are asked to find the value of $\sqrt{a^{}+b^{}}$

STEP 2

Since the given pairs are coordinates, we can equate the x-coordinates and the y-coordinates separately.
For the x-coordinates, we have $a +2 = -1$

STEP 3

olving the equation for $a$ , we get $a = -1 -2$

STEP 4

Calculate the value of $a$ .
$a = -1 -2 = -3$

STEP 5

For the y-coordinates, we have $63 = b^{3} -1$

STEP 6

olving the equation for $b^{3}$ , we get $b^{3} =63 +1$

STEP 7

Calculate the value of $b^{3}$ .
$b^{3} =63 +1 =64$

STEP 8

To find the value of $b$ , we take the cube root of $b^{3}$ .
$b = \sqrt[3]{b^{3}}$

STEP 9

Substitute the value of $b^{3}$ into the equation to find $b$ .
$b = \sqrt[3]{64}$

STEP 10

Calculate the value of $b$ .
$b = \sqrt[3]{64} =4$

STEP 11

Now that we have the values of $a$ and $b$ , we can substitute these into the equation $\sqrt{a^{}+b^{}}$ .
$\sqrt{a^{}+b^{}} = \sqrt{(-3)^{}+4^{}}$

STEP 12

implify the equation.
$\sqrt{a^{2}+b^{2}} = \sqrt{9+16}$

STEP 13

Calculate the value inside the square root.
$\sqrt{a^{2}+b^{2}} = \sqrt{25}$

STEP 14

Finally, calculate the square root of25.
$\sqrt{a^{2}+b^{2}} = \sqrt{25} =$ So, $\sqrt{a^{2}+b^{2}} =$ . The correct answer is (c).

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QuestionIf $(a+2,63)=\left(-1, b^{3}-1\right)$ , find $\sqrt{a^{2}+b^{2}}=\ldots$ (a) 25 (b) 7 (c) 5 (d) $\pm 5$

Studdy Solution

STEP 1

Assumptions1. The given coordinates are $(a+,63)$ and $(-1, b^{3}-1)$ . We are asked to find the value of $\sqrt{a^{}+b^{}}$

STEP 2

Since the given pairs are coordinates, we can equate the x-coordinates and the y-coordinates separately.
For the x-coordinates, we have $a +2 = -1$

STEP 3

olving the equation for $a$ , we get $a = -1 -2$

STEP 4

Calculate the value of $a$ .
$a = -1 -2 = -3$

STEP 5

For the y-coordinates, we have $63 = b^{3} -1$

STEP 6

olving the equation for $b^{3}$ , we get $b^{3} =63 +1$

STEP 7

Calculate the value of $b^{3}$ .
$b^{3} =63 +1 =64$

STEP 8

To find the value of $b$ , we take the cube root of $b^{3}$ .
$b = \sqrt[3]{b^{3}}$

STEP 9

Substitute the value of $b^{3}$ into the equation to find $b$ .
$b = \sqrt[3]{64}$

STEP 10

Calculate the value of $b$ .
$b = \sqrt[3]{64} =4$

STEP 11

Now that we have the values of $a$ and $b$ , we can substitute these into the equation $\sqrt{a^{}+b^{}}$ .
$\sqrt{a^{}+b^{}} = \sqrt{(-3)^{}+4^{}}$

STEP 12

implify the equation.
$\sqrt{a^{2}+b^{2}} = \sqrt{9+16}$

STEP 13

Calculate the value inside the square root.
$\sqrt{a^{2}+b^{2}} = \sqrt{25}$

STEP 14

Finally, calculate the square root of25.
$\sqrt{a^{2}+b^{2}} = \sqrt{25} =$ So, $\sqrt{a^{2}+b^{2}} =$ . The correct answer is (c).

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QuestionIf (a+2,63)=(−1,b3−1)(a+2,63)=\left(-1, b^{3}-1\right)(a+2,63)=(−1,b3−1), find a2+b2=…\sqrt{a^{2}+b^{2}}=\ldotsa2+b2​=… (a) 25 (b) 7 (c) 5 (d) ±5\pm 5±5

Studdy Solution

STEP 1

Assumptions1. The given coordinates are (a+,63)(a+,63)(a+,63) and (−1,b3−1)(-1, b^{3}-1)(−1,b3−1). We are asked to find the value of a+b\sqrt{a^{}+b^{}}a+b​

STEP 2

Since the given pairs are coordinates, we can equate the x-coordinates and the y-coordinates separately.For the x-coordinates, we havea+2=−1a +2 = -1a+2=−1

STEP 3

olving the equation for aaa, we geta=−1−2a = -1 -2a=−1−2

STEP 4

Calculate the value of aaa.a=−1−2=−3a = -1 -2 = -3a=−1−2=−3

STEP 5

For the y-coordinates, we have63=b3−163 = b^{3} -163=b3−1

STEP 6

olving the equation for b3b^{3}b3, we getb3=63+1b^{3} =63 +1b3=63+1

STEP 7

Calculate the value of b3b^{3}b3.b3=63+1=64b^{3} =63 +1 =64b3=63+1=64

STEP 8

To find the value of bbb, we take the cube root of b3b^{3}b3.b=b33b = \sqrt[3]{b^{3}}b=3b3​

STEP 9

Substitute the value of b3b^{3}b3 into the equation to find bbb.b=643b = \sqrt[3]{64}b=364​

STEP 10

Calculate the value of bbb.b=643=4b = \sqrt[3]{64} =4b=364​=4

STEP 11

Now that we have the values of aaa and bbb, we can substitute these into the equation a+b\sqrt{a^{}+b^{}}a+b​.a+b=(−3)+4\sqrt{a^{}+b^{}} = \sqrt{(-3)^{}+4^{}}a+b​=(−3)+4​

STEP 12

implify the equation.a2+b2=9+16\sqrt{a^{2}+b^{2}} = \sqrt{9+16}a2+b2​=9+16​

STEP 13

Calculate the value inside the square root.a2+b2=25\sqrt{a^{2}+b^{2}} = \sqrt{25}a2+b2​=25​

STEP 14

Finally, calculate the square root of25.a2+b2=25=\sqrt{a^{2}+b^{2}} = \sqrt{25} =a2+b2​=25​=So, a2+b2=\sqrt{a^{2}+b^{2}} =a2+b2​=. The correct answer is (c).

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QuestionIf $(a+2,63)=\left(-1, b^{3}-1\right)$ , find $\sqrt{a^{2}+b^{2}}=\ldots$ (a) 25 (b) 7 (c) 5 (d) $\pm 5$

Assumptions1. The given coordinates are $(a+,63)$ and $(-1, b^{3}-1)$ . We are asked to find the value of $\sqrt{a^{}+b^{}}$

Since the given pairs are coordinates, we can equate the x-coordinates and the y-coordinates separately.
For the x-coordinates, we have $a +2 = -1$

olving the equation for $a$ , we get $a = -1 -2$

Calculate the value of $a$ .
$a = -1 -2 = -3$

For the y-coordinates, we have $63 = b^{3} -1$

olving the equation for $b^{3}$ , we get $b^{3} =63 +1$

Calculate the value of $b^{3}$ .
$b^{3} =63 +1 =64$

To find the value of $b$ , we take the cube root of $b^{3}$ .
$b = \sqrt[3]{b^{3}}$

Substitute the value of $b^{3}$ into the equation to find $b$ .
$b = \sqrt[3]{64}$

Calculate the value of $b$ .
$b = \sqrt[3]{64} =4$

Now that we have the values of $a$ and $b$ , we can substitute these into the equation $\sqrt{a^{}+b^{}}$ .
$\sqrt{a^{}+b^{}} = \sqrt{(-3)^{}+4^{}}$

implify the equation.
$\sqrt{a^{2}+b^{2}} = \sqrt{9+16}$

Calculate the value inside the square root.
$\sqrt{a^{2}+b^{2}} = \sqrt{25}$

Finally, calculate the square root of25.
$\sqrt{a^{2}+b^{2}} = \sqrt{25} =$ So, $\sqrt{a^{2}+b^{2}} =$ . The correct answer is (c).