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3 years ago

SOMEONE PLISS ANSWER THIS... I NEED TO HAND IT IN NOW

Mathematics

1 answer:

lana [24]3 years ago

8 0

Answer:

a) 151

b) 160₉

c)i) 30

c)ii) 21

Step-by-step explanation:

(a) The value of the digits 3 and 4 is the value of the number with the other digits set to zero.

304₇ = 3×7² +0×7¹ +4×7⁰ = 3×49 +4 = 151₁₀

The total value of the digits 3 and 4 is 151.

(b) We can make a short table for conversions.

10₁₀ = 11₉

100₁₀ = (11₉)² = 121₉

So,

135₁₀ = 1×121₉ +3×11₉ +5 = 121₉ +33₉ +5 = 154₉ +5 = 160₉

The base-9 equivalent of 135 is 160₉.

(ci) 110₅ = 1×5² +1×5¹ +0×5⁰ = 25 +5 = 30₁₀

Lisa is 30 in base 10.

(cii) 110₅ -14₅ = 41₅ = 4×5¹ +1×5⁰ = 20+1 = 21₁₀

Lisa's younger brother is 21 in base 10.

Alternate solution

14₅ = 5+4 = 9₁₀ . . . now we can find the difference in base 10:

30 -9 = 21

_____

Additional comment

It helps to understand place value and how the digit positions relate to the powers of the number base. I find "expanded form using exponents" to be helpful, as you can see in the answers above. The relationships between place value and powers of the base hold true for any number base, and they extend to either side of the "decimal point."

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In two or more complete sentences, explain how to solve the cube root equation, <img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B3

nordsb [41]

Isolate the root expression:

$\sqrt[3]{x+1}+2=0\implies\sqrt[3]{x+1}=-2$

Take the third power of both sides:

$\sqrt[3]{x+1}=-2\implies(\sqrt[3]{x+1})^3=(-2)^3$

Simplify:

$(\sqrt[3]{x+1})^3=(-2)^3\implies x+1=-8$

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Since the cube root function is bijective, we know this won't be an extraneous solution, but it doesn't hurt to verify that this is correct. When $x=-9$ , we have

$\sqrt[3]{-9+1}=\sqrt[3]{-8}=\sqrt[3]{(-2)^3}=-2$

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3 years ago

The sum of the first two terms of an infinite GP is 6 and each term is 5 times the sum of the succeeding termsthen the second te

kozerog [31]

Answer:

$\dfrac{6}{7}$ .

Step-by-step explanation:

It is given that the sum of first two terms of an infinite GP is 6.

nth term of a GP is

$a_n=ar^{n-1}$

$a+ar=6$ ...(1)

Each term is 5 times the sum of the succeeding terms.

$a_n=5(a_{n+1}+a_{n+2}+...+\infty)$

$ar^{n-1}=5(ar^n+ar^{n+1}+...+\infty)$

$ar^{n-1}=5ar^n(1+r+r^2+...+\infty)$

Divide both sides by a.

$r^{n-1}=5r^n(\dfrac{1}{1-r})$ $[\text{Sum of infinite GP}=\dfrac{a}{1-r}]$

$\dfrac{r^n}{r}=\dfrac{5r^n}{1-r}$

$\dfrac{1}{r}=\dfrac{5}{1-r}$

$1-r=5r$

$1=6r$

$\dfrac{1}{6}=r$

The common ratio is 1/6.

Put r=1/6 in (1).

$a+a(\dfrac{1}{6})=6$

$6a+a=36$

$7a=36$

$a=\dfrac{36}{7}$

Second term $ar=\dfrac{36}{7}\times \dfrac{1}{6}=\dfrac{6}{7}$

Therefore, the second term is $\dfrac{6}{7}$ .

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3 years ago

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2 years ago

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We can use PEMDAS to help us:

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E - Exponents

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3 years ago

SOMEONE PLISS ANSWER THIS... I NEED TO HAND IT IN NOW​

SOMEONE PLISS ANSWER THIS... I NEED TO HAND IT IN NOW