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2 years ago

Plz help quick giving 14 point if it’s right :)

Mathematics

1 answer:

Ivanshal [37]2 years ago

5 0

Through the muscle contractions, or A

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Consider the expression 25 – 10 ÷ 2 + 3.

Tems11 [23]

Given:

The expression is:

$25-10\div 2+3$

To find:

Part A: The expression using parentheses so that the expression equals 23.

Part B: The expression using parentheses so that the expression equals 3.

Solution:

Part A:

In option A,

$(25-10)\div 2+3=15\div 2+3$

$(25-10)\div 2+3=7.5+3$ [Using BODMAS]

$(25-10)\div 2+3=10.5$

In option B,

$25-10\div (2+3)=25-10\div 5$

$25-10\div (2+3)=25-2$ [Using BODMAS]

$25-10\div (2+3)=23$

In option C,

$(25-10)\div (2+3)=15\div 5$

$(25-10)\div (2+3)=3$

In option D,

$25-(10\div 2)+3=25-5+3$

$25-(10\div 2)+3=28-5$ [Using BODMAS]

$25-(10\div 2)+3=23$

After the calculation, we have $25-10\div (2+3)=23$ and $25-(10\div 2)+3=23$ .

Therefore, the correct options are B and D.

Part B: From part A, it is clear that

$(25-10)\div (2+3)=3$

Therefore, the correct option is C.

8 0

2 years ago

Remember to show work and explain. Use the math font.

MrMuchimi

Answer:

$\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}$

Step-by-step explanation:

$\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================$

$1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}$

$\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)$

$--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)$

$--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}$