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3 years ago

Help mee plzz will give brainliest

Mathematics

2 answers:

Alinara [238K]3 years ago

8 0

Answer: X=12
Explanation: 180-72= 108. 108 divided by 9 is 12

Hitman42 [59]3 years ago

5 0

Answer:

Answer: X=12

Explanation: 180-72= 108. 108/9 = 12

Step-by-step explanation:

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Solve for n <br> n + 6 = 12

zepelin [54]

Answer:

n=6

Step-by-step explanation:

12-6=6

4 0

3 years ago

Read 2 more answers

Number 1. Write an Equation for this situation Amanda is training for a marathon she sets the pace so that every mile takes eigh

guajiro [1.7K]

ANSWER!

Number 1: t=8m

Number 2: 2.50d=x

7 0

3 years ago

Michaela works in a bank she is paid £19200 per annum how much does she get per month

Step2247 [10]

in a year she earns=19200

So in a month she earns=19200/12

=1600

So she earns 1600 every month.

If the answer is correct then make me brain list.please

4 0

3 years ago

Given: m∠EYL=72°<br> Find: mEHL , mLVE

White raven [17]

Answer:

Part 1) The measure of arc EHL is $108\°$

Part 2) The measure of angle LVE is $54\°$

Step-by-step explanation:

step 1

Let

x-----> the measure of arc EHL

y----> the measure of arc EVL

we know that

The measurement of the outer angle is the semi-difference of the arcs it encompasses.

$m$

we have

$m$

substitute

$72\°=\frac{1}{2}(y-x)$

$144\°=(y-x)$

$y=144\°+x$ ------> equation A

Remember that

$x+y=360\°$ -----> equation B ( complete circle)

substitute equation A in equation B and solve for x

$x+(144\°+x)=360\°$

$2x=360\°-144\°$

$x=216\°/2=108\°$

Find the value of y

$y=144\°+x$

$y=144\°+108\°=252\°$

therefore

The measure of arc EHL is $108\°$

The measure of arc EVL is $252\°$

step 2

Find the measure of angle LVE

we know that

The inscribed angle measures half that of the arc comprising

Let

x-----> the measure of arc EHL

$m$

we have

$x=108\°$

substitute

$m$

3 0

3 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:

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

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

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

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

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

6 0

3 years ago