Pls help me with this

Mathematics

1 answer:

JulijaS [17]3 years ago

4 0

ayaw mag add eh ate oh kuya Step-by-step explanation:

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I really need help. Find a. Round to the nearest tenth.

valina [46]

Answer:

a ≈ 1.8

Step-by-step explanation:

a / sin (180 - 105 - 15)° = 2 /sin 105°

a = (2 /sin 105°) x sin 60°

a = (2 / 0.97) x 0.87

a = 1.79 (≈ 1.8)

6 0

3 years ago

Sarah borrowed some money from her father at a simple interest rate of 5.5% to pay for her college tuition. At the end of the ye

gavmur [86]

$\bf ~~~~~~ \textit{Simple Interest Earned Amount}
\\\\
A=P(1+rt)\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &\$2555\\
P=\textit{original amount deposited}\\
r=rate\to 5.5\%\to \frac{5.5}{100}\to &0.055\\
t=years\to &1
\end{cases}
\\\\\\
2555=P[1+(0.055)(1)]\implies \cfrac{2555}{1+(0.055)(1)}=P
\\\\\\
\cfrac{2555}{1.055}=P\implies 2421.800947867 \approx P$

4 0

3 years ago

What is the value of the shown expression? 33 x 3-2 A) -54 B) -243 C) 243 D) 3

alex41 [277]

We have the following general rule for exponents:

$a^n\cdot a^m=a^{n+m}$

then, in this case we have the following:

$3^3\cdot3^{-2}=3^{3+(-2)}=3^{3-2}=3^1=3$

therefore, the answer is 3

4 0

1 year ago

Use implicit differentiation to find the points where the parabola defined by x2−2xy+y2+4x−8y+20=0 has horizontal and vertical t

Komok [63]

Answer:

The parabola has a horizontal tangent line at the point (2,4)

The parabola has a vertical tangent line at the point (1,5)

Step-by-step explanation:

Ir order to perform the implicit differentiation, you have to differentiate with respect to x. Then, you have to use the conditions for horizontal and vertical tangent lines.

-To obtain horizontal tangent lines, the condition is:

$\frac{dy}{dx}=0$ (The slope is zero)

--To obtain vertical tangent lines, the condition is:

$\frac{dy}{dx}=\frac{1}{0}$ (The slope is undefined, therefore the denominator is set to zero)

Derivating respect to x:

$\frac{d(x^{2}-2xy+y^{2}+4x-8y+20)}{dx} = \frac{d(x^{2})}{dx}-2\frac{d(xy)}{dx}+\frac{d(y^{2})}{dx}+4\frac{dx}{dx}-8\frac{dy}{dx}+\frac{d(20)}{dx}$ = $2x -2(y+x\frac{dy}{dx})+2y\frac{dy}{dx}+4-8\frac{dy}{dx}= 0$