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BigorU [14]
3 years ago
13

Please help i’ll give brainliest if you help thanks

Mathematics
1 answer:
Sphinxa [80]3 years ago
6 0
The answer is d.

Hope this helps
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Graph the line with 3/4 slope<br> passing through the point (2,-2).
S_A_V [24]

Answer:

Step-by-step explanation

x1 = 2, y1 = -2 and m = 3/4

y - y1 = m(x - x1)

y - (-2) = 3/4(x - 2)

y + 2 = 3/4(x - 2)

Multiply each term by 4

4y + 8 = 3(x - 2)

4y + 8 = 3x - 6

4y = 3x - 6 - 8

4y = 3x - 14

5 0
3 years ago
What is the area of a regular hexagon with an apothem 18.5 inches long and a side 21 inches
earnstyle [38]

The area of a regular hexagon with an apothem 18.5 inches long and a side 21 inches is 1, 165. 5 In²

<h3>How to calculate the area of a regular hexagon</h3>

The formula is given thus;

Area of hexagon = (1/2) × a × P

where a = the length of the apothem

P = perimeter of the hexagon

Given a = 18. 5 inches

Note that Perimeter, p = 6a with 'a' as side

p = 6 × 21 = 126 inches

Substitute values into the formula

Area, A = 1 ÷2 × 18. 5 × 126 = 1 ÷2 × 2331 = 1, 165. 5 In²

Thus, the area of the regular hexagon is 1, 165. 5 In²

Learn more about the area of a hexagon here:

brainly.com/question/15424654

#SPJ1

3 0
2 years ago
To solve x-8=16 cassie added 8 to both side of the equation which property justifies this step?
JulijaS [17]

Answer: Addition Property Of Equality


Step-by-step explanation:


4 0
3 years ago
Which of the following sets of side lengths could be those of triangle LMN?
allsm [11]
Answer is D
LMN is similar to PQR
its side are just 3 times greater than PQR scale factor is 3<span />
3 0
4 years ago
54. Find <img src="https://tex.z-dn.net/?f=y%5E%7B%5Cprime%7D" id="TexFormula1" title="y^{\prime}" alt="y^{\prime}" align="absmi
Stella [2.4K]

Step-by-step explanation:

Take the natural log of both sides:

ln ({x}^{y} ) = ln ({y}^{x} )

Logarithm rules allow you to bring down the exponents:

yln(x) = xln(y)

Now differentiate. We will have to implicitly differentiate 'y' since it is a function of 'x'. Both sides require the product rule:

\frac{dy}{dx} ln(x) +  \frac{y}{x}  = ln(y) +  \frac{x}{y}  \frac{dy}{dx}

Isolate the terms that have y' since that is what we want:

\frac{dy}{dx} ln(x) -  \frac{x}{y}  \frac{dy}{dx}= ln(y) - \frac{y}{x}

Factor out y' to get:

\frac{dy}{dx}( ln(x) -   \frac{x}{y})= ln(y) - \frac{y}{x}

Therefore:

\frac{dy}{dx} =  \frac{ln(y) -  \frac{y}{x} }{ln(x) -  \frac{x}{y} }

7 0
2 years ago
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