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2 years ago

The diameter of a peacan pie is 7 inches. What is the circumference?

Mathematics

1 answer:

Zina [86]2 years ago

3 0

Answer:

21.98/C

Step-by-step explanation:

The equation for circumference is pi(d).

so 3.14*7 is about 21.98

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Slope intercept form of a line that has a slope of -3/4 and passes through (4, 3)

tia_tia [17]

<h3>Explanation</h3>

First Method

We have the given slope value and the coordinate point that the graph passes through.

$y = mx + b$

where m = slope and b = y-intercept. Substitute the value of slope in the equation.

$y = - \frac{3}{4} x + b$

We have the given coordinate point as well. After we substitute the slope, we substitute the coordinate point value in the equation.

$3 = - \frac{3}{4} (4) + b \\$

Solve the equation for b-term

$3 = - 3 + b \\ 3 + 3 = b \\ 6 = b$

The value of b is 6. We substitute the value of b in the equation.

$y = - \frac{3}{4} x + 6$

Second Method

We can also use the Point-Slope form to solve the question.

$y - y_1 = m(x - x_1)$

Given the y1 and x1 = the coordinate point value.

Substitute the slope and coordinate point value in the point slope form.

$y - 3 = - \frac{3}{4} (x - 4)$

Simplify/Convert into Slope-intercept

$y = - \frac{3}{4} (x - 4) + 3 \\ y = - \frac{3}{4} x + \frac{12}{4} + 3 \\ y = - \frac{3}{4} x + 3 + 3 \\y = - \frac{3}{4} x + 6$

<h3>Answer</h3>

 $\large \boxed {y = - \frac{3}{4} x + 6}$ 

6 0

2 years ago

Find 2 a for a = 3 1/4 1) 6 1/2 2) 6 1/4 3) 23 1/4 4) 5 1/2

Nataly_w [17]

Answer:

1) 6 1/2

Step-by-step explanation:

Put 3 1/4 where "a" is and do the arithmetic.

2(3 1/4) = 2·3 + 2·(1/4) = 6 2/4 = 6 1/2

4 0

3 years ago

X2 - 6x + 12 and y = 2x - 4, algebraically are

olga_2 [115]

The solution is (4, 4)

Solution:

Given system of equations are:

$y = x^2 - 6x + 12 ------ eqn\ 1\\\\y = 2x - 4 ---------- eqn\ 2$

Substitute eqn 2 in eqn 1

$x^2 - 6x + 12 = 2x - 4$

Make the right side of equation 0

$x^2 - 6x + 12 - 2x + 4 = 0\\\\x^2 -8x + 16 = 0$

Solve by quadratic equation

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=-8,\:c=16:\\\\x=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:16}}{2\cdot \:1}\\\\x=\frac{-\left(-8\right)\pm \sqrt{0}}{2\cdot \:1}\\\\x = \frac{8}{2}\\\\x = 4$