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2 years ago

Which equation, in slope-intercept form, passes through (–2, 4) and has a

Mathematics

1 answer:

kipiarov [429]2 years ago

8 0

Answer:

y=3x+10

Step-by-step explanation:

First, put the equation into point-slope form.

y-y1=m(x-x1)

y-4=3(x+2)

Next, distribute 3 to (x+2) and simplify.

y-4=3x+6

y=3x+10

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Triangle ABC with coordinates A (3,-2), B (5,5), and C (-4, 2) is reflected across

son4ous [18]

Step-by-step explanation:

the formula of coordinates (x, y) that reflected across the x-axis : (x, y) => (x, -y)

so,

A(3, -2) => A'(3, 2)

B(5, 5) => B'(5, -5)

C(-4, 2) => C'(-4, -2)

5 0

2 years ago

Read 2 more answers

What is the solution? y = 3x - 5 <br> y = 1\3*+3

PSYCHO15rus [73]

Answer:

Step-by-step explanation:

y = 3x - 5

y = 1/3x + 3

3x - 5 = 1/3x + 3....multiply everything by 3 to get rid of the fraction

9x - 15 = x + 9

9x - x = 9 + 15

8x = 24

x = 24/8

x = 3

y = 3x - 5

y = 3(3) - 5

y = 9 - 5

y = 4

solution is : x = 3 and y = 4...or (3,4) <==

3 0

2 years ago

What is 3(2x+1)+2(x+3) Help please x

katrin [286]

Answer:

8x + 9

Step-by-step explanation:

Treat the question as 2 separate brackets

First expand 3(2x + 1) :

6x + 3

Then expand 2(x + 3) :

2x + 6

Now add the two answers together and simplify :

= 6x + 3 + 2x + 6

= 8x + 9

hope this helps :)

8 0

2 years ago

Read 2 more answers

Directions: Calculate the area of a circle using 3.14x the radius

Leokris [45]

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

As we know ~

Area of the circle is :

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

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<h3>Problem 1</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 4.4\div 2$

$\qquad \sf \dashrightarrow \:r = 2.2 \: mm$

Now find the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(2.2)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {4.84}^{}$

$\qquad \sf \dashrightarrow \:area \approx 15.2 \: \: mm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 2</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 3.7 \div 2$

$\qquad \sf \dashrightarrow \:r = 1.85 \: \: cm$

Bow, calculate the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (1.85) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 3.4225 {}^{}$

$\qquad \sf \dashrightarrow \:area \approx 10.75 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 3 </h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (8.3) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 68.89$

$\qquad \sf \dashrightarrow \:area \approx216.31 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 4</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 5.8 \div 2$

$\qquad \sf \dashrightarrow \:r = 2.9 \: \: yd$

now, let's calculate area ~

$\qquad \sf \dashrightarrow \:3.14 \times {(2.9)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 8.41$

$\qquad \sf \dashrightarrow \:area \approx26.41 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 5</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 1 \div 2$

$\qquad \sf \dashrightarrow \:r = 0.5 \: \: yd$

Now, let's calculate area ~

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (0.5) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 0.25$

$\qquad \sf \dashrightarrow \:area \approx0.785 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 6</h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(8)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 64$

$\qquad \sf \dashrightarrow \:area = 200.96 \: \: yd {}^{2}$

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8 0

2 years ago

The solution to 3x2 – 12x 24 = 0 is

zzz [600]

$3x^2 - 12x + 24 = 0 \\ x^2 - 4x + 8 = 0 \\ x= \frac{-b \pm \sqrt{ b^{2} -4ac} }{2a} 
$ ; where a = 1, b = -4 and c = 8
$x= \frac{-(-4) \pm \sqrt{ (-4)^{2} -4 \times 1 \times 8} }{2 \times 1} \\ = \frac{4 \pm \sqrt{ 16 -32} }{2} \\ =\frac{4 \pm \sqrt{ -16} }{2} \\ =\frac{4 \pm 4i }{2} \\ =2+2i \ or \ 2-2i$

7 0

3 years ago

Read 2 more answers