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2 years ago

What is the equation that has a slope of -4 and passes through the coordinates (-4,14)

Mathematics

1 answer:

Agata [3.3K]2 years ago

4 0

Answer:

y=-4x-2

Step-by-step explanation:

y-y1=m(x-x1)

y-14=-4(x-(-4))

y-14=-4(x+4)

y=-4x-16+14

y=-4x-2

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Help plz ty gl please

denpristay [2]

It was difficult but finally i get there where 4 womens because there where 29 mens 9 swiming 6 plaiyng squash the other 14 in the gym so
18-14=4 the number of womens in the gym

5 0

2 years ago

Can someone solve this with steps cause idk how to solve the radicals

IceJOKER [234]

$\bf 343^{\frac{2}{3}}+36^{\frac{1}{2}}-256^{\frac{3}{4}}\qquad \begin{cases}
343=7\cdot 7\cdot 7\\
\qquad 7^3\\
36=6\cdot 6\\
\qquad 6^2\\
256=4\cdot 4\cdot 4\cdot 4\\
\qquad 4^4
\end{cases}\\\\\\ (7^3)^{\frac{2}{3}}+(6^2)^{\frac{1}{2}}-(4^4)^{\frac{3}{4}}
\\\\\\
\sqrt[3]{(7^3)^2}+\sqrt[2]{(6^2)^1}-\sqrt[4]{(4^4)^3}\implies \sqrt[3]{(7^2)^3}+\sqrt[2]{(6^1)^2}-\sqrt[4]{(4^3)^4}
\\\\\\
7^2+6-4^3\implies 49+6-64\implies -9$

to see what you can take out of the radical, you can always do a quick "prime factoring" of the values, that way you can break it in factors to see who is what.

8 0

3 years ago

The central angle of a circle is equal in measure to one radian when the corresponding arc length is equal to which of the follo

irinina [24]

The formula to find the arc length L is
L = r*theta
where r is the radius and theta is the central angle in radians (this formula will not work if theta is in degrees)

If the central angle is 1 radian, then theta = 1 and
L = r*theta
L = r*1
L = r
So the arc length is the same as the radius

Answer: Choice A) The radius of the circle

5 0

3 years ago

Read 2 more answers

Q1. A fluorescent lamp takes in energy from the electricity supply at the rate of 18 joules each second. It gives out 3 joules o

Dmitrij [34]

Answer:

15 joules

Step-by-step explanation:

SUBTRACT 3 FROM 18

18 JOULES-3 JOULES=15 JOULES

HOPE IT HELPS

8 0

2 years ago

Sam entered three functions into a graphing calculator, y1 = x + 2, y2 = x2 + 2, and y3 = 2x. A portion of the table created by

V125BC [204]

For this case we have the following functions:
$y1 = x + 2

y2 = x ^ 2 + 2

y3 = 2 ^ x$

When x = 0 we have:
For y1:
$y1 = 0 + 2

y1 = 2$
For y2:
$y2 = 0 ^ 2 + 2

y2 = 0 + 2

y2 = 2$
Therefore, we have to:
$y1 = y2
$

When x = 5 we have:
For y2:
$y2 = 5 ^ 2 + 2

y2 = 25 + 2

y2 = 27$
For y3:
$y3 = 2 ^ 5

y3 = 32$
Therefore, we have to:
$y2 \ \textless \ y3
$

When x = -1 we have:
For y1:
$y1 = -1 + 2

y1 = 1$
For y2:
$y2 = (-1) ^ 2 + 2

y2 = 1 + 2

y2 = 3
$
For y3:
$y3 = 2 ^ {-1}

y3 = 1/2

y3 = 0.5$
Therefore, we have to:
$y3 \ \textless \ y1 \ \textless \ y2
$

Answer:
When x = 0, y1 = y2
When x = 5, y2 <y3
When x = -1, y3 <y1 <y2

4 0

3 years ago