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Masja [62]
3 years ago
15

What is the equation of the line that passes through the points (2,-4) (-4,5) SHOW WORK

Mathematics
1 answer:
aliya0001 [1]3 years ago
8 0

Answer:

y=−3/2x−1

Step-by-step explanation:

hope this helps brainliest appreciated

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If the sin P = √2/2, what is the value of cos P
ArbitrLikvidat [17]

Answer:

√2/2

Step-by-step explanation:

If sinP = √2/2,

Rationalize;

√2/2

= √2/2,*√2/√2

√4/2√2

= 2/2√2

= 1/√2

Hence sinP = 1/√2

Take sin inverse of both sies

arcsinP = arcsin1/√2

P = 45 degrees

cos P =  Cos 45

= 1/√2

= 1/√2 * √2/√2

= √2/2

Hence the value of cosP is √2/2

5 0
3 years ago
Solve the equation for all values of x.<br> (8x+9)(49x^2-100)=0
Liono4ka [1.6K]

Answer:

x = \frac{-9}{8} , \frac{-10}{7} , \frac{10}{7}

Step-by-step explanation:

8x+9=0

8x=-9

x = \frac{-9}{8}

or

49x^{2}-100 = 0

(7x)^{2} -10^{2} =0,

(7x+10)(7x-10)=0

7x+10=0, 7x-10=0

7x=-10, 7x=10

x=-\frac{-10}{7}, x = \frac{10}{7}

4 0
3 years ago
In a class, 42 students, there are 8 more boys than girls write and solve a system of 3 equations
zhenek [66]

Answer:

x = 17

Step-by-step explanation:

x + x + 8 = 42     because x = boys and x + 8 = girls

2x = 34      because 2x = 42 - 8

x = 17            



i hope this helps!  you don't need the 2x = 42 - 8 to be an equation      ;)

8 0
3 years ago
A circle has a circumference of 907.46907.46 units. What is the diameter of the circle?
astra-53 [7]

question is wrong sir

there is no any value with two points

6 0
3 years ago
Find the derivative of the function at P 0 in the direction of A. ​f(x,y,z) = 3 e^x cos(yz)​, P0 (0, 0, 0), A = - i + 2 j + 3k
Alik [6]

The derivative of f(x,y,z) at a point p_0=(x_0,y_0,z_0) in the direction of a vector \vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k is

\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}

We have

f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k

and

\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}

Then the derivative at p_0 in the direction of \vec a is

3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}

3 0
4 years ago
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