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Svetllana [295]
2 years ago
6

0/1 What is the equation of the line in y=mx+b form? (Photo)

Mathematics
1 answer:
laiz [17]2 years ago
6 0

Answer:

y=-15x+100

Step-by-step explanation:

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Order these values from least to greatest.
liubo4ka [24]

Answer:

1 . 16

2 . 4

3 . 64

4 . -4

The fourth one goes first and then the second one and then the first one and then the third one.

3 0
3 years ago
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Graph the description of each line A line has a slope of -3 and a y-intercept of 3.
Basile [38]

Answer:

The equation is y = -3 + 3

Step-by-step explanation:

It will go through (0,3) and it wil go up 3 right one.

5 0
2 years ago
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If f(x) = 2(x)^2+5 sqrt (x-2). complete the following statement. <br><br> f(2)=_____
dangina [55]

Answer:

The value of function at x = 2 is 8.

Step-by-step explanation:

Given : Function f(x)=2x^2+5\sqrt{x-2}

We have to find the value of given function at x = 2 , that is f(2)

Consider the given function  f(x)=2x^2+5\sqrt{x-2}

To find the value f(2) put x = 2 in given function ,

We have,

f(2)=2(2)^2+5\sqrt{2-2}

Simplify, we have,

f(2)=8+5\sqrt{0}

Simplify further, we have,

f(2)=8

Thus, the value of function at x = 2 is 8.

4 0
4 years ago
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A hyperbola centered at the origin has verticies at (add or subtract square root of 61,0 and foci at (add or subtract square roo
deff fn [24]

Answer:

\frac{x^2}{61}-\frac{y^2}{37}  =1

Step-by-step explanation:

The standard equation of a hyperbola is given by:

\frac{(x-h)^2}{a^2} -\frac{(y-k)^2}{b^2} =1

where (h, k) is the center, the vertex is at (h ± a, k), the foci is at (h ± c, k) and c² = a² + b²

Since the hyperbola is centered at the origin, hence (h, k) = (0, 0)

The vertices is (h ± a, k) = (±√61, 0). Therefore a = √61

The foci is (h ± c, k) = (±√98, 0). Therefore c = √98

Hence:

c² = a² + b²

(√98)² = (√61)² + b²

98 = 61 + b²

b² = 37

b = √37

Hence the equation of the hyperbola is:

\frac{x^2}{61}-\frac{y^2}{37}  =1

6 0
3 years ago
34560×23458÷34 <br><br>yo what's the answer
Phoenix [80]

Answer:

23,163,099.42857

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7 0
3 years ago
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