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2 years ago

Multiple correct answers

Mathematics

1 answer:

tatuchka [14]2 years ago

5 0

Answer: the answers I got were "B" 5, and "D" 7.

Hope this helps

Step-by-step explanation:

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A line passes through $A(1,1)$ and $B(100,1000)$. How many other points with integer coordinates are on the line and strictly be

jonny [76]

Answer:

Step-by-step explanation:

Integers are whole numbers or opposite of whole numbers.

The slope is 1.

How?

Change of y is 100-1=99.

Change of x is 100-1=99.

The slope is 99/99=1 or 1/1.

So if we start at (1,1) and we rise 1 and run 1 right, we get (2,2).

If we do that again from (2,2) we get (3,3).

Following the pattern of going up 1 and right from each new location discovered we get all the of these points:

Start (1,1)

(2,2)

(3,3)

(4,4)

(5,5)

(6,6)

....

(96,96)

(97,97)

(98,98)

(99,99)

end (100,100)

So we just need to count all the numbers from 2 to 99.

99-2+1

97+1

7 0

2 years ago

What is the slope of the line that passes through the points (13,-8) and (-11,4)

pogonyaev

Answer:

- $\frac{1}{2}$

Step-by-step explanation:

Find slope using the slope formula : $\frac{y2-y1}{x2-x1}$

Plug in the given points : $\frac{4- (-8)}{-11-13}$

Add the numbers : $\frac{12}{-11-13}$

Calculate the difference : $\frac{12}{-24}$

Reduce the fraction : - $\frac{1}{2}$

Solution : - $\frac{1}{2}$

3 0

2 years ago

Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).

fomenos

Answer:

$\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1$

Step-by-step explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

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

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is $2a=|4--10|$ .

$\implies 2a=|14|$

$\implies 2a=14$

$\implies a=7$

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is $(\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)$

We need to use the relation $a^2+b^2=c^2$ to find $b^2$ .

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

$c=|6--3|=9$

$\implies 7^2+b^2=9^2$

$\implies b^2=9^2-7^2$

$\implies b^2=81-49$

$\implies b^2=32$

We substitute these values into the standard equation of the hyperbola to obtain:

$\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1$

$\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1$

7 0

2 years ago

Solve the quadratic equation by completing the square.<br> QUESTION ATTCHED BELOW!!!!!!!!!!

Mariulka [41]

Answer:

x ≈ - 9.41, x ≈ - 6.59

Step-by-step explanation:

Given

x² + 16x + 62 = 0 ( subtract 62 from both sides )

x² + 16x = - 62

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(8)x + 64 = - 62 + 64

(x + 8)² = 2 ( take the square root of both sides )

x + 8 = ± $\sqrt{2}$ ( subtract 8 from both sides )

x = - 8 ± $\sqrt{2}$

Hence

x = - 8 - $\sqrt{2}$ ≈ - 9.41

x = - 8 + $\sqrt{2}$ ≈ - 6.59

4 0

2 years ago

Read 2 more answers

Graph x^2 +8x+4y+8=0

Afina-wow [57]

Answer:

x y

− 6 1

− 5 7 /4

− 4 2

− 3 7 /4

− 2 1

Step-by-step explanation:

5 0

2 years ago