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3 years ago

What is the slope of the line that contains the points (-6, 2) and (4,-5)?

Mathematics

2 answers:

Dominik [7]3 years ago

8 0

Answer:

-7 / 10

Step-by-step explanation:

use the slope formula

slope= (y2 - y1) / (x2- x1)

which for our case is:

slope= (-5 - 2) / (4- - 6) = - 7/10

Mademuasel [1]3 years ago

7 0

Answer:

Opcion C

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PLEASE HELP! TWO PROBLEMS!!

MrRissso [65]

1) x=−2 2) y=−6 Hopefully that helps you ❤.

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3 years ago

NO LINKS!! Please help me with this problem

iogann1982 [59]

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

The given figure shows a vertical hyperbola with its centre at origin, and as we observe the figure, we can conclude that :

Length of transverse axis is :

$\qquad \sf \dashrightarrow \: 2b = 12$

$\qquad \sf \dashrightarrow \: b = 6$

length of conjugate axis is :

$\qquad \sf \dashrightarrow \: 2a = 8$

$\qquad \sf \dashrightarrow \: a = 4$

Equation of hyperbola ~

$\qquad \sf \dashrightarrow \: \cfrac{ {y}^{2} }{ {b}^{2} } - \cfrac{ {x}^{2} }{ {a}^{2} } = 1$

plug in the values ~

$\qquad \sf \dashrightarrow \: \cfrac{ {y}^{2} }{ {6}^{2} } - \cfrac{ {x}^{2} }{ {4}^{2} } = 1$

$\qquad \sf \dashrightarrow \: \cfrac{ {y}^{2} }{ {36}^{} } - \cfrac{ {x}^{2} }{ {16}^{} } = 1$

5 0

2 years ago

Read 2 more answers

One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

scZoUnD [109]

Answer:

$c^2 = 9dp$

Step-by-step explanation:

Given

$dx^2 + cx + p = 0$

Let the roots be $\alpha$ and $\beta$

So:

$\alpha = 2\beta$

Required

Determine the relationship between d, c and p

$dx^2 + cx + p = 0$

Divide through by d

$\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0$

$x^2 + \frac{c}{d}x + \frac{p}{d} = 0$

A quadratic equation has the form:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

So:

$x^2 - (2\beta+ \beta)x + \beta*\beta = 0$

$x^2 - (3\beta)x + \beta^2 = 0$

So, we have:

$\frac{c}{d} = -3\beta$ -- (1)

and

$\frac{p}{d} = \beta^2$ -- (2)

Make $\beta$ the subject in (1)

$\frac{c}{d} = -3\beta$

$\beta = -\frac{c}{3d}$

Substitute $\beta = -\frac{c}{3d}$ in (2)

$\frac{p}{d} = (-\frac{c}{3d})^2$

$\frac{p}{d} = \frac{c^2}{9d^2}$

Multiply both sides by d

$d * \frac{p}{d} = \frac{c^2}{9d^2}*d$

$p = \frac{c^2}{9d}$

Cross Multiply

$9dp = c^2$

$c^2 = 9dp$

Hence, the relationship between d, c and p is: $c^2 = 9dp$

8 0

3 years ago

Shea and her best friend Kelly both babysit during the summer. Shea charges $4 per hour plus a $3 service fee, while Kelly charg

Evgen [1.6K]

6 hours

The First hour- Shea made $7 and Kelly made $9.50
The Second hour- Shea made $11 and Kelly made $13
The third hour- Shea made $15 and Kelly made $16.50
The fourth hour- Shea made $19 and Kelly made $20
The fifth hour- Shea made $23 and Kelly made $23.50
And the sixth hour- Shea made $27 and Kelly made $27

Hope that was helpful

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3 years ago

What is the slope of the line that passes through the points (-10, -4)(−10,−4) and (10, -29) ?(10,−29)? Write your answer in sim

irga5000 [103]

Answer:

The answer is -1.25

Hope that helps. x

Step-by-step explanation:

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2 years ago