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3 years ago

13

What is the slope of the line that passes through the points (-2,-3), and (5, 4)?

1 answer:

Ratling [72]3 years ago

3 0

Answer:

I believe its A

Step-by-step explanation:

A. 3

Hope this helps

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A ramp with a constant incline is made to connect a driveway to a front door. at a point 4 feet from the driveway, the height of

ziro4ka [17]

The rate of change:
( y2 - y1 ) / (x2 - x1) = ( 18 in - 12 in ) / ( 6 ft - 4 ft ) =
= 6 in / 2 ft = 3 in / 1 ft
Answer: D ) 3 inches up per foot across

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3 years ago

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a bag contains eleven counters. five are white. a counter is taken out the bag and is not replaced. a second counter is taken ou

Fed [463]

Answer:

$\displaystyle P(A)=\frac{6}{11}$

Step-by-step explanation:

<u>Probabilities</u>

The question describes an event where two counters are taken out of a bag that originally contains 11 counters, 5 of which are white.

Let's call W to the event of picking a white counter in any of the two extractions, and N when the counter is not white. The sample space of the random experience is

$\Omega=\{WW,WN,NW,NN\}$

We are required to compute the probability that only one of the counters is white. It means that the favorable options are

$A=\{WN,NW\}$

Let's calculate both probabilities separately. At first, there are 11 counters, and 5 of them are white. Thus the probability of picking a white counter is

$\displaystyle \frac{5}{11}$

Once a white counter is out, there are only 4 of them and 10 counters in total. The probability to pick a non-white counter is now

$\displaystyle \frac{6}{10}$

Thus the option WN has the probability

$\displaystyle P(WN)=\frac{5}{11}\cdot \frac{6}{10}=\frac{30}{110}=\frac{3}{11}$

Now for the second option NW. The initial probability to pick a non-white counter is

$\displaystyle \frac{6}{11}$

The probability to pick a white counter is

$\displaystyle \frac{5}{10}$

Thus the option NW has the probability

$\displaystyle P(NW)=\frac{6}{11}\cdot \frac{5}{10}=\frac{30}{110}=\frac{3}{11}$

The total probability of event A is the sum of both

$\displaystyle P(A)=\frac{3}{11}+\frac{3}{11}=\frac{6}{11}$

$\boxed{\displaystyle P(A)=\frac{6}{11}}$

7 0

3 years ago

Quadrilateral RUST has a vertex at R(2,3). What are the coordinates of R' after dilation by a scale factor of 3, centered at the

coldgirl [10]

Concept of this answer is (3,5) in the y(x)=-3

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3 years ago

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What are the solutions of 2x2 + 3x = -5

Daniel [21]

$2x^2 + 3x = -5\\2x^2 + 3x +5=0\\ \Delta=3^2-4\cdot2\cdot5=9-40=-31\\\\\Delta$

7 0

3 years ago

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PLs help 50 PTS!!!!! PLEASE ILL GIVE BRAINLIEST!!!!!

Nookie1986 [14]

Answer:

$\large\boxed{y=\dfrac{1}{4}x^2-x-4}$

Step-by-step explanation:

The equation of a parabola in vertex form:

$y=a(x-h)^2+k$

<em>(h, k)</em><em> - vertex</em>

The focus is

$\left(h,\ k+\dfrac{1}{4a}\right)$

We have the vertex (2, -5) and the focus (2, -4).

Calculate the value of <em>a</em> using $k+\dfrac{1}{4a}$

<em>k = -5</em>

$-5+\dfrac{1}{4a}=-4$ <em>add 5 to both sides</em>

$\dfrac{1}{4a}=1$ <em>multiply both sides by 4</em>

$4\!\!\!\!\diagup^1\cdot\dfrac{1}{4\!\!\!\!\diagup_1a}=4$

$\dfrac{1}{a}=4\to a=\dfrac{1}{4}$

Substitute

$a=\dfrac{1}{4},\ h=2,\ k=-5$

to the vertex form of an equation of a parabola:

$y=\dfrac{1}{4}(x-2)^2-5$

The standard form:

$y=ax^2+bx+c$

Convert using

$(a-b)^2=a^2-2ab+b^2$

$y=\dfrac{1}{4}(x^2-2(x)(2)+2^2)-5\\\\y=\dfrac{1}{4}(x^2-4x+4)-5$

<em>use the distributive property: a(b+c)=ab+ac</em>

$y=\left(\dfrac{1}{4}\right)(x^2)+\left(\dfrac{1}{4}\right)(-4x)+\left(\dfrac{1}{4}\right)(4)-5\\\\y=\dfrac{1}{4}x^2-x+1-5\\\\y=\dfrac{1}{4}x^2-x-4$

3 0

4 years ago