If p varies directly with q, and p = 8 when q = 2, what is the value of p when q = 7? A. p = 40 B. p = 28 C. p = 1 D. p = 100

HOW TO CONSENTRATE ON STUDY?

Ad libitum [116K]

Answer:

Listen to music or watch a video about what your about to study. Or do what I do... eat while studying

Step-by-step explanation:

4 0

3 years ago

Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given poi

lara [203]

For each curve, plug in the given point $(x,y)$ and check if the equality holds. For example:

(I) (2, 3) does lie on $x^2+xy-y^2=1$ since 2^2 + 2*3 - 3^2 = 4 + 6 - 9 = 1.

For part (a), compute the derivative $\frac{\mathrm dy}{\mathrm dx}$ , and evaluate it for the given point $(x,y)$ . This is the slope of the tangent line at the point. For example:

(I) The derivative is

$x^2+xy-y^2=1\overset{\frac{\mathrm d}{\mathrm dx}}{\implies}2x+x\dfrac{\mathrm dy}{\mathrm dx}+y-2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x+y}{2y-x}$

so the slope of the tangent at (2, 3) is

$\dfrac{\mathrm dy}{\mathrm dx}(2,3)=\dfrac74$

and its equation is then

$y-3=\dfrac74(x-2)\implies y=\dfrac74x-\dfrac12$

For part (b), recall that normal lines are perpendicular to tangent lines, so their slopes are negative reciprocals of the slopes of the tangents, $-\frac1{\frac{\mathrm dy}{\mathrm dx}}$ . For example:

(I) The tangent has slope 7/4, so the normal has slope -4/7. Then the normal line has equation

$y-3=-\dfrac47(x-2)\implies y=-\dfrac47x+\dfrac{29}7$

3 0

4 years ago

Which calculation results in the best estimate of 148 % of 203

adoni [48]

203---- 100%
x-----148%
x=203*148/100=203*1.48=300.44

6 0

3 years ago

Read 2 more answers

the number of three-digit numbers with distinct digits that be formed using the digits 1,2,3,5,8 and 9 is . The probability that

jolli1 [7]

Answer:

a)120

b)6.67%

Step-by-step explanation:

Given:

No. of digits given= 6

Digits given= 1,2,3,5,8,9

Number to be formed should be 3-digits, as we have to choose 3 digits from given 6-digits so the no. of combinations will be

6P3= 6!/3!

= 6*5*4*3*2*1/3*2*1

=6*5*4

=120

Now finding the probability that both the first digit and the last digit of the three-digit number are even numbers:

As the first and last digits can only be even

then the form of number can be

a)2n8 or

b)8n2

where n can be 1,3,5 or 9

4*2=8

so there can be 8 three-digit numbers with both the first digit and the last digit even numbers

And probability = 8/120

= 0.0667

=6.67%

The probability that both the first digit and the last digit of the three-digit number are even numbers is 6.67% !

5 0

4 years ago

Read 2 more answers

Can someone please help

Vsevolod [243]

It may be convenient here to write the equation of the line as

... ∆y(x -x0) -∆x(y -y0) = 0

where (x0, y0) is a point on the line, and (∆x, ∆y) is the difference between two points on the line.

Using the first two points, we have

... (∆x, ∆y) = (-9, 10) - (-15, 5) = (6, 5)

So, the equation of the line is

... 5(x +9) -6(y -10) = 0 . . . . . . using the point (x0, y0) = (-9, 10)

... 5x -6y = -105 . . . . . . . . . . . simplify to standard form

Now, dividing by the constant on the right, we can put this into intercept form

... x/(-21) +y/(17.5) = 1

This tells us the y-intercept is (0, 17.5) and the x-intercept is (-21, 0).

_____

You can work directly with the values in the table. The y-value increases by 5 from one line to another, while the x-value increases by 6. Thus, to make the y-value decrease by 5 from the first point, we need to decrease that point's x-value by 6 to -21. That is, the x-intercept is (-21, 0).

Similarly, the x-value need to increase by only 3 from the last point. That amounts to half of the usual increase of 6. Thus the y-intercept will be the last point plus have the usual y-increase of 5, or 15+2.5 = 17.5. That is, the y-intercept is (0, 17.5).

5 0

4 years ago