What linear function represents the line given by the point slope equation y + 1 = -3(x

GUYS I NEED HELP PLSSSSSS! Ive been working on this ALL day!

Vika [28.1K]

First, let's write what we know.

We can represent the number of students in the play from each class as L, G, and C. We know that L = 7, and if Gardener has 4 more students than Cho, then G = C + 4.

Then, taking the third line, we can write an inequality:

C < L < G

C < 7 < C + 4

C - 4 < 3 < C

3 < C < 7

If C is greater than 3 and less than 7 and is an integer, than means C is 4, 5, or 6.

Then, we need to find how many students are in the play.

C + L + G

C + 7 + C + 4

2C + 11

So we have our expression for the number of students in the play. Then, we need to find the total number of students. We know that 2C + 11 will be 30% of the total, so if T is the total, we can find T.

0.3T = 2C + 11

(Divide by 3/10 or multiply by 10/3 on both sides)

T = 20/3 C + 110/3

We know from before that C is 4, 5 or 6. We can plug these into our equation here to find which one produces a whole number.

T = 20/3 * 4 + 110/3

T = 190/3

T = 20/3 * 5 + 110/3

T = 210/3

T = 70

T = 20/3 * 6 + 110/3

T = 230/3

We can see here that only when C is 5 will the total be a whole number. That means Mrs. Cho has 5 students in the play. If Mrs. Gardner has 4 more than that, she has 9 students in the play in her class. We now need to figure out the number of student in her class.

The total students are Cho's, Logan's, and Gardner's classes added together. We know that Logan's class is 23 students, so if we subtract that from the total, we can see that Cho's and Gardner's class have 47 students. If Mrs. Cho has 24 students in her class, we can subtract that from the 47, so we know that Mrs. Gardner has 23 students in her class.

4 0

3 years ago

Suppose u and v are functions of x that are differentiable at x=0 and that u(0)= 7,u'(0)=-5,v(0)= -1, v'(0)= -4.

andriy [413]

This question is incomplete, the complete question is;

Suppose u and v are functions of x that are differentiable at x=0

and that { u(0) = 7, u'(0) = -5 } { v(0)= -1, v'(0) = -4 }

Find the values of the following derivatives at x = 0.

a) $\frac{d}{dx}$ ( uv )

b) $\frac{d}{dx}$ ( $\frac{u}{v}$ )

c) $\frac{d}{dx}$ ( $\frac{v}{u}$ )

Answer:

a) $\frac{d}{dx}$ ( uv ) = -23

b) $\frac{d}{dx}$ ( $\frac{u}{v}$ ) = 33

c) $\frac{d}{dx}$ ( $\frac{v}{u}$ ) = -32/49 or - 0.6531

Step-by-step explanation:

Given that;

{ u(0) = 7, u'(0) = -5 } { v(0)= -1, v'(0) = -4 }

a)

$\frac{d}{dx}$ ( uv )

we differentiate

$\frac{d}{dx}$ ( uv ) = uv' + vu'

at x = (0), we substitute our values

$\frac{d}{dx}$ ( uv ) = ( 7 × -4 ) + ( -1 × -5)

$\frac{d}{dx}$ ( uv ) = -28 + 5

$\frac{d}{dx}$ ( uv ) = -23

b)

$\frac{d}{dx}$ ( $\frac{u}{v}$ )

we differentiate

$\frac{d}{dx}$ ( $\frac{u}{v}$ ) = ( vu' - uv' ) / v²

at x=0, we substitute our values

$\frac{d}{dx}$ ( $\frac{u}{v}$ ) = ( (-1 × -5) - (7 × -4 ) ) / (-1)²

$\frac{d}{dx}$ ( $\frac{u}{v}$ ) = (( 5 - ( -28 )) / 1

$\frac{d}{dx}$ ( $\frac{u}{v}$ ) = 33 / 1

$\frac{d}{dx}$ ( $\frac{u}{v}$ ) = 33

c) $\frac{d}{dx}$ ( $\frac{v}{u}$ )

we differentiate

$\frac{d}{dx}$ ( $\frac{v}{u}$ ) = ( uv' - vu' ) / u²

at x=0, we substitute our values

$\frac{d}{dx}$ ( $\frac{v}{u}$ ) = ( (7 × -4) - (-1 × -4) ) / (7)²

$\frac{d}{dx}$ ( $\frac{v}{u}$ ) = ( -28 - ( 4 ) ) / 49

$\frac{d}{dx}$ ( $\frac{v}{u}$ ) = ( -28 - 4 ) /49

$\frac{d}{dx}$ ( $\frac{v}{u}$ ) = -32 / 49

$\frac{d}{dx}$ ( $\frac{v}{u}$ ) = -32/49 or - 0.6531

6 0

3 years ago

What are the answer to these? PLEASE ANSWER (don’t mind my work)

SSSSS [86.1K]

uno is the answer i believe

5 0

4 years ago

Solve for x, rounding to the nearest tenth if necessary.

Natalija [7]

Answer:

99

Step-by-step explanation:

hope this helps

6 0

3 years ago

Leading coefficient of the polynomial 4x^2−x^6+2x−6

ICE Princess25 [194]

Answer: The leading coefficient is -1.

Step-by-step explanation: Coefficients are the numbers than procede each term with a variable (in this instance,<em> </em>x) in the equation.

$-x^6+4x^2+2x-6\\$

There are four terms in this equation, which has been reordered in the correct form, which is from the highest term to the lowest term (based on exponent values).

Thus, there are three coefficients in this equation, which, in order, are as follows:

$-1, 4, 2$

6 0

3 years ago

What linear function represents the line given by the point slope equation y + 1 = -3(x - 5)​

What linear function represents the line given by the point slope equation y + 1 = -3(x - 5)